PHYS 101 - ELEMENTS OF PHYSICS
- Lens and mirror equations
Question Bank - Set 5
Liberty University
Question 1
Question
An object is placed −15 cm to the left of a convex lens with a focal length of
20 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
Solution
Step 1: Identify the given quantities and the focal length of the lens. The object
distance (u) is −15 cm and the focal length (f) is 20 cm.
Step 2: Determine the image distance using the lens equation 1
f=1
v+1
u.
Substitute the given values to find the image distance (v):
1
20 =1
v+1
−15
1
v=1
20 −1
15
1
v=3−4
60 =−1
60
v=−60 cm
Step 3: Calculate the magnification (m) using the formula m=−v
u. Sub-
stitute the values of vand u:
m=−−60
−15 = 4
Step 4: Determine whether the image is real or virtual based on the sign of
v. Since vis negative, the image is virtual.
Therefore, the image distance is −60 cm, the magnification is 4, and the
image is virtual.
Question 2
Question
An object is placed 15 cm from a concave mirror with a focal length of 10 cm.
Determine the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values. The object distance (do) is 15 cm, the focal
length (f) is -10 cm (since the mirror is concave), and the magnification formula
is m=−di
do.
Step 2: Apply the mirror equation. The mirror equation is 1
f=1
do+1
di.
Substituting the known values gives us 1
−10 =1
15 +1
di.
Step 3: Solve for the image distance. Solving the equation for di, we get
1
di=1
−10 −1
15 =−1
30 . Therefore, di=−30 cm.
Step 4: Calculate the magnification. Using the formula m=−di
do, we sub-
stitute di=−30 cm and do= 15 cm to find m=−−30
15 = 2.
Step 5: Interpret the results. The image is formed at a distance of 30 cm from
the mirror (behind the mirror), and the image is magnified with a magnification
factor of 2.
Question 3
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Determine the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values and the mirror equation. The given values are:
Object distance, do=−30 cm (since the object is in front of the mirror, the
distance is negative) Focal length, f=−20 cm (negative for concave mirror)
The mirror equation is: 1
f=1
do
+1
di
Step 2: Substitute the values into the mirror equation. Substitute f=
−20 cm and do=−30 cm into the mirror equation to find di.
1
−20 =1
−30 +1
di
Step 3: Solve for the image distance, di.
1
di
=1
−20 −1
−30 =3
60 =1
20
2
di= 20 cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−di
do
Substitute di= 20 cm and do=−30 cm into the magnification formula.
M=−20
−30 =2
3
Therefore, the image is formed 20 cm from the mirror on the same side as
the object, and the magnification of the image is 2
3.
Question 4
Question
An object is placed 40 cm from a concave mirror with a focal length of 30 cm.
Determine the image distance and magnification of the image formed by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
f=−30 cm
p=−40 cm
Mirror equation: 1
f=1
p+1
q
Step 2: Substitute the given values into the mirror equation.
1
−30 =1
−40 +1
q
Step 3: Solve for q.
−1
30 =−1
40 +1
q
1
q=−1
30 +1
40
1
q=−4+3
120
1
q=−1
120
3
q=−120 cm
Step 4: Calculate the magnification using the magnification equation.
M=−q
p
M=−−120
−40
M= 3
Answer: The image distance is -120 cm and the magnification of the image
formed by the mirror is 3.
Question 5
Question
An object is placed 30 cm in front of a concave mirror of focal length 15 cm.
Determine the position and nature of the image formed.
Solution
Step 1: Identify the given values and the mirror formula.
Object distance (u) = -30 cm (since it is in front of the mirror, we take it
as negative)
Focal length (f) = -15 cm (since it is a concave mirror, its focal length is
negative)
Mirror formula: 1
f=1
v+1
u
Step 2: Substitute the given values in the mirror formula and solve for the
image distance (v).
1
−15 =1
v+1
−30
−1
15 =1
v−1
30
−1
15 =2−v
2v
−2v= 2v−30
4v= 30
v= 7.5 cm
Step 3: Determine the position of the image. Since the image distance (v)
is positive, the image is formed on the same side as the object, i.e., in front of
the mirror.
Step 4: Determine the nature of the image. To determine the nature of the
image, we compare the object distance (u) and the focal length (f).
4
If f > |v|(image distance), the image is real and inverted.
If f < |v|(image distance), the image is virtual and erect.
Since |v|= 7.5 cm and f= 15 cm, we have f < |v|, so the image is virtual and
erect.
Therefore, the image is formed 7.5 cm in front of the concave mirror and is
virtual and erect.
Question 6
Question
A concave mirror has a focal length of 15 cm. An object is placed 30 cm in
front of the mirror. Determine the image distance and magnification produced
by the mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Given: f=−15 cm (concave mirror focal length), do=−30 cm (object
distance).
Mirror equation: 1
f=1
do
+1
di
.
Step 2: Substitute the given values into the mirror equation and solve for
di.
1
−15 =1
−30 +1
di
−2
30 =−1
di
di=−15 cm
Therefore, the image is formed 15 cm in front of the mirror.
Step 3: Calculate the magnification using the magnification equation.
Magnification equation: m=−di
do
.
Step 4: Substitute the values of diand dointo the magnification equation
and solve for m.
m=−−15
−30
m=1
2
Therefore, the image produced by the concave mirror is real, formed 15 cm
in front of the mirror, and is half the size of the object.
5
Question 7
Question
An object is placed 23 cm in front of a concave mirror of focal length -20 cm.
Determine the position and size of the image formed by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given: Object
distance, do=−23 cm (negative because it is in front of the mirror), Focal
length, f=−20 cm.
The mirror equation is: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation.
1
−20 =1
−23 +1
di
Step 3: Solve for the image distance, di.
1
−20 =1
−23 +1
di
−3
60 =1
di
di=−60
3
di=−20 cm
So, the image is formed 20 cm in front of the mirror.
Step 4: Use the magnification formula to find the size of the image. The
magnification formula is:
m=−di
do
Step 5: Substitute the known values into the magnification formula.
m=−−20
−23
m=20
23
Step 6: The negative sign in the magnification equation indicates that the
image is inverted. The magnification factor also tells us that the image is smaller
than the object since |m|<1.
Therefore, the image formed by the concave mirror is 20 cm in front of the
mirror and smaller than the object.
6
Question 8
Question
An object is placed 15 cm in front of a concave mirror with a focal length of
10 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
Solution
Step 1: Identify the given values. The object distance ois 15 cm (in front of
the mirror), and the focal length fof the concave mirror is -10 cm (since it is a
concave mirror, the focal length is negative). Given: o= 15 cm, f=−10 cm.
Step 2: Use the mirror equation to find the image distance. The mirror
equation is given by: 1
f=1
o+1
i
Substitute the given values:
1
−10 =1
15 +1
i
Solving for i:
−1
10 =1
15 +1
i
−1
10 −1
15 =1
i
−3
30 −2
30 =1
i
−5
30 =1
i
−i= 6
i=−6 cm
Therefore, the image distance iis -6 cm.
Step 3: Calculate the magnification. The magnification Mis given by:
M=−i
o
Substitute the values obtained:
M=−−6
15
M=2
5
Thus, the magnification is 2
5.
Step 4: Determine if the image is real or virtual. Since the image distance
is negative, the image is formed on the same side as the object. Therefore, the
image is virtual.
7
Question 9
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Identify the given values and the type of mirror. Given:
Object distance, u=−20 cm (negative because the object is in front of
the mirror)
Focal length, f=−15 cm (negative for a concave mirror)
Since the focal length is negative, the mirror is a concave mirror.
Step 2: Apply the mirror equation to find the image distance. The mirror
equation is: 1
f=1
v+1
u
Substitute the given values: f=−15 cm, u=−20 cm
1
−15 =1
v+1
−20
Step 3: Solve for image distance, v.
−1
15 =1
v−1
20
−1
15 +1
20 =1
v
4
60 −3
60 =1
v
1
60 =1
v
v= 60 cm
Therefore, the image is formed 60 cm behind the mirror.
Step 4: Determine the nature of the image. To determine the nature of the
image, we consider the sign of v: - If vis positive, the image is real and located
on the opposite side of the object. - If vis negative, the image is virtual and
located on the same side as the object.
Since vis positive, the image formed is real and located 60 cm behind the
mirror.
8
Question 10
Question
A concave mirror with a focal length of 10 cm forms an image that is four times
the size of the object. If the object is placed 20 cm from the mirror, at what
distance is the image formed from the mirror?
Solution
Let fbe the focal length of the concave mirror, pbe the object distance, and q
be the image distance.
Step 1: Given that the mirror equation is 1
f=1
p+1
qand the magnification
equation is m=−q
p, we can find the image distance q.
First, we know that m=h′
h=−q
p=−4, where h′and hare the heights of
the image and object respectively.
Step 2: Substituting m=−4 into m=−q
p, we get −4 = −q
p. Therefore,
q= 4p.
Step 3: Now, substitute f= 10 cm and p= 20 cm into the mirror equation
1
f=1
p+1
qto find q.
1
10 =1
20 +1
q
1
10 −1
20 =1
q
2
20 −1
20 =1
q
1
20 =1
q
q= 20 cm
Step 4: Therefore, the image is formed at a distance of 20 cm from the
concave mirror.
Question 11
Question
A converging lens with a focal length of 20 cm is placed 30 cm in front of a
diverging mirror with a focal length of 10 cm. Determine the final image distance
from the mirror, whether the image is real or virtual, and the magnification of
the system.
9
Solution
Step 1: Let’s label the given focal lengths and object distance as fl= 20 cm (for
the converging lens), fm=−10 cm (for the diverging mirror), and do=−30 cm
(object distance in front of the lens) respectively.
Step 2: Calculate the image distance from the lens using the lens formula
1
fl
=1
do
+1
di
.
Substitute the given values into the lens formula:
1
20 =1
−30 +1
di
Solve for di:
di=1
1
20 −1
30
=−60 cm
Therefore, the image distance from the lens is di=−60 cm.
Step 3: Calculate the image distance from the mirror using the mirror for-
mula 1
fm
=1
di
+1
dm
.
Since the image distance from the lens is -60 cm, we have the following
equation: 1
−10 =1
−60 +1
dm
Solve for dm:
dm=1
1
−10 −1
−60
= 15 cm
Therefore, the final image distance from the mirror is dm= 15 cm.
Step 4: Determine whether the image is real or virtual by calculating the
magnification. The magnification Mof the system is given by di
do
.
Substitute the values of diand do:
M=−60
−30 = 2
Since M > 0, the image is upright. Therefore, the image is virtual.
Step 5: The magnification of the system is calculated as M= 2, indicating
that the image is upright and magnified by a factor of 2.
Question 12
Question
A concave mirror has a focal length of −15 cm. An object is placed 25 cm in front
of the mirror. Determine the image distance, image height, and magnification.
10
Solution
Step 1: Identify the given values and the mirror equation.
Given values: Focal length of mirror, f=−15 cm Object distance, do=−25
cm
The mirror equation for concave mirrors is:
1
f=1
do
+1
di
Step 2: Substitute the given values into the mirror equation to find the image
distance, di.
1
−15 =1
−25 +1
di
−1
15 =−1
25 +1
di
1
di
=−1
15 +1
25
1
di
=−5+3
75
1
di
=−2
75
di=−75
2
di=−37.5 cm
Therefore, the image distance is −37.5 cm.
Step 3: Use the magnification formula to calculate the magnification, M.
The magnification formula is:
M=−di
do
Substitute the values of diand dointo the formula:
M=−(−37.5)
−25
M=37.5
25
M= 1.5
Therefore, the magnification is 1.5.
11
Step 4: Use the magnification to find the image height, hi, given the object
height, ho(assuming ho= 1 unit).
The image height is related to the object height by:
hi=M×ho
Substitute the values of Mand hointo the formula:
hi= 1.5×1
hi= 1.5
Thus, the image height is 1.5 units.
Question 13
Question
An object is placed 20 cm from a concave mirror with a focal length of 10 cm.
Calculate the image distance and the magnification of the image formed.
Solution
Step 1: Identify the given and unknown quantities. Given: f=−10 cm (focal
length of the concave mirror) do=−20 cm (object distance)
Unknown: di(image distance) M(magnification)
Step 2: Apply the lens/mirror equation for concave mirrors:
1
f=1
do
+1
di
Step 3: Substitute the given values and solve for di:
1
−10 =1
−20 +1
di
1
di
=1
−10 −1
−20
1
di
=−1
10 +1
20
1
di
=−2−1
20
1
di
=−1
20
di=−20 cm
12
Step 4: Calculate the magnification using the magnification formula:
M=−di
do
M=−−20
−20
M= 1
Answer: The image distance is -20 cm and the magnification is 1.
Question 14
Question
An object is placed 10 cm in front of a concave mirror with a focal length of 20
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. The given values are:
- Object distance, do=−10 cm (negative since it is in front of the mirror) -
Focal length, f=−20 cm (negative for a concave mirror) The mirror equation
for concave mirrors is given by:
−1
f=1
di
+1
do
where di= image distance
Step 2: Substitute the given values into the mirror equation.
−1
−20 =1
di
+1
−10
Step 3: Solve for di.1
20 =1
di
−1
10
1
di
=1
20 +1
10
1
di
=1
20 +2
20
1
di
=3
20
di=20
3= 6.67 cm
The image is formed 6.67 cm behind the mirror which is a real image.
13
Step 4: Determine the magnification. The magnification, m, is given by the
formula:
m=−di
do
Substitute the values of diand do:
m=−6.67
−10
m= 0.67
The negative magnification indicates that the image is inverted.
Step 5: Determine the nature of the image. Since the image distance is
positive, the image is real. The positive magnification value indicates that the
image is upright.
Therefore, the image formed by the concave mirror is real, inverted, and
upright. It is located 6.67 cm behind the mirror.
Question 15
Question
An object is placed 30 cm in front of a convex lens with a focal length of 20 cm.
Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values and the lens formula: Given: - Object distance
u=−30 cm (negative since it is in front of the lens), - Focal length f= 20 cm
for a convex lens.
The lens formula relates object distance (u), image distance (v), and focal
length (f): 1
f=1
v+1
u
Step 2: Substitute the values into the lens formula:
1
20 =1
v+1
−30
Step 3: Solve for v:1
20 =1
v−1
30
1
v=1
20 +1
30
1
v=3
60 +2
60
14
1
v=5
60
v=60
5= 12 cm
Therefore, the image distance vis 12 cm.
Step 4: Calculate the magnification: The magnification (m) of the lens is
given by the formula:
m=−v
u
Step 5: Substitute the values of vand uinto the magnification formula:
m=−12
−30 =2
5
Hence, the magnification produced by the lens is 2
5.
Question 16
Question
An object is placed 15 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and magnification of the image formed by
the mirror.
Solution
Step 1: Recall the mirror equation:
1
f=1
do
+1
di
where fis the focal length of the mirror, dois the object distance, and diis the
image distance.
Step 2: Substituting the given values, we get:
1
10 =1
15 +1
di
Step 3: Solve for di:
1
di
=1
10 −1
15 =3−2
30 =1
30
⇒di= 30 cm
Step 4: Next, calculate the magnification using the formula:
M=−di
do
15
Step 5: Substitute the values:
M=−30
15 =−2
Step 6: Therefore, the image distance is 30 cm and the magnification of the
image formed by the mirror is -2.
Question 17
Question
An object is placed 15 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and the magnification of the image.
Solution
Step 1: Identify known values and the given information.
The given values are:
Object distance do=−15 cm (negative since it is in front of the mirror)
Focal length f= 10 cm (negative for a concave mirror)
Step 2: Apply the mirror equation to find the image distance di.
The mirror equation is given by:
1
f=1
di
+1
do
Plug in the known values and solve for di:
1
−10 =1
di
+1
−15
−1
10 +1
15 =1
di
3
30 −2
30 =1
di
1
30 =1
di
di= 30 cm
Step 3: Calculate the magnification M.
The magnification is given by:
M=−di
do
16
Plug in the calculated values:
M=−30
−15 = 2
Therefore, the image distance is 30 cm and the magnification of the image
is 2.
Question 18
Question
An object is placed 20 cm in front of a converging lens with a focal length of
10 cm. The image produced by the lens is then used as an object for a concave
mirror with a focal length of 15 cm. Determine the final image distance from
the mirror.
Solution
Step 1: Calculate the image distance produced by the lens using the lens equa-
tion. 1
flens
=1
do
+1
di
1
10 =1
20 +1
di
Solving for di:1
di
=1
10 −1
20
1
di
=2
20 −1
20
1
di
=1
20
di= 20 cm
Step 2: Use the mirror equation to find the final image distance from the
mirror. 1
fmirror
=1
d′
i
+1
d′
o
Given that fmirror =−15 cm and d′
o=−20 cm (since the image produced by
the lens is a virtual object for the mirror), and d′
i=?,
1
−15 =1
d′
i
+1
−20
Solving for d′
i:
1
d′
i
=−1
15 +1
20
17
1
d′
i
=−4
60 +3
60
1
d′
i
=−1
60
d′
i=−60 cm
Therefore, the final image distance from the mirror is −60 cm.
Question 19
Question
An object is placed 20 cm in front of a convex lens of focal length 15 cm. If the
image formed is virtual, upright, and 5 cm in front of the lens, determine the
magnification of the image.
Solution
Step 1: Identify the given values and the formula for magnification. Let do=
−20 cm (object distance), f= 15 cm (focal length), di=−5 cm (image dis-
tance), and mbe the magnification. The formula for magnification is given
by:
m=−di
do
Step 2: Convert all distances to positive values. Since the image is virtual,
the image distance diis considered positive. Thus, |di|= 5 cm.
Step 3: Solve for the magnification. Substitute the given values into the
formula for magnification:
m=−5
−20 =1
4
Step 4: Interpret the result. The magnification of the image is 1
4, indicating
that the image is one-fourth the size of the object.
Question 20
Question
A concave mirror with a focal length of 20 cm creates a virtual image that is
15 cm tall. If the object height is 5 cm, determine the object distance from the
mirror.
18
Solution
Let fbe the focal length of the concave mirror, hibe the height of the image,
and hobe the height of the object. Let dobe the object distance from the
mirror. Using the mirror equation and the magnification equation, we can solve
for do.
Step 1: Write down the mirror equation:
1
f=1
do
+1
di
where diis the image distance from the mirror. Given that f=−20 cm and
hi=−15 cm, we can find di:
1
−20 =1
do
+1
di
=⇒1
−20 =1
do
−1
15
Step 2: Solve for di:
1
−20 =1
do
−1
15 =⇒1
do
−1
15 =1
−20 =⇒1
do
=1
15 −1
20
Step 3: Find dousing the magnification equation:
hi
ho
=−di
do
=⇒−15
5=−1
do
=⇒ −3 = −1
do
Step 4: Solve for do:
−3 = −1
do
=⇒1
do
= 3 =⇒do=1
3= 3 cm
Therefore, the object distance from the mirror is 3 cm.
Question 21
Question
An object is placed 15 cm to the left of a converging lens of focal length 10 cm.
Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the lens equation. Given: Object distance,
u=−15 cm (negative sign indicates object is to the left of the lens) Focal length,
f= 10 cm
The lens formula relates object distance (u), image distance (v), and focal
length (f): 1
f=1
v+1
u
19
Step 2: Plug in the known values into the lens equation.
1
10 =1
v+1
−15
Step 3: Solve for image distance (v).
1
v=1
10 +1
15 =3+2
30 =5
30 =1
6
v= 6 cm
Therefore, the image distance is 6 cm.
Step 4: Calculate the magnification (m) using the formula:
m=−v
u
Step 5: Plug in the values to calculate magnification.
m=−6
−15 =2
5
Therefore, the magnification is 2
5.
Question 22
Question
An object is placed 20 cm in front of a concave lens with a focal length of 15
cm. Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values. The object distance uis 20 cm and the focal
length fof the lens is -15 cm (concave lens, so focal length is negative).
Step 2: Apply the lens equation. The lens equation is given by:
1
f=1
v+1
u
Where: - fis the focal length, - vis the image distance, and - uis the object
distance.
Step 3: Substitute the given values into the lens equation.
1
−15 =1
v+1
20
Step 4: Solve for the image distance v.
−1
15 =1
v+1
20
20
1
v=−1
15 −1
20
1
v=−4
60 −3
60 =−7
60
v=60
−7=−8.57 cm
Step 5: Calculate the magnification m. The magnification is given by:
m=−v
u
Step 6: Substitute the values of vand uinto the magnification formula.
m=−−8.57
20 = 0.4285
Therefore, the image distance is -8.57 cm and the magnification is 0.4285.
Question 23
Question
A converging lens with a focal length of 15 cm is placed 10 cm in front of a
diverging lens with a focal length of 20 cm. Determine the position and nature
of the final image formed by the system.
Solution
Step 1: Determine the position of the image formed by the converging lens.
Using the lens equation for the converging lens:
1
flens
=1
do
+1
di
where flens is the focal length of the lens, dois the object distance, and diis the
image distance. Plugging in the values for the converging lens:
1
15 =1
10 +1
di
Solving for di:
di=1
1
15 −1
10
= 30 cm
Step 2: Determine the position of the image formed by the diverging lens.
The object distance for the diverging lens is the image distance of the converging
lens, do= 30 cm. Using the lens equation for the diverging lens:
1
flens
=1
do
+1
di
21
Plugging in the values for the diverging lens:
1
20 =1
30 +1
d′
i
Solving for d′
i:
d′
i=1
1
20 −1
30
=−60 cm
Step 3: Determine the nature of the final image. The negative value of
d′
iindicates that the image formed by the diverging lens is virtual. Since the
final image is formed on the same side as the object, it is also virtual. The
magnification of the system can be calculated using the magnification equation:
M=d′
i
do
=−60
30 =−2
The negative magnification indicates that the image is inverted. Therefore, the
final image formed by the system of lenses is virtual, inverted, and magnified.
Question 24
Question
An object is placed 20 cm in front of a converging lens with a focal length of
10 cm. Determine the position and magnification of the image formed by the
lens.
Solution
Step 1: Identify the given values and known quantities.
Object distance (p) = −20 cm (negative sign indicates object is placed in
front of the lens)
Focal length (f) = 10 cm
Step 2: Apply the lens equation to find the image distance (q). The lens
equation is given by: 1
f=1
p+1
q
Substitute the known values:
1
10 =1
−20 +1
q
Solve for q:1
q=1
10 −1
−20 =1
10 +1
20 =3
20
22
⇒q=20
3cm
Step 3: Calculate the magnification (M) of the image. The magnification is
given by:
M=−q
p
Substitute the values of pand q:
M=−
20
3
−20 =20
3×20 =1
3
Step 4: Analyze the results. The image is formed at a distance of 20
3cm
from the lens on the same side as the object. The magnification of the image is
1
3, indicating that the image is reduced in size.
Question 25
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Given that the object distance (u) is -20 cm and the focal length (f)
is -15 cm (the negative sign indicates that the object distance and focal length
are measured on the side of the mirror opposite to the incoming light).
Step 2: Using the mirror formula 1
f=1
u+1
v, where vis the image distance:
Step 3: Plugging in the values, we get 1
−15 =1
−20 +1
v.
Step 4: Simplifying, we have −1
15 =−1
20 +1
v.
Step 5: Rearranging the equation, we find 1
v=−1
15 +1
20 .
Step 6: Finding a common denominator, we get 1
v=−4
60 +3
60 .
Step 7: Adding the fractions gives 1
v=−1
60 .
Step 8: Therefore, v=−60 cm.
Step 9: The negative sign indicates that the image is formed on the same
side as the object, meaning it is a virtual image.
Step 10: Since the image is formed on the same side as the object, it is erect
and diminished compared to the object.
Question 26
Question
An object is placed 20 cm in front of a convex lens with a focal length of 10 cm.
Determine the image distance and magnification produced by the lens.
23
Solution
Step 1: Use the lens equation 1
f=1
do+1
diwhere fis the focal length of the
lens, dois the object distance, and diis the image distance.
Given that f= 10 cm and do=−20 cm, we can solve for di.
1
10 =1
−20 +1
di
1
di
=1
10 +1
20
1
di
=3
20
di=20
3cm = 6.67 cm
Step 2: Calculate the magnification using the formula M=−di
do.
Given that di= 6.67 cm and do=−20 cm, we can find the magnification.
M=−6.67
−20
M= 0.333
Therefore, the image distance produced by the lens is 6.67 cm and the mag-
nification is 0.333.
Question 27
Question
A student wants to set up a simple optical system using a convex lens and a
concave mirror. The student wants to form an image of an object located 40
cm from the lens on the same side as the object, and the final image should be
virtual and 25 cm from the lens. If the focal length of the lens is 20 cm and the
focal length of the mirror is 15 cm, determine the required distance between the
lens and the mirror for the system to work.
Solution
1. Let’s denote the distance between the lens and the mirror as d. We need to
use the lens equation and the mirror equation to solve for d.
2. The lens equation is given by:
1
flens
=1
do
+1
di
where flens is the focal length of the lens, dois the object distance, and diis the
image distance.
24
3. Substituting the given values into the lens equation:
1
20 =1
40 +1
−25
1
20 =1
40 −1
25
1
20 =5
200 −8
200
1
20 =−3
200
4. We find that this is not possible, which means the initial setup is not
possible.
5. It’s important to note that for the lens to form a virtual image on the
same side as the object, the object distance must be less than the focal length
of the lens. In this case, the object distance is greater than the focal length of
the lens.
6. Therefore, it is not possible to create a virtual image as described in the
initial problem statement using a convex lens and a concave mirror with the
given focal lengths.
Question 28
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and magnification produced by the mirror.
Solution
Step 1: Recall the mirror formula for concave mirrors: 1
f=1
v+1
u, where fis
the focal length, vis the image distance, and uis the object distance.
Step 2: Given that f=−10 cm (since it is a concave mirror) and u=−20
cm (since the object is placed in front of the mirror), we can plug these values
into the mirror formula to solve for v.
1
−10 =1
v+1
−20
Step 3: Simplifying the equation gives:
−1
10 =1
v−1
20
Step 4: Adding 1/20 to both sides and taking the reciprocal of both sides
gives: −10
11 =v
25
Step 5: Therefore, the image distance vis −10
11 cm.
Step 6: To find the magnification m, we use the formula m=−v
u.
m=−−10
11
−20 =10
22 =5
11
Step 7: Thus, the image distance produced by the mirror is −10
11 cm and the
magnification is 5
11 .
Question 29
Question
A converging lens with a focal length of 15 cm is placed 20 cm to the left of
a converging mirror with a focal length of 20 cm. Determine the final image
distance relative to the lens, the magnification, and whether the final image is
real or virtual.
Solution
Step 1: Calculate the image distance relative to the lens using the lens equation:
1
flens
=1
do
+1
di
Given that flens = 15 cm and do=−20 cm (because the object is placed to
the left of the lens), we can substitute these values into the equation to solve
for di:1
15 =1
−20 +1
di
Solving for di, we get:
di=1
1
15 −1
−20
di=20 ×15
15 + 20
di=300
35
di= 8.57 cm
Step 2: Determine the magnification of the system using the formula:
M=−di
do
Substitute the values of di= 8.57 cm and do=−20 cm into the formula to find
the magnification, M:
M=−8.57
−20
26
M= 0.429
Step 3: Determine if the final image is real or virtual based on the sign of
the image distance: Since diis positive, the final image is real.
Therefore, the final image distance relative to the lens is 8.57 cm, the mag-
nification is 0.429, and the final image is real.
Question 30
Question
An object is placed 10 cm in front of a concave mirror of focal length 15 cm.
Calculate the image distance, magnification, and describe the nature of the
image formed.
Solution
Step 1: Identify the given values: The object distance o= 10 cm (in front of
the mirror) and the focal length f= -15 cm (since it’s a concave mirror, the
focal length is negative).
Step 2: Apply the mirror formula to find the image distance i: The mirror
formula is 1
f=1
o+1
i. Plugging in the values, we get: 1
−15 =1
10 +1
i. Solving
for i, we get: −1
15 =1
i−1
10
1
i=1
10 −1
15
1
i=3−2
30 =1
30
Therefore, i= 30 cm.
Step 3: Calculate the magnification m: The magnification is given by m=
−i
o. Plugging in the values, we get: m=−30
10 =−3.
Step 4: Determine the nature of the image: Since the magnification is nega-
tive, the image is inverted. Also, since the image distance is positive, the image
is real.
Therefore, the image is real, inverted, and magnified.
Question 31
Question
A concave mirror with a focal length of 15 cm is used to form an image of an
object placed 10 cm from the mirror. Determine the position, magnification,
and nature of the image formed.
Solution
Step 1: Given that the focal length of the mirror is f=−15 cm and the object
distance is do=−10 cm. We will use the mirror formula to calculate the image
27
distance. The mirror formula is given by:
1
f=1
di
+1
do
where fis the focal length, diis the image distance, and dois the object distance.
Step 2: Substituting the given values into the mirror formula, we get:
1
−15 =1
di
+1
−10
Step 3: Solving for di:1
di
=1
−15 −1
−10
1
di
=2
30
di=−15 cm
Step 4: The negative sign for diindicates that the image is formed on the
same side as the object. Now we will calculate the magnification using the
magnification formula, which is given by:
m=−di
do
where mis the magnification, diis the image distance, and dois the object
distance.
Step 5: Substituting the values of diand dointo the magnification formula,
we get:
m=−−15
−10 =3
2
Step 6: The positive magnification indicates that the image is upright. Fi-
nally, let’s determine the nature of the image. Since the magnification is greater
than 1, the image is magnified. And since the image is formed on the same
side as the object, it is a virtual image. Hence, the image formed is a virtual,
magnified, and upright image.
Question 32
Question
An object is placed 30 cm to the left of a convex lens with a focal length of 20
cm. Calculate the image distance and magnification of the lens system.
28
Solution
Step 1: Identify the given values and the type of lens. Given: - Object distance
(do) = -30 cm (negative sign indicates object is on the left side of the lens) -
Focal length (f) = 20 cm Since the focal length is positive for a convex lens, it
is a converging lens.
Step 2: Apply the lens equation to find the image distance. The lens equation
relates the object distance (do), image distance (di), and focal length (f) of the
lens: 1
f=1
do
+1
di
Substitute the given values into the lens equation:
1
20 =1
−30 +1
di
Step 3: Solve for the image distance. Solving the equation for di:
1
20 =1
−30 +1
di
1
di
=1
20 −1
−30
1
di
=3
60 +2
60
1
di
=5
60
di=60
5
di= 12 cm
Step 4: Calculate the magnification. The magnification (M) of the lens
system is given by the formula:
M=−di
do
Substitute the values of diand do:
M=−12
−30
M= 0.4
Therefore, the image distance is 12 cm and the magnification of the lens
system is 0.4.
29
Question 33
Question
An object is placed 15 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance, magnification, and describe the nature of
the image.
Solution
Step 1: Identify the given values and mirror equation.
Given: - Object distance, do=−15 cm (negative sign indicates object is
in front of the mirror) - Focal length, f=−10 cm (negative sign for concave
mirror) - Mirror equation: 1
f=1
do+1
di
Step 2: Calculate the image distance. Using the mirror equation:
1
−10 =1
−15 +1
di
−1
10 =−1
15 +1
di
1
di
=1
15 −1
10
1
di
=2
30 −3
30
1
di
=−1
30
di=−30 cm
Therefore, the image distance is −30 cm.
Step 3: Calculate the magnification. Using the magnification formula:
m=−di
do
m=−−30
−15
m=−2
Therefore, the magnification is −2.
Step 4: Determine the nature of the image. Since the magnification is nega-
tive, the image is inverted. Additionally, since the image distance is larger than
the object distance, the image is real.
Therefore, the image is real, inverted, and magnified with a magnification of
−2 and located 30 cm in front of the mirror.
30
Question 34
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Identify the given values and mirror equation. Given: - Object distance,
p=−30 cm (negative since the object is in front of the mirror) - Focal length
of the concave mirror, f=−20 cm (negative for concave mirror)
The mirror equation relates the object distance (p), image distance (q), and
focal length (f) of a mirror: 1
f=1
p+1
q
Step 2: Calculate the image distance. Substitute the given values into the
mirror equation: 1
−20 =1
−30 +1
q
Solving for q:1
q=1
−20 −1
−30 =−1
60
q=−60 cm
Therefore, the image distance, q, is -60 cm, which means the image forms
on the same side as the object (in front of the mirror).
Step 3: Determine the nature of the image. Since the image distance (q)
is negative, according to the sign conventions for mirrors, the image is virtual.
The negative sign indicates that the image is formed on the same side as the
object (in front of the mirror).
Further, since the image distance is greater in magnitude compared to the
object distance, the image is magnified compared to the object.
Question 35
Question
A convex lens has a focal length of 15 cm. An object is placed 30 cm from the
lens. Calculate the image distance from the lens and determine if the image is
real or virtual.
31
Solution
Step 1: Identify the given values and assign the sign conventions.
Given that the focal length of the convex lens, f= 15 cm.
The object distance from the lens, u=−30 cm (since the object is placed
on the opposite side of the incident light).
The focal length of a convex lens is positive.
Distances on the opposite side of the incident light are negative.
Step 2: Apply the lens formula to find the image distance, v.
1
f=1
v−1
u
Substitute the given values into the formula:
1
15 =1
v−1
−30
Step 3: Solve for image distance, v.
1
v=1
15 −1
30
1
v=2
30 −1
30
1
v=1
30
v= 30 cm
Step 4: Analyze if the image is real or virtual. Since the image distance vis
positive, the image is formed on the same side as the outgoing light. Therefore,
the image is virtual.
Therefore, the image distance from the lens is 30 cm and the image is virtual.
32
Question 2
Question
An object is placed 15 cm from a concave mirror with a focal length of 10 cm.
Determine the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values. The object distance (do) is 15 cm, the focal
length (f) is -10 cm (since the mirror is concave), and the magnification formula
is m=−di
do.
Step 2: Apply the mirror equation. The mirror equation is 1
f=1
do+1
di.
Substituting the known values gives us 1
−10 =1
15 +1
di.
Step 3: Solve for the image distance. Solving the equation for di, we get
1
di=1
−10 −1
15 =−1
30 . Therefore, di=−30 cm.
Step 4: Calculate the magnification. Using the formula m=−di
do, we sub-
stitute di=−30 cm and do= 15 cm to find m=−−30
15 = 2.
Step 5: Interpret the results. The image is formed at a distance of 30 cm from
the mirror (behind the mirror), and the image is magnified with a magnification
factor of 2.
Question 3
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Determine the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values and the mirror equation. The given values are:
Object distance, do=−30 cm (since the object is in front of the mirror, the
distance is negative) Focal length, f=−20 cm (negative for concave mirror)
The mirror equation is: 1
f=1
do
+1
di
Step 2: Substitute the values into the mirror equation. Substitute f=
−20 cm and do=−30 cm into the mirror equation to find di.
1
−20 =1
−30 +1
di
Step 3: Solve for the image distance, di.
1
di
=1
−20 −1
−30 =3
60 =1
20
2
di= 20 cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−di
do
Substitute di= 20 cm and do=−30 cm into the magnification formula.
M=−20
−30 =2
3
Therefore, the image is formed 20 cm from the mirror on the same side as
the object, and the magnification of the image is 2
3.
Question 4
Question
An object is placed 40 cm from a concave mirror with a focal length of 30 cm.
Determine the image distance and magnification of the image formed by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
f=−30 cm
p=−40 cm
Mirror equation: 1
f=1
p+1
q
Step 2: Substitute the given values into the mirror equation.
1
−30 =1
−40 +1
q
Step 3: Solve for q.
−1
30 =−1
40 +1
q
1
q=−1
30 +1
40
1
q=−4+3
120
1
q=−1
120
3
q=−120 cm
Step 4: Calculate the magnification using the magnification equation.
M=−q
p
M=−−120
−40
M= 3
Answer: The image distance is -120 cm and the magnification of the image
formed by the mirror is 3.
Question 5
Question
An object is placed 30 cm in front of a concave mirror of focal length 15 cm.
Determine the position and nature of the image formed.
Solution
Step 1: Identify the given values and the mirror formula.
Object distance (u) = -30 cm (since it is in front of the mirror, we take it
as negative)
Focal length (f) = -15 cm (since it is a concave mirror, its focal length is
negative)
Mirror formula: 1
f=1
v+1
u
Step 2: Substitute the given values in the mirror formula and solve for the
image distance (v).
1
−15 =1
v+1
−30
−1
15 =1
v−1
30
−1
15 =2−v
2v
−2v= 2v−30
4v= 30
v= 7.5 cm
Step 3: Determine the position of the image. Since the image distance (v)
is positive, the image is formed on the same side as the object, i.e., in front of
the mirror.
Step 4: Determine the nature of the image. To determine the nature of the
image, we compare the object distance (u) and the focal length (f).
4
If f > |v|(image distance), the image is real and inverted.
If f < |v|(image distance), the image is virtual and erect.
Since |v|= 7.5 cm and f= 15 cm, we have f < |v|, so the image is virtual and
erect.
Therefore, the image is formed 7.5 cm in front of the concave mirror and is
virtual and erect.
Question 6
Question
A concave mirror has a focal length of 15 cm. An object is placed 30 cm in
front of the mirror. Determine the image distance and magnification produced
by the mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Given: f=−15 cm (concave mirror focal length), do=−30 cm (object
distance).
Mirror equation: 1
f=1
do
+1
di
.
Step 2: Substitute the given values into the mirror equation and solve for
di.
1
−15 =1
−30 +1
di
−2
30 =−1
di
di=−15 cm
Therefore, the image is formed 15 cm in front of the mirror.
Step 3: Calculate the magnification using the magnification equation.
Magnification equation: m=−di
do
.
Step 4: Substitute the values of diand dointo the magnification equation
and solve for m.
m=−−15
−30
m=1
2
Therefore, the image produced by the concave mirror is real, formed 15 cm
in front of the mirror, and is half the size of the object.
5
Question 7
Question
An object is placed 23 cm in front of a concave mirror of focal length -20 cm.
Determine the position and size of the image formed by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given: Object
distance, do=−23 cm (negative because it is in front of the mirror), Focal
length, f=−20 cm.
The mirror equation is: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation.
1
−20 =1
−23 +1
di
Step 3: Solve for the image distance, di.
1
−20 =1
−23 +1
di
−3
60 =1
di
di=−60
3
di=−20 cm
So, the image is formed 20 cm in front of the mirror.
Step 4: Use the magnification formula to find the size of the image. The
magnification formula is:
m=−di
do
Step 5: Substitute the known values into the magnification formula.
m=−−20
−23
m=20
23
Step 6: The negative sign in the magnification equation indicates that the
image is inverted. The magnification factor also tells us that the image is smaller
than the object since |m|<1.
Therefore, the image formed by the concave mirror is 20 cm in front of the
mirror and smaller than the object.
6
Question 8
Question
An object is placed 15 cm in front of a concave mirror with a focal length of
10 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
Solution
Step 1: Identify the given values. The object distance ois 15 cm (in front of
the mirror), and the focal length fof the concave mirror is -10 cm (since it is a
concave mirror, the focal length is negative). Given: o= 15 cm, f=−10 cm.
Step 2: Use the mirror equation to find the image distance. The mirror
equation is given by: 1
f=1
o+1
i
Substitute the given values:
1
−10 =1
15 +1
i
Solving for i:
−1
10 =1
15 +1
i
−1
10 −1
15 =1
i
−3
30 −2
30 =1
i
−5
30 =1
i
−i= 6
i=−6 cm
Therefore, the image distance iis -6 cm.
Step 3: Calculate the magnification. The magnification Mis given by:
M=−i
o
Substitute the values obtained:
M=−−6
15
M=2
5
Thus, the magnification is 2
5.
Step 4: Determine if the image is real or virtual. Since the image distance
is negative, the image is formed on the same side as the object. Therefore, the
image is virtual.
7
Question 9
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Identify the given values and the type of mirror. Given:
Object distance, u=−20 cm (negative because the object is in front of
the mirror)
Focal length, f=−15 cm (negative for a concave mirror)
Since the focal length is negative, the mirror is a concave mirror.
Step 2: Apply the mirror equation to find the image distance. The mirror
equation is: 1
f=1
v+1
u
Substitute the given values: f=−15 cm, u=−20 cm
1
−15 =1
v+1
−20
Step 3: Solve for image distance, v.
−1
15 =1
v−1
20
−1
15 +1
20 =1
v
4
60 −3
60 =1
v
1
60 =1
v
v= 60 cm
Therefore, the image is formed 60 cm behind the mirror.
Step 4: Determine the nature of the image. To determine the nature of the
image, we consider the sign of v: - If vis positive, the image is real and located
on the opposite side of the object. - If vis negative, the image is virtual and
located on the same side as the object.
Since vis positive, the image formed is real and located 60 cm behind the
mirror.
8
Question 10
Question
A concave mirror with a focal length of 10 cm forms an image that is four times
the size of the object. If the object is placed 20 cm from the mirror, at what
distance is the image formed from the mirror?
Solution
Let fbe the focal length of the concave mirror, pbe the object distance, and q
be the image distance.
Step 1: Given that the mirror equation is 1
f=1
p+1
qand the magnification
equation is m=−q
p, we can find the image distance q.
First, we know that m=h′
h=−q
p=−4, where h′and hare the heights of
the image and object respectively.
Step 2: Substituting m=−4 into m=−q
p, we get −4 = −q
p. Therefore,
q= 4p.
Step 3: Now, substitute f= 10 cm and p= 20 cm into the mirror equation
1
f=1
p+1
qto find q.
1
10 =1
20 +1
q
1
10 −1
20 =1
q
2
20 −1
20 =1
q
1
20 =1
q
q= 20 cm
Step 4: Therefore, the image is formed at a distance of 20 cm from the
concave mirror.
Question 11
Question
A converging lens with a focal length of 20 cm is placed 30 cm in front of a
diverging mirror with a focal length of 10 cm. Determine the final image distance
from the mirror, whether the image is real or virtual, and the magnification of
the system.
9
Solution
Step 1: Let’s label the given focal lengths and object distance as fl= 20 cm (for
the converging lens), fm=−10 cm (for the diverging mirror), and do=−30 cm
(object distance in front of the lens) respectively.
Step 2: Calculate the image distance from the lens using the lens formula
1
fl
=1
do
+1
di
.
Substitute the given values into the lens formula:
1
20 =1
−30 +1
di
Solve for di:
di=1
1
20 −1
30
=−60 cm
Therefore, the image distance from the lens is di=−60 cm.
Step 3: Calculate the image distance from the mirror using the mirror for-
mula 1
fm
=1
di
+1
dm
.
Since the image distance from the lens is -60 cm, we have the following
equation: 1
−10 =1
−60 +1
dm
Solve for dm:
dm=1
1
−10 −1
−60
= 15 cm
Therefore, the final image distance from the mirror is dm= 15 cm.
Step 4: Determine whether the image is real or virtual by calculating the
magnification. The magnification Mof the system is given by di
do
.
Substitute the values of diand do:
M=−60
−30 = 2
Since M > 0, the image is upright. Therefore, the image is virtual.
Step 5: The magnification of the system is calculated as M= 2, indicating
that the image is upright and magnified by a factor of 2.
Question 12
Question
A concave mirror has a focal length of −15 cm. An object is placed 25 cm in front
of the mirror. Determine the image distance, image height, and magnification.
10
Solution
Step 1: Identify the given values and the mirror equation.
Given values: Focal length of mirror, f=−15 cm Object distance, do=−25
cm
The mirror equation for concave mirrors is:
1
f=1
do
+1
di
Step 2: Substitute the given values into the mirror equation to find the image
distance, di.
1
−15 =1
−25 +1
di
−1
15 =−1
25 +1
di
1
di
=−1
15 +1
25
1
di
=−5+3
75
1
di
=−2
75
di=−75
2
di=−37.5 cm
Therefore, the image distance is −37.5 cm.
Step 3: Use the magnification formula to calculate the magnification, M.
The magnification formula is:
M=−di
do
Substitute the values of diand dointo the formula:
M=−(−37.5)
−25
M=37.5
25
M= 1.5
Therefore, the magnification is 1.5.
11
Step 4: Use the magnification to find the image height, hi, given the object
height, ho(assuming ho= 1 unit).
The image height is related to the object height by:
hi=M×ho
Substitute the values of Mand hointo the formula:
hi= 1.5×1
hi= 1.5
Thus, the image height is 1.5 units.
Question 13
Question
An object is placed 20 cm from a concave mirror with a focal length of 10 cm.
Calculate the image distance and the magnification of the image formed.
Solution
Step 1: Identify the given and unknown quantities. Given: f=−10 cm (focal
length of the concave mirror) do=−20 cm (object distance)
Unknown: di(image distance) M(magnification)
Step 2: Apply the lens/mirror equation for concave mirrors:
1
f=1
do
+1
di
Step 3: Substitute the given values and solve for di:
1
−10 =1
−20 +1
di
1
di
=1
−10 −1
−20
1
di
=−1
10 +1
20
1
di
=−2−1
20
1
di
=−1
20
di=−20 cm
12
Step 4: Calculate the magnification using the magnification formula:
M=−di
do
M=−−20
−20
M= 1
Answer: The image distance is -20 cm and the magnification is 1.
Question 14
Question
An object is placed 10 cm in front of a concave mirror with a focal length of 20
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. The given values are:
- Object distance, do=−10 cm (negative since it is in front of the mirror) -
Focal length, f=−20 cm (negative for a concave mirror) The mirror equation
for concave mirrors is given by:
−1
f=1
di
+1
do
where di= image distance
Step 2: Substitute the given values into the mirror equation.
−1
−20 =1
di
+1
−10
Step 3: Solve for di.1
20 =1
di
−1
10
1
di
=1
20 +1
10
1
di
=1
20 +2
20
1
di
=3
20
di=20
3= 6.67 cm
The image is formed 6.67 cm behind the mirror which is a real image.
13
Step 4: Determine the magnification. The magnification, m, is given by the
formula:
m=−di
do
Substitute the values of diand do:
m=−6.67
−10
m= 0.67
The negative magnification indicates that the image is inverted.
Step 5: Determine the nature of the image. Since the image distance is
positive, the image is real. The positive magnification value indicates that the
image is upright.
Therefore, the image formed by the concave mirror is real, inverted, and
upright. It is located 6.67 cm behind the mirror.
Question 15
Question
An object is placed 30 cm in front of a convex lens with a focal length of 20 cm.
Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values and the lens formula: Given: - Object distance
u=−30 cm (negative since it is in front of the lens), - Focal length f= 20 cm
for a convex lens.
The lens formula relates object distance (u), image distance (v), and focal
length (f): 1
f=1
v+1
u
Step 2: Substitute the values into the lens formula:
1
20 =1
v+1
−30
Step 3: Solve for v:1
20 =1
v−1
30
1
v=1
20 +1
30
1
v=3
60 +2
60
14
1
v=5
60
v=60
5= 12 cm
Therefore, the image distance vis 12 cm.
Step 4: Calculate the magnification: The magnification (m) of the lens is
given by the formula:
m=−v
u
Step 5: Substitute the values of vand uinto the magnification formula:
m=−12
−30 =2
5
Hence, the magnification produced by the lens is 2
5.
Question 16
Question
An object is placed 15 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and magnification of the image formed by
the mirror.
Solution
Step 1: Recall the mirror equation:
1
f=1
do
+1
di
where fis the focal length of the mirror, dois the object distance, and diis the
image distance.
Step 2: Substituting the given values, we get:
1
10 =1
15 +1
di
Step 3: Solve for di:
1
di
=1
10 −1
15 =3−2
30 =1
30
⇒di= 30 cm
Step 4: Next, calculate the magnification using the formula:
M=−di
do
15
Step 5: Substitute the values:
M=−30
15 =−2
Step 6: Therefore, the image distance is 30 cm and the magnification of the
image formed by the mirror is -2.
Question 17
Question
An object is placed 15 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and the magnification of the image.
Solution
Step 1: Identify known values and the given information.
The given values are:
Object distance do=−15 cm (negative since it is in front of the mirror)
Focal length f= 10 cm (negative for a concave mirror)
Step 2: Apply the mirror equation to find the image distance di.
The mirror equation is given by:
1
f=1
di
+1
do
Plug in the known values and solve for di:
1
−10 =1
di
+1
−15
−1
10 +1
15 =1
di
3
30 −2
30 =1
di
1
30 =1
di
di= 30 cm
Step 3: Calculate the magnification M.
The magnification is given by:
M=−di
do
16
Plug in the calculated values:
M=−30
−15 = 2
Therefore, the image distance is 30 cm and the magnification of the image
is 2.
Question 18
Question
An object is placed 20 cm in front of a converging lens with a focal length of
10 cm. The image produced by the lens is then used as an object for a concave
mirror with a focal length of 15 cm. Determine the final image distance from
the mirror.
Solution
Step 1: Calculate the image distance produced by the lens using the lens equa-
tion. 1
flens
=1
do
+1
di
1
10 =1
20 +1
di
Solving for di:1
di
=1
10 −1
20
1
di
=2
20 −1
20
1
di
=1
20
di= 20 cm
Step 2: Use the mirror equation to find the final image distance from the
mirror. 1
fmirror
=1
d′
i
+1
d′
o
Given that fmirror =−15 cm and d′
o=−20 cm (since the image produced by
the lens is a virtual object for the mirror), and d′
i=?,
1
−15 =1
d′
i
+1
−20
Solving for d′
i:
1
d′
i
=−1
15 +1
20
17
1
d′
i
=−4
60 +3
60
1
d′
i
=−1
60
d′
i=−60 cm
Therefore, the final image distance from the mirror is −60 cm.
Question 19
Question
An object is placed 20 cm in front of a convex lens of focal length 15 cm. If the
image formed is virtual, upright, and 5 cm in front of the lens, determine the
magnification of the image.
Solution
Step 1: Identify the given values and the formula for magnification. Let do=
−20 cm (object distance), f= 15 cm (focal length), di=−5 cm (image dis-
tance), and mbe the magnification. The formula for magnification is given
by:
m=−di
do
Step 2: Convert all distances to positive values. Since the image is virtual,
the image distance diis considered positive. Thus, |di|= 5 cm.
Step 3: Solve for the magnification. Substitute the given values into the
formula for magnification:
m=−5
−20 =1
4
Step 4: Interpret the result. The magnification of the image is 1
4, indicating
that the image is one-fourth the size of the object.
Question 20
Question
A concave mirror with a focal length of 20 cm creates a virtual image that is
15 cm tall. If the object height is 5 cm, determine the object distance from the
mirror.
18
Solution
Let fbe the focal length of the concave mirror, hibe the height of the image,
and hobe the height of the object. Let dobe the object distance from the
mirror. Using the mirror equation and the magnification equation, we can solve
for do.
Step 1: Write down the mirror equation:
1
f=1
do
+1
di
where diis the image distance from the mirror. Given that f=−20 cm and
hi=−15 cm, we can find di:
1
−20 =1
do
+1
di
=⇒1
−20 =1
do
−1
15
Step 2: Solve for di:
1
−20 =1
do
−1
15 =⇒1
do
−1
15 =1
−20 =⇒1
do
=1
15 −1
20
Step 3: Find dousing the magnification equation:
hi
ho
=−di
do
=⇒−15
5=−1
do
=⇒ −3 = −1
do
Step 4: Solve for do:
−3 = −1
do
=⇒1
do
= 3 =⇒do=1
3= 3 cm
Therefore, the object distance from the mirror is 3 cm.
Question 21
Question
An object is placed 15 cm to the left of a converging lens of focal length 10 cm.
Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the lens equation. Given: Object distance,
u=−15 cm (negative sign indicates object is to the left of the lens) Focal length,
f= 10 cm
The lens formula relates object distance (u), image distance (v), and focal
length (f): 1
f=1
v+1
u
19
Step 2: Plug in the known values into the lens equation.
1
10 =1
v+1
−15
Step 3: Solve for image distance (v).
1
v=1
10 +1
15 =3+2
30 =5
30 =1
6
v= 6 cm
Therefore, the image distance is 6 cm.
Step 4: Calculate the magnification (m) using the formula:
m=−v
u
Step 5: Plug in the values to calculate magnification.
m=−6
−15 =2
5
Therefore, the magnification is 2
5.
Question 22
Question
An object is placed 20 cm in front of a concave lens with a focal length of 15
cm. Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values. The object distance uis 20 cm and the focal
length fof the lens is -15 cm (concave lens, so focal length is negative).
Step 2: Apply the lens equation. The lens equation is given by:
1
f=1
v+1
u
Where: - fis the focal length, - vis the image distance, and - uis the object
distance.
Step 3: Substitute the given values into the lens equation.
1
−15 =1
v+1
20
Step 4: Solve for the image distance v.
−1
15 =1
v+1
20
20
1
v=−1
15 −1
20
1
v=−4
60 −3
60 =−7
60
v=60
−7=−8.57 cm
Step 5: Calculate the magnification m. The magnification is given by:
m=−v
u
Step 6: Substitute the values of vand uinto the magnification formula.
m=−−8.57
20 = 0.4285
Therefore, the image distance is -8.57 cm and the magnification is 0.4285.
Question 23
Question
A converging lens with a focal length of 15 cm is placed 10 cm in front of a
diverging lens with a focal length of 20 cm. Determine the position and nature
of the final image formed by the system.
Solution
Step 1: Determine the position of the image formed by the converging lens.
Using the lens equation for the converging lens:
1
flens
=1
do
+1
di
where flens is the focal length of the lens, dois the object distance, and diis the
image distance. Plugging in the values for the converging lens:
1
15 =1
10 +1
di
Solving for di:
di=1
1
15 −1
10
= 30 cm
Step 2: Determine the position of the image formed by the diverging lens.
The object distance for the diverging lens is the image distance of the converging
lens, do= 30 cm. Using the lens equation for the diverging lens:
1
flens
=1
do
+1
di
21
Plugging in the values for the diverging lens:
1
20 =1
30 +1
d′
i
Solving for d′
i:
d′
i=1
1
20 −1
30
=−60 cm
Step 3: Determine the nature of the final image. The negative value of
d′
iindicates that the image formed by the diverging lens is virtual. Since the
final image is formed on the same side as the object, it is also virtual. The
magnification of the system can be calculated using the magnification equation:
M=d′
i
do
=−60
30 =−2
The negative magnification indicates that the image is inverted. Therefore, the
final image formed by the system of lenses is virtual, inverted, and magnified.
Question 24
Question
An object is placed 20 cm in front of a converging lens with a focal length of
10 cm. Determine the position and magnification of the image formed by the
lens.
Solution
Step 1: Identify the given values and known quantities.
Object distance (p) = −20 cm (negative sign indicates object is placed in
front of the lens)
Focal length (f) = 10 cm
Step 2: Apply the lens equation to find the image distance (q). The lens
equation is given by: 1
f=1
p+1
q
Substitute the known values:
1
10 =1
−20 +1
q
Solve for q:1
q=1
10 −1
−20 =1
10 +1
20 =3
20
22
⇒q=20
3cm
Step 3: Calculate the magnification (M) of the image. The magnification is
given by:
M=−q
p
Substitute the values of pand q:
M=−
20
3
−20 =20
3×20 =1
3
Step 4: Analyze the results. The image is formed at a distance of 20
3cm
from the lens on the same side as the object. The magnification of the image is
1
3, indicating that the image is reduced in size.
Question 25
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Given that the object distance (u) is -20 cm and the focal length (f)
is -15 cm (the negative sign indicates that the object distance and focal length
are measured on the side of the mirror opposite to the incoming light).
Step 2: Using the mirror formula 1
f=1
u+1
v, where vis the image distance:
Step 3: Plugging in the values, we get 1
−15 =1
−20 +1
v.
Step 4: Simplifying, we have −1
15 =−1
20 +1
v.
Step 5: Rearranging the equation, we find 1
v=−1
15 +1
20 .
Step 6: Finding a common denominator, we get 1
v=−4
60 +3
60 .
Step 7: Adding the fractions gives 1
v=−1
60 .
Step 8: Therefore, v=−60 cm.
Step 9: The negative sign indicates that the image is formed on the same
side as the object, meaning it is a virtual image.
Step 10: Since the image is formed on the same side as the object, it is erect
and diminished compared to the object.
Question 26
Question
An object is placed 20 cm in front of a convex lens with a focal length of 10 cm.
Determine the image distance and magnification produced by the lens.
23
Solution
Step 1: Use the lens equation 1
f=1
do+1
diwhere fis the focal length of the
lens, dois the object distance, and diis the image distance.
Given that f= 10 cm and do=−20 cm, we can solve for di.
1
10 =1
−20 +1
di
1
di
=1
10 +1
20
1
di
=3
20
di=20
3cm = 6.67 cm
Step 2: Calculate the magnification using the formula M=−di
do.
Given that di= 6.67 cm and do=−20 cm, we can find the magnification.
M=−6.67
−20
M= 0.333
Therefore, the image distance produced by the lens is 6.67 cm and the mag-
nification is 0.333.
Question 27
Question
A student wants to set up a simple optical system using a convex lens and a
concave mirror. The student wants to form an image of an object located 40
cm from the lens on the same side as the object, and the final image should be
virtual and 25 cm from the lens. If the focal length of the lens is 20 cm and the
focal length of the mirror is 15 cm, determine the required distance between the
lens and the mirror for the system to work.
Solution
1. Let’s denote the distance between the lens and the mirror as d. We need to
use the lens equation and the mirror equation to solve for d.
2. The lens equation is given by:
1
flens
=1
do
+1
di
where flens is the focal length of the lens, dois the object distance, and diis the
image distance.
24
3. Substituting the given values into the lens equation:
1
20 =1
40 +1
−25
1
20 =1
40 −1
25
1
20 =5
200 −8
200
1
20 =−3
200
4. We find that this is not possible, which means the initial setup is not
possible.
5. It’s important to note that for the lens to form a virtual image on the
same side as the object, the object distance must be less than the focal length
of the lens. In this case, the object distance is greater than the focal length of
the lens.
6. Therefore, it is not possible to create a virtual image as described in the
initial problem statement using a convex lens and a concave mirror with the
given focal lengths.
Question 28
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and magnification produced by the mirror.
Solution
Step 1: Recall the mirror formula for concave mirrors: 1
f=1
v+1
u, where fis
the focal length, vis the image distance, and uis the object distance.
Step 2: Given that f=−10 cm (since it is a concave mirror) and u=−20
cm (since the object is placed in front of the mirror), we can plug these values
into the mirror formula to solve for v.
1
−10 =1
v+1
−20
Step 3: Simplifying the equation gives:
−1
10 =1
v−1
20
Step 4: Adding 1/20 to both sides and taking the reciprocal of both sides
gives: −10
11 =v
25
Step 5: Therefore, the image distance vis −10
11 cm.
Step 6: To find the magnification m, we use the formula m=−v
u.
m=−−10
11
−20 =10
22 =5
11
Step 7: Thus, the image distance produced by the mirror is −10
11 cm and the
magnification is 5
11 .
Question 29
Question
A converging lens with a focal length of 15 cm is placed 20 cm to the left of
a converging mirror with a focal length of 20 cm. Determine the final image
distance relative to the lens, the magnification, and whether the final image is
real or virtual.
Solution
Step 1: Calculate the image distance relative to the lens using the lens equation:
1
flens
=1
do
+1
di
Given that flens = 15 cm and do=−20 cm (because the object is placed to
the left of the lens), we can substitute these values into the equation to solve
for di:1
15 =1
−20 +1
di
Solving for di, we get:
di=1
1
15 −1
−20
di=20 ×15
15 + 20
di=300
35
di= 8.57 cm
Step 2: Determine the magnification of the system using the formula:
M=−di
do
Substitute the values of di= 8.57 cm and do=−20 cm into the formula to find
the magnification, M:
M=−8.57
−20
26
M= 0.429
Step 3: Determine if the final image is real or virtual based on the sign of
the image distance: Since diis positive, the final image is real.
Therefore, the final image distance relative to the lens is 8.57 cm, the mag-
nification is 0.429, and the final image is real.
Question 30
Question
An object is placed 10 cm in front of a concave mirror of focal length 15 cm.
Calculate the image distance, magnification, and describe the nature of the
image formed.
Solution
Step 1: Identify the given values: The object distance o= 10 cm (in front of
the mirror) and the focal length f= -15 cm (since it’s a concave mirror, the
focal length is negative).
Step 2: Apply the mirror formula to find the image distance i: The mirror
formula is 1
f=1
o+1
i. Plugging in the values, we get: 1
−15 =1
10 +1
i. Solving
for i, we get: −1
15 =1
i−1
10
1
i=1
10 −1
15
1
i=3−2
30 =1
30
Therefore, i= 30 cm.
Step 3: Calculate the magnification m: The magnification is given by m=
−i
o. Plugging in the values, we get: m=−30
10 =−3.
Step 4: Determine the nature of the image: Since the magnification is nega-
tive, the image is inverted. Also, since the image distance is positive, the image
is real.
Therefore, the image is real, inverted, and magnified.
Question 31
Question
A concave mirror with a focal length of 15 cm is used to form an image of an
object placed 10 cm from the mirror. Determine the position, magnification,
and nature of the image formed.
Solution
Step 1: Given that the focal length of the mirror is f=−15 cm and the object
distance is do=−10 cm. We will use the mirror formula to calculate the image
27
distance. The mirror formula is given by:
1
f=1
di
+1
do
where fis the focal length, diis the image distance, and dois the object distance.
Step 2: Substituting the given values into the mirror formula, we get:
1
−15 =1
di
+1
−10
Step 3: Solving for di:1
di
=1
−15 −1
−10
1
di
=2
30
di=−15 cm
Step 4: The negative sign for diindicates that the image is formed on the
same side as the object. Now we will calculate the magnification using the
magnification formula, which is given by:
m=−di
do
where mis the magnification, diis the image distance, and dois the object
distance.
Step 5: Substituting the values of diand dointo the magnification formula,
we get:
m=−−15
−10 =3
2
Step 6: The positive magnification indicates that the image is upright. Fi-
nally, let’s determine the nature of the image. Since the magnification is greater
than 1, the image is magnified. And since the image is formed on the same
side as the object, it is a virtual image. Hence, the image formed is a virtual,
magnified, and upright image.
Question 32
Question
An object is placed 30 cm to the left of a convex lens with a focal length of 20
cm. Calculate the image distance and magnification of the lens system.
28
Solution
Step 1: Identify the given values and the type of lens. Given: - Object distance
(do) = -30 cm (negative sign indicates object is on the left side of the lens) -
Focal length (f) = 20 cm Since the focal length is positive for a convex lens, it
is a converging lens.
Step 2: Apply the lens equation to find the image distance. The lens equation
relates the object distance (do), image distance (di), and focal length (f) of the
lens: 1
f=1
do
+1
di
Substitute the given values into the lens equation:
1
20 =1
−30 +1
di
Step 3: Solve for the image distance. Solving the equation for di:
1
20 =1
−30 +1
di
1
di
=1
20 −1
−30
1
di
=3
60 +2
60
1
di
=5
60
di=60
5
di= 12 cm
Step 4: Calculate the magnification. The magnification (M) of the lens
system is given by the formula:
M=−di
do
Substitute the values of diand do:
M=−12
−30
M= 0.4
Therefore, the image distance is 12 cm and the magnification of the lens
system is 0.4.
29
Question 33
Question
An object is placed 15 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance, magnification, and describe the nature of
the image.
Solution
Step 1: Identify the given values and mirror equation.
Given: - Object distance, do=−15 cm (negative sign indicates object is
in front of the mirror) - Focal length, f=−10 cm (negative sign for concave
mirror) - Mirror equation: 1
f=1
do+1
di
Step 2: Calculate the image distance. Using the mirror equation:
1
−10 =1
−15 +1
di
−1
10 =−1
15 +1
di
1
di
=1
15 −1
10
1
di
=2
30 −3
30
1
di
=−1
30
di=−30 cm
Therefore, the image distance is −30 cm.
Step 3: Calculate the magnification. Using the magnification formula:
m=−di
do
m=−−30
−15
m=−2
Therefore, the magnification is −2.
Step 4: Determine the nature of the image. Since the magnification is nega-
tive, the image is inverted. Additionally, since the image distance is larger than
the object distance, the image is real.
Therefore, the image is real, inverted, and magnified with a magnification of
−2 and located 30 cm in front of the mirror.
30
Question 34
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Determine the position and nature of the image formed by the mirror.
Solution
Step 1: Identify the given values and mirror equation. Given: - Object distance,
p=−30 cm (negative since the object is in front of the mirror) - Focal length
of the concave mirror, f=−20 cm (negative for concave mirror)
The mirror equation relates the object distance (p), image distance (q), and
focal length (f) of a mirror: 1
f=1
p+1
q
Step 2: Calculate the image distance. Substitute the given values into the
mirror equation: 1
−20 =1
−30 +1
q
Solving for q:1
q=1
−20 −1
−30 =−1
60
q=−60 cm
Therefore, the image distance, q, is -60 cm, which means the image forms
on the same side as the object (in front of the mirror).
Step 3: Determine the nature of the image. Since the image distance (q)
is negative, according to the sign conventions for mirrors, the image is virtual.
The negative sign indicates that the image is formed on the same side as the
object (in front of the mirror).
Further, since the image distance is greater in magnitude compared to the
object distance, the image is magnified compared to the object.
Question 35
Question
A convex lens has a focal length of 15 cm. An object is placed 30 cm from the
lens. Calculate the image distance from the lens and determine if the image is
real or virtual.
31
Solution
Step 1: Identify the given values and assign the sign conventions.
Given that the focal length of the convex lens, f= 15 cm.
The object distance from the lens, u=−30 cm (since the object is placed
on the opposite side of the incident light).
The focal length of a convex lens is positive.
Distances on the opposite side of the incident light are negative.
Step 2: Apply the lens formula to find the image distance, v.
1
f=1
v−1
u
Substitute the given values into the formula:
1
15 =1
v−1
−30
Step 3: Solve for image distance, v.
1
v=1
15 −1
30
1
v=2
30 −1
30
1
v=1
30
v= 30 cm
Step 4: Analyze if the image is real or virtual. Since the image distance vis
positive, the image is formed on the same side as the outgoing light. Therefore,
the image is virtual.
Therefore, the image distance from the lens is 30 cm and the image is virtual.
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