PHYS 101 - ELEMENTS OF PHYSICS
- Lens and mirror equations
Question Bank - Set 3
Liberty University
Question 1
Question
An object is placed 20 cm in front of a concave mirror of focal length 15 cm.
Determine the image distance and magnification.
Solution
Step 1: Identify the given quantities. The object distance pis given as p=−20
cm (since it lies in front of the mirror, p is negative). The focal length of the
concave mirror fis given as f=−15 cm (since it is a concave mirror).
Step 2: Determine the image distance using the mirror formula:
1
f=1
p+1
q
Substitute the known values:
1
−15 =1
−20 +1
q
Solve for q:
−1
15 =−1
20 +1
q
1
q=−1
15 +1
20
1
q=−4
60 +3
60
1
q=−1
60
q=−60 cm
Therefore, the image distance qis −60 cm.
Step 3: Calculate the magnification using the magnification formula:
m=−q
p
Substitute the values of pand q:
m=−
−60
−20
m= 3
Hence, the image distance is −60 cm and the magnification is 3.
Question 2
Question
A concave mirror has a focal length of 10 cm. An object is placed 15 cm away
from the mirror along its principal axis. Determine the position and nature of
the image formed by the mirror.
Solution
Step 1: Identify the given values and mirror equation.
Given: f=−10 cm, do= 15 cm.
The mirror equation is given by:
1
f=1
di
+1
do
Step 2: Substitute the given values into the mirror equation.
Plugging in f=−10 cm and do= 15 cm, we get:
1
−10 =1
di
+1
15
Step 3: Solve for di.
Solving for di, we get: 1
−10 =1
di
+1
15
1
di
=1
15 −1
10 =2
30 −3
30 =−1
30
di=−30 cm
2
Step 4: Analyze the result.
Since diis negative, the image is formed on the same side as the object, in-
dicating a virtual image. The negative sign also indicates that the image is
upright.
Step 5: Determine the magnification.
The magnification is given by:
M=−di
do
=−
−30
15 = 2
Step 6: Conclusion.
The image is formed 30 cm from the mirror on the same side as the object. It
is virtual, upright, and magnified with a magnification of 2.
Question 3
Question
A concave mirror has a focal length of 15 cm. An object is placed 30 cm from
the mirror. Determine the magnification of the image, the image distance, and
the image height.
Solution
Step 1: Identify the given values and the known equation relating them: Given:
Focal length, f=−15 cm (negative for concave mirror) Object distance, do=
−30 cm (negative since object is in front of mirror) The mirror equation for
concave mirrors is: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation and solve for
the image distance, di:1
−15 =1
−30 +1
di
−2
30 =−1
di
−1
15 =−1
di
di= 15 cm
Step 3: Calculate the magnification, m, using the formula:
m=−di
do
m=−15
−30
3
m=1
2
Step 4: Calculate the image height using the magnification: The image
height is given by:
hi=m×ho
Since the height of the object is usually considered positive, the image height
will be upside down and negative. Let’s assume the object height, ho, is 1 unit
for simplicity:
hi=1
2×1
hi=−0.5 units
Therefore, the magnification of the image is 1
2, the image distance is 15 cm,
and the image height is -0.5 units.
Question 4
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Calculate the image distance and magnification.
Solution
Step 1: Identify the given values. The object distance pis 30 cm and the focal
length fis -20 cm (negative because it is a concave mirror).
Step 2: Apply the mirror equation to find the image distance q. The mirror
equation is given by: 1
f=1
p+1
q
Substitute p= 30 cm and f=−20 cm into the equation:
1
−20 =1
30 +1
q
Step 3: Solve for the image distance.
−0.05 = 0.033 + 1
q
1
q=−0.083
q=−12 cm
Step 4: Calculate the magnification. The magnification Mis given by:
M=−q
p
4
Substitute q=−12 cm and p= 30 cm into the equation:
M=−
−12
30
M= 0.4
Therefore, the image distance is -12 cm and the magnification is 0.4.
Question 5
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Calculate the image distance and magnification.
Solution
Step 1: Recall the mirror equation for concave mirrors, which relates the object
distance (do), image distance (di), and focal length (f):
1
f=1
do
+1
di
Step 2: Substitute the given values into the mirror equation:
1
10 =1
20 +1
di
Step 3: Solve for di:
1
10 −1
20 =1
di
=⇒1
di
=1
10 −1
20
Step 4: Simplify the equation:
1
di
=2−1
20 =⇒1
di
=1
20
Step 5: Therefore, the image distance is di= 20 cm.
Step 6: To find the magnification (M), we use the formula:
M=−di
do
Step 7: Substitute the values for diand do:
M=−20
20 =−1
Step 8: The magnification is −1, indicating that the image is the same size
as the object but inverted.
5
Question 6
Question
An object is placed 20 cm in front of a convex lens of focal length 10 cm.
Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values and the information provided by the problem.
Object distance (do) = -20 cm (negative because it is in front of the lens)
Focal length (f) = 10 cm
Step 2: Apply the lens equation to find the image distance (di). The lens
equation is given by: 1
f=1
di
+1
do
Plugging in the values: 1
10 =1
di
+1
−20
Solving for digives:
1
10 +1
20 =1
di
=⇒3
20 =1
di
=⇒di=20
3cm ≈6.67 cm
Step 3: Calculate the magnification (M) produced by the lens. The magni-
fication is given by:
M=−di
do
Substitute the values in:
M=−
20
3
−20 =20
3×20 =1
3
Therefore, the image distance is approximately 6.67 cm and the magnifica-
tion produced by the lens is 1
3.
Question 7
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Determine the image distance and the magnification produced by the lens.
6
Solution
Step 1: Identify the given values and the lens/mirror equation.
Object distance, do=−20 cm (negative because the object is in front of
the lens)
Focal length, f= 15 cm
Lens equation: 1
f=1
do+1
di
Step 2: Substitute the given values into the lens equation to find the image
distance, di.
1
15 =1
−20 +1
di
1
15 =−1
20 +1
di
1
15 +1
20 =1
di
4
60 +3
60 =1
di
7
60 =1
di
di=60
7cm
Step 3: Calculate the magnification produced by the lens using the formula
m=−di
do.
m=−
60
7
−20
m=−60
7×
−1
20
m=3
7≈0.43
Therefore, the image distance is 60
7cm and the magnification produced by
the lens is approximately 0.43.
Question 8
Question
An object is placed 30 cm in front of a concave mirror with a focal length of
20 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
7
Solution
Step 1: Identify the given values and the mirror formula. Given: - Object
distance (u) = -30 cm (negative sign indicates object is in front of mirror) -
Focal length (f) = -20 cm (negative sign for concave mirror) Mirror formula:
1
f=1
v+1
u
Step 2: Solve for the image distance (v). Substitute the given values into
the mirror formula: 1
−20 =1
v+1
−30
Solve for v: 1
v=−1
20 +1
30 =−3
60 +2
60 =−1
60
v=−60 cm
Step 3: Calculate the magnification (m). Magnification formula:
m=−v
u
Substitute the values of v and u:
m=−60
−30 = 2
Step 4: Determine if the image is real or virtual. For concave mirrors, the
image is real if the image distance (v) is positive. Otherwise, the image is
virtual. In this case, v is -60 cm, so the image is virtual.
Therefore, the image distance is -60 cm, the magnification is 2, and the
image is virtual.
Question 9
Question
A concave mirror with a focal length of 10 cm forms an image that is 2/3 the
size of the object and located 15 cm from the mirror. Determine the object
distance from the mirror.
Solution
Let the object distance be denoted by p, the image distance by q, the focal
length by f, the magnification by M, the object height by ho, and the image
height by hi. We are given that f=−10 cm, hi=−2
3ho, and q=−15 cm.
Step 1: Use the magnification formula to relate the object distance to the
image distance:
−hi
ho
=q
p
8
−2
3=−15
p
Solving for p:
p=15
2×3 = 22.5 cm
Therefore, the object distance from the mirror is 22.5 cm.
Question 10
Question
A concave mirror with a focal length of 10 cm forms an image that is 3 times the
size of the object. If the object is placed 20 cm in front of the mirror, determine
the position of the image relative to the mirror.
Solution
Step 1: Given that the concave mirror has a focal length of -10 cm and the object
distance do=−20 cm. Step 2: We can use the mirror equation 1
f=1
do+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Step 3: Substituting the values into the mirror equation, we have 1
−10 =1
−20 +1
di.
Step 4: Simplifying the equation gives 1
−10 =1
−20 +1
di. Step 5: Solving for di,
we have 1
di=1
−10 −1
−20 . Step 6: Hence, 1
di=2+1
20 =3
20 . Step 7: Therefore,
di=20
3cm. Step 8: Since the image is formed on the same side as the object
for a concave mirror, the position of the image is negative. Step 9: Thus, the
position of the image relative to the mirror is di=−20
3cm.
Question 11
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
15 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
Solution
Step 1: Identify the given values and mirror equation.
Object distance, p=−20 cm (negative sign indicates object is in front of
mirror)
Focal length, f= 15 cm
Mirror equation: 1
f=1
p+1
q
9
Step 2: Substitute the given values into the mirror equation to find the image
distance, q.
1
15 =1
−20 +1
q
1
15 =−1
20 +1
q
1
15 +1
20 =1
q
q=1
1
15 +1
20
q=60
7≈8.57 cm
Step 3: Calculate the magnification, m.
m=−q
p
m=−8.57
−20
m≈0.43
Step 4: Determine if the image is real or virtual by the sign of q. Since qis
positive, the image is virtual.
Therefore, the image distance is approximately 8.57 cm, the magnification
is approximately 0.43, and the image formed is virtual.
Question 12
Question
A concave mirror with a focal length of 15 cm forms an image of an object
located 30 cm from the mirror. Calculate the position and nature of the image
using the mirror equation.
Solution
Step 1: Identify the given values and mirror equation.
Given: f=−15 cm, do=−30 cm (negative values indicate distances on the
same side as the object), and using the mirror equation:
1
f=1
do
+1
di
Step 2: Substitute the values into the mirror equation.
1
−15 =1
−30 +1
di
10
Step 3: Solve for the image distance (di).
−1
15 =−1
30 +1
di
−1
15 +1
30 =1
di
2
30 −1
30 =1
di
1
30 =1
di
di= 30 cm
Step 4: Calculate the position and nature of the image.
The image distance di= 30 cm, which is positive, indicating that the image is
formed on the opposite side of the mirror (real image). Therefore, the position
of the image is 30 cm from the mirror, and the nature of the image is real.
Question 13
Question
An object is placed in front of a concave mirror with a focal length of 15 cm.
The image is formed 25 cm in front of the mirror. Determine the magnification
of the image.
Solution
Let’s denote the object distance as do, the image distance as di, and the focal
length as f. The magnification, denoted by M, is given by the formula:
M=−di
do
Given that f= 15 cm and di= 25 cm, we need to find dofirst.
Step 1: Find the object distance From the mirror equation:
1
f=1
do
+1
di
Substitute f= 15 cm and di= 25 cm:
1
15 =1
do
+1
25
1
do
=1
15 −1
25
11
1
do
=5−3
75
1
do
=2
75
do=75
2
do= 37.5 cm
So, do= 37.5 cm.
Step 2: Find the magnification Now, we can substitute do= 37.5 cm
and di= 25 cm into the formula for magnification:
M=−25
37.5
M=−2
3
Thus, the magnification of the image is −2
3.
Question 14
Question
An object is placed 8 cm in front of a concave mirror with a focal length of 12
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the mirror equation. Given: - Object
distance, do=−8 cm (since it is in front of the mirror), - Focal length, f= 12
cm.
The mirror equation is given by:
1
f=1
do
+1
di
Step 2: Plug in the given values to find the image distance. Substitute
f= 12 cm and do=−8 cm into the mirror equation:
1
12 =1
−8+1
di
Step 3: Solve for image distance.
1
di
=1
12 −1
−8
12
1
di
=2−3
24
1
di
=−1
24
di=−24 cm
Therefore, the image distance is −24 cm.
Step 4: Calculate the magnification using the formula m=−di
do. Substitute
di=−24 cm and do=−8 cm into the magnification formula:
m=−
−24
−8=−3
Therefore, the magnification is −3.
Question 15
Question
An object is placed 20 cm to the left of a converging lens with a focal length of
15 cm. Determine the position and magnification of the image formed by the
lens.
Solution
Step 1: Identify the given quantities and the lens equation. Given: - Object
distance (u) = -20 cm (negative because it is to the left of the lens) - Focal length
(f) = 15 cm The lens equation relates the object distance, image distance, and
focal length: 1
f=1
u+1
v
Step 2: Substitute the known values into the lens equation and solve for the
image distance (v). Plugging in the values:
1
15 cm =1
−20 cm +1
v
Solving for v:
v=1
1
15 −1
20
Step 3: Calculate the image distance (v). Simplify the expression:
v=1
4
60 −3
60
=60
1= 60 cm
13
Step 4: Determine the position and nature of the image formed. Since the
object distance is negative, the image distance is positive, indicating that the
image is formed on the opposite side where the object is located. Therefore, the
image is real and located 60 cm to the right of the lens.
Step 5: Calculate the magnification. The magnification can be calculated
using the formula:
M=−v
u
Substitute the values:
M=−60 cm
−20 cm = 3
Step 6: Final answer The image is real and located 60 cm to the right of the
lens, with a magnification of 3.
Question 16
Question
An object is placed 18 cm in front of a concave mirror with a focal length of
12 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
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 (distance
of the object from the mirror), - diis the image distance (distance of the image
from the mirror).
Given f=−12 cm, do=−18 cm, and di=?.
Step 2: Substitute the values into the mirror equation and solve for di:
1
−12 =1
−18 +1
di
−1
12 =−1
18 +1
di
−1
12 +1
18 =1
di
3
36 −2
36 =1
di
1
36 =1
di
14
di= 36 cm
Step 3: Calculate the magnification Musing the formula:
M=−di
do
M=−36
−18
M= 2
Step 4: Determine if the image is real or virtual based on the sign of di. A
negative diindicates a virtual image, while a positive diindicates a real image.
Since diis positive, the image is real.
Therefore, the image distance is 36 cm, the magnification is 2, and the image
formed is real.
Question 17
Question
A convex lens has a focal length of 10 cm and is placed 20 cm from an object.
If the image formed is virtual and upright, find the distance of the image from
the lens.
Solution
Step 1: Given that the focal length of the convex lens, f= 10 cm, the object
distance, u=−20 cm (negative sign indicates the object is placed to the left of
the lens), and the image distance, vis to be determined.
Step 2: Using the lens formula 1
f=1
v−1
u, we can substitute f= 10 cm and
u=−20 cm to solve for v.1
10 =1
v−1
−20
Step 3: Simplifying the above equation, we get
1
10 =1
v+1
20
Step 4: Combining the fractions on the right side, we have
1
10 =2 + v
20v
Step 5: Cross multiplying, we get
20 + 2v= 10v
15
Step 6: Simplifying the above equation gives
20 = 8v
Step 7: Dividing by 8 on both sides, we find
v=20
8= 2.5 cm
Step 8: Therefore, the distance of the virtual upright image from the lens is
2.5 cm.
Question 18
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Calculate the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values and the type of lens.
Object distance, do=−20 cm (negative as the object is placed to the left
of the lens)
Focal length, f= 15 cm (positive for a converging lens)
Step 2: Use the lens formula to find the image distance. The lens formula is
given by: 1
f=1
do
+1
di
Substitute the given values into the lens formula:
1
15 =1
−20 +1
di
Step 3: Solve for the image distance, di.
1
di
=1
15 −1
−20
1
di
=20 −15
300
di=300
5= 60 cm
Step 4: Calculate the magnification, m. The magnification is given by:
m=−di
do
16
Substitute the values of diand dointo the formula:
m=−60
−20
m= 3
Therefore, the image distance is 60 cm and the magnification of the image
formed is 3.
Question 19
Question
A convex lens with a focal length of 20 cm is placed 40 cm to the left of a
concave mirror with a focal length of 15 cm. Determine the final image position
and magnification when an object is placed 10 cm to the left of the lens.
Solution
Step 1: Calculate the image position for the lens using the lens equation:
Given: - flens = 20 cm - dobject,lens =−10 cm
The lens equation is given by:
1
flens
=1
dobject,lens
+1
dimage,lens
Substitute in the values:
1
20 =1
−10 +1
dimage,lens
Solve for dimage,lens:
1
dimage,lens
=1
20 +1
10 =3
20
dimage,lens =20
3cm
Step 2: Calculate the object position for the mirror:
Using the mirror equation:
1
fmirror
=1
dobject,mirror
+1
dimage,mirror
Given: - fmirror =−15 cm - dobject,mirror = 40 cm - dimage,mirror =−20/3 cm
Substitute in the values:
1
−15 =1
40 +1
−20/3
17
−3
20 =1
40 −3
20
−3
20 =−2
40 −3
20
From this equation we find dobject,mirror = 40 cm.
Step 3: Calculate the image position for the mirror:
Use the mirror equation again with the newly found object position dobject,mirror =
40 cm: 1
−15 =1
40 +1
dimage,mirror
−3
20 =1
40 +1
dimage,mirror
−3
20 −1
40 =1
dimage,mirror
Solving for dimage,mirror gives:
dimage,mirror =−60 cm
Step 4: Calculate the final image position:
Since the mirror forms an image to the left, the final image position is the
sum of the object position for the mirror and the distance from the mirror to
the final image:
dfinal image =dobject,mirror +dimage,mirror
dfinal image = 40 + (−60) = −20 cm
Step 5: Calculate the magnification:
The magnification is given by:
M=himage
hobject
=−dimage,mirror
dobject,mirror
Substitute the values:
M=−
−60
40 =3
2
Therefore, the final image position is −20 cm to the left of the mirror and
the magnification is 3
2.
Question 20
Question
An object is placed 30 cm from a concave mirror with a focal length of 10 cm.
Find the image distance and magnification.
18
Solution
Step 1: Identify the given values.
Object distance, do=−30 cm (negative because it is in front of the mirror)
Focal length, f=−10 cm (negative for a concave mirror)
Step 2: Use the mirror equation to find the image distance, di.
1
f=1
do
+1
di
Substitute do=−30 cm and f=−10 cm into the mirror equation:
1
−10 =1
−30 +1
di
Step 3: Solve for di.
−1
30 +1
di
=−1
10
1
di
=−1
10 +1
30
1
di
=−3
30 +1
30
1
di
=−2
30
di=−15 cm
The image distance is −15 cm, which means the image is formed on the same
side as the object (virtual image).
Step 4: Calculate the magnification, M.
M=−di
do
M=−
−15
−30
M=1
2
The magnification is 1
2, indicating that the image is half the size of the
object.
Question 21
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
10 cm. If the image is virtual, upright, and magnified, determine the image
distance.
19
Solution
Step 1: Identify the given values and the mirror equation. Given: Object
distance, do=−20 cm (negative because it is in front of the mirror) Focal
length, f=−10 cm (negative because the mirror is concave) Image distance,
di=?
The mirror equation is: 1
f=1
di
+1
do
Step 2: Substitute the given values into the mirror equation.
1
−10 =1
di
+1
−20
Step 3: Solve the equation for the image distance.
−1
10 =1
di
−1
20
−1
10 +1
20 =1
di
1
20 =1
di
di= 20 cm
Step 4: Verify the sign of the image distance. The positive value of the
image distance indicates that the image is virtual and formed on the same side
as the object, consistent with the given information.
Therefore, the image distance is 20 cm.
Question 22
Question
An object is placed 15 cm in front of a concave mirror of focal length 10 cm.
Determine the magnification of the image and the distance of the image from
the mirror.
Solution
Step 1: Identify the given variables.
The given variables are:
u=−15 cm (object distance)
f=−10 cm (focal length)
20
Step 2: Calculate the image distance using the mirror equation. The mirror
equation is given by:
−1
f=1
v+1
u
Substitute f=−10 cm and u=−15 cm into the equation:
−1
−10 =1
v+1
−15
0.1 = 1
v−1
15
1
v= 0.1 + 1
15
1
v=1
10 +1
15
1
v=3+2
30
1
v=5
30
v=30
5= 6 cm
Therefore, the image distance vis 6 cm.
Step 3: Calculate the magnification of the image. The magnification mis
given by:
m=−v
u
Substitute v= 6 cm and u=−15 cm into the equation:
m=−6
−15 = 0.4
Therefore, the magnification of the image is 0.4.
Question 23
Question
An object is placed 15 cm in front of a convex lens with a focal length of 10 cm.
Determine the image distance and magnification produced by the lens.
21
Solution
Step 1: Identify the given values and the unknowns.
Given:
Object distance, do=−15 cm (negative since the object is in front of the
lens)
Focal length, f= 10 cm
Unknowns:
Image distance, di
Magnification, m
Step 2: Apply the lens equation to calculate the image distance.
The lens equation is given by:
1
f=1
di
+1
do
Substitute f= 10 cm and do=−15 cm:
1
10 =1
di
+1
−15
Solve for di:1
di
=1
10 +1
15 =3+2
30 =5
30 =1
6
di= 6 cm
Step 3: Calculate the magnification using the magnification formula.
The magnification mis given by:
m=−di
do
Substitute di= 6 cm and do=−15 cm:
m=−6
−15 =2
5
Therefore, the image distance is 6 cm and the magnification is 2
5.
Question 24
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
10 cm. Determine the position and magnification of the image formed by the
mirror.
22
Solution
Step 1: Identify the given values and variables. Given: - Object distance (do)
= -20 cm (negative since the object is in front of the mirror) - Focal length (f)
= 10 cm - Since the mirror is concave, the focal length is negative. - Using the
sign convention: object distance (do) is negative for real objects placed in front
of the mirror.
Step 2: Apply the mirror formula to find the image distance. The mirror
equation is given by:
−1
f=1
di
+1
do
Substitute the given values into the equation:
−1
10 =1
di
+1
−20
Solve for di:
−1
10 =1
di
−1
20
−1
10 +1
20 =1
di
1
20 =1
di
di= 20 cm
Step 3: Calculate the magnification of the image. The magnification (m) is
given by:
m=−di
do
Substitute the values of diand do:
m=−20
−20
m=−1
Step 4: Analyze the results. The image is located 20 cm behind the mirror,
which indicates that it is a real image. The negative value of magnification
indicates that the image is inverted.
Therefore, the image is formed 20 cm behind the mirror with a magnification
of -1.
Question 25
Question
A concave mirror with a focal length of 20 cm forms an image of an object
located 30 cm from the mirror. Determine the position and nature of the image
formed.
23
Solution
Step 1: Given that the focal length of the concave mirror is 20 cm and the
object distance is 30 cm. Step 2: Using the mirror equation 1
f=1
do+1
di, where
fis the focal length, dois the object distance, and diis the image distance.
Step 3: Substitute the given values into the mirror equation to solve for di:
1
20 =1
30 +1
di. Step 4: Simplifying the equation, we get 1
di=1
20 −1
30 . Step 5:
Find a common denominator: 1
di=3
60 −2
60 . Step 6: Therefore, 1
di=1
60 . Step 7:
Solving for di, we get di= 60 cm. Step 8: Since the image distance is positive,
the image is formed on the same side as the object (real image). Step 9: The
position of the image is 60 cm from the mirror surface. Step 10: The nature of
the image is real and inverted.
Therefore, the image formed by the concave mirror is real, inverted, and
located 60 cm from the mirror surface.
Question 26
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Determine the image distance and magnification using the mirror equation.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
Object distance do=−20 cm (negative because it is in front of the mirror)
Focal length f= 15 cm
Mirror equation: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation to find the
image distance.
Substitute f= 15 cm and do=−20 cm into the mirror equation:
1
15 =1
−20 +1
di
Step 3: Solve for the image distance.
1
15 =−1
20 +1
di
1
di
=1
15 +1
20
1
di
=4
60 +3
60
24
1
di
=7
60
di=60
7≈8.57 cm
Therefore, the image distance is approximately 8.57 cm.
Step 4: Calculate the magnification using the magnification equation.
Magnification equation: m=−di
do
Step 5: Substitute the image distance and object distance to find the mag-
nification.
Substitute di= 8.57 cm and do=−20 cm into the magnification equation:
m=−8.57
−20
m= 0.4285
Therefore, the magnification is 0.4285.
Question 27
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
15 cm. Calculate the image distance and verify whether the image is real or
virtual.
Solution
Step 1: Identify the given values. The object distance ois 20 cm, and the focal
length fis 15 cm. The focal length is negative for a concave mirror.
Step 2: Apply the mirror equation. The mirror equation is given by 1
f=
1
o+1
i, where fis the focal length, ois the object distance, and iis the image
distance.
Step 3: Substitute the given values into the mirror equation. Substitute
f=−15 cm and o= 20 cm:
−1
15 =1
20 +1
i
Step 4: Solve for i.
−1
15 =1
20 +1
i
−1
15 −1
20 =1
i
−4
60 −3
60 =1
i
25
−7
60 =1
i
Solving for igives:
i=−60
7cm ≈ −8.57 cm
Step 5: Determine if the image is real or virtual. For concave mirrors, the
image is real if iis negative (in front of the mirror) and virtual if iis positive
(behind the mirror). Since iis negative in this case, the image is real.
Therefore, the image distance is approximately -8.57 cm, and the image
formed is real.
Question 28
Question
An object is placed 20 cm in front of a converging lens with a focal length of 10
cm. Determine the image distance and height of the image formed by the lens.
Solution
Step 1: Identify the given values and the lens equation. Given: f= 10 cm (focal
length), o=−20 cm (object distance).
The lens equation is given by:
1
f=1
o+1
i
Step 2: Substitute the given values into the lens equation. Substitute f= 10
cm and o=−20 cm into the lens equation to get:
1
10 =1
−20 +1
i
Step 3: Solve for i. Solving the equation above for 1/i, we get:
1
i=1
10 −1
20 =2−1
20 =1
20
Thus, i= 20 cm.
Step 4: Calculate the magnification. The magnification (M) is given by:
M=−i
o
Substitute i= 20 cm and o=−20 cm into the magnification equation:
M=−20
−20 = 1
26
Step 5: Calculate the image height. The height of the image can be deter-
mined using the magnification. Since the magnification is 1, the image is the
same size as the object. Therefore, the height of the image is the same as the
height of the object.
Step 6: Write the final answer. The image distance is 20 cm and the height
of the image is the same as the height of the object.
Question 29
Question
A converging lens forms an image of an object located 20 cm from the lens. The
image is real, inverted, and four times the size of the object. Find the focal
length of the lens.
Solution
Step 1: Identify the given values and the sign conventions. Step 2: Use the lens
equation to find the focal length of the lens.
Step 1: Given values and sign conventions: The object distance p=−20
cm (since the object is located on the opposite side of the incoming light).
The image distance qis positive for a real image. The magnification M=−4
(negative sign indicates an inverted image).
Step 2: Applying the lens equation: The lens equation is given by:
1
f=1
p+1
q
Substitute the given values into the lens equation:
1
f=1
−20 +1
q
Since M=−q
p, we have q=−4p. Substitute q=−4pinto the equation:
1
f=1
−20 +1
−4p
Simplify the equation:
1
f=−4−20
(−20)(−4) =−24
80 =−3
10
Therefore, the focal length of the lens is f=−10
3cm.
Question 30
Question
A convex lens has a focal length of 15 cm. An object is placed 25 cm in front of
the lens. Calculate the image distance and magnification produced by the lens.
27
Solution
Step 1: Identify the given values: The focal length of the convex lens, f= 15
cm, The object distance, u=−25 cm (negative because the object is located to
the left of the lens), The image distance, v=?, The magnification, m=?.
Step 2: Apply the lens equation to find the image distance: The lens equation
is given by: 1
f=1
v+1
u
Substitute f= 15 cm and u=−25 cm:
1
15 =1
v+1
−25
Solving for v:1
v=1
15 +1
25 =5+3
75 =8
75
v=75
8= 9.375 cm
Step 3: Calculate the magnification: The magnification can be calculated
using the formula:
m=−v
u
Substitute u=−25 cm and v= 9.375 cm:
m=−9.375
−25 = 0.375
Step 4: Interpret the results: The image distance is 9.375 cm in front of
the lens, and the magnification is 0.375. The positive value of magnification
indicates the image is upright relative to the object and diminished in size.
Question 31
Question
A converging lens has a focal length of 15 cm. An object is placed 30 cm in
front of the lens. Calculate the position and nature (real or virtual) of the image
formed by the lens.
Solution
Step 1: Given data The focal length of the lens, f= 15 cm
The object distance, u=−30 cm (negative because the object is placed in front
of the lens)
28
Step 2: Using the lens equation The lens equation relates the object distance
(u), the image distance (v), and the focal length (f) of the lens:
1
f=1
v+1
u
Step 3: Substitute the given values into the lens equation
1
15 =1
v+1
−30
Step 4: Solve for v1
v=1
15 −1
30
1
v=2
30 −1
30
1
v=1
30
v= 30 cm
Step 5: Interpret the result The positive value of vindicates that the image
is formed on the opposite side of the lens from the object (real image).
Therefore, the image is formed 30 cm in front of the lens and the nature of
the image is real.
Question 32
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.
Solution
Step 1: Identify the given values and the known lens equation. Given: - Object
distance, p=−20 cm (negative sign indicates that the object is located to the
left of the lens). - Focal length, f= 10 cm. The lens equation is:
1
f=1
p+1
q
Step 2: Substitute the known values into the lens equation.
1
10 =1
−20 +1
q
Step 3: Solve for the image distance, q.
1
10 =−1
20 +1
q
29
1
10 +1
20 =1
q
3
20 =1
q
q=20
3cm
Therefore, the image distance is q=20
3cm.
Step 4: Calculate the magnification using the magnification formula. The
magnification, M, is given by:
M=−q
p
Substitute the values of qand pto find the magnification.
M=−
20
3
−20
M=20
3×20
M=1
3
So, the magnification produced by the lens is 1
3.
Question 33
Question
A concave mirror with a focal length of 20 cm forms a real image that is mag-
nified 3 times. If the object is located 30 cm from the mirror, what is the
magnification produced by the mirror?
Solution
Step 1: First, we need to determine the image distance using the mirror equa-
tion: 1
f=1
do
+1
di
Given that f=−20 cm (since it is a concave mirror) and do=−30 cm (since
the object is placed in front of the mirror), we can rearrange the mirror equation
to solve for di:1
−20 =1
−30 +1
di
−3
20 =−3
30 +1
di
−3
20 +1
3=1
di
30
−9
60 +20
60 =1
di
11
60 =1
di
di=60
11
di≈5.45 cm
Step 2: Next, we can calculate the magnification produced by the mirror
using the magnification equation:
M=−di
do
Substitute the values of diand dointo the formula:
M=−5.45
−30
M=5.45
30
M= 0.183
Therefore, the magnification produced by the mirror is 0.183.
Question 34
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Calculate the image distance and determine if the image is real or virtual.
Solution
Step 1: Use the mirror equation to find the image distance.
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: Substitute the given values into the equation.
1
10 cm =1
20 cm +1
di
Step 3: Solve for di.1
di
=1
10 cm −1
20 cm
1
di
=2
20 cm −1
20 cm =1
20 cm
di= 20 cm
Step 4: Since the image distance is positive, the image is real.
31
Question 35
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the position and nature of the image formed.
Solution
Step 1: Given that the object distance (do) is -20 cm (since the object is in front
of the mirror) and the focal length (f) is -10 cm (since the mirror is concave),
we can use the mirror equation:
1
f=1
do
+1
di
Substitute the known values:
1
−10 =1
−20 +1
di
Step 2: Solve for the image distance (di):
−1
10 =−1
20 +1
di
−1
10 +1
20 =1
di
1
di
=2−1
20 =1
20
Thus, di= 20 cm.
Step 3: To determine the nature of the image, we can use the magnification
formula:
m=−di
do
Substitute the values:
m=−20
−20 = 1
Since the magnification is positive, the image is upright.
Step 4: Since the image distance is positive, the image is formed on the
same side as the object (in front of the mirror). Therefore, the image formed is
virtual, upright, and magnified.
32
q=−60 cm
Therefore, the image distance qis −60 cm.
Step 3: Calculate the magnification using the magnification formula:
m=−q
p
Substitute the values of pand q:
m=−
−60
−20
m= 3
Hence, the image distance is −60 cm and the magnification is 3.
Question 2
Question
A concave mirror has a focal length of 10 cm. An object is placed 15 cm away
from the mirror along its principal axis. Determine the position and nature of
the image formed by the mirror.
Solution
Step 1: Identify the given values and mirror equation.
Given: f=−10 cm, do= 15 cm.
The mirror equation is given by:
1
f=1
di
+1
do
Step 2: Substitute the given values into the mirror equation.
Plugging in f=−10 cm and do= 15 cm, we get:
1
−10 =1
di
+1
15
Step 3: Solve for di.
Solving for di, we get: 1
−10 =1
di
+1
15
1
di
=1
15 −1
10 =2
30 −3
30 =−1
30
di=−30 cm
2
Step 4: Analyze the result.
Since diis negative, the image is formed on the same side as the object, in-
dicating a virtual image. The negative sign also indicates that the image is
upright.
Step 5: Determine the magnification.
The magnification is given by:
M=−di
do
=−
−30
15 = 2
Step 6: Conclusion.
The image is formed 30 cm from the mirror on the same side as the object. It
is virtual, upright, and magnified with a magnification of 2.
Question 3
Question
A concave mirror has a focal length of 15 cm. An object is placed 30 cm from
the mirror. Determine the magnification of the image, the image distance, and
the image height.
Solution
Step 1: Identify the given values and the known equation relating them: Given:
Focal length, f=−15 cm (negative for concave mirror) Object distance, do=
−30 cm (negative since object is in front of mirror) The mirror equation for
concave mirrors is: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation and solve for
the image distance, di:1
−15 =1
−30 +1
di
−2
30 =−1
di
−1
15 =−1
di
di= 15 cm
Step 3: Calculate the magnification, m, using the formula:
m=−di
do
m=−15
−30
3
m=1
2
Step 4: Calculate the image height using the magnification: The image
height is given by:
hi=m×ho
Since the height of the object is usually considered positive, the image height
will be upside down and negative. Let’s assume the object height, ho, is 1 unit
for simplicity:
hi=1
2×1
hi=−0.5 units
Therefore, the magnification of the image is 1
2, the image distance is 15 cm,
and the image height is -0.5 units.
Question 4
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Calculate the image distance and magnification.
Solution
Step 1: Identify the given values. The object distance pis 30 cm and the focal
length fis -20 cm (negative because it is a concave mirror).
Step 2: Apply the mirror equation to find the image distance q. The mirror
equation is given by: 1
f=1
p+1
q
Substitute p= 30 cm and f=−20 cm into the equation:
1
−20 =1
30 +1
q
Step 3: Solve for the image distance.
−0.05 = 0.033 + 1
q
1
q=−0.083
q=−12 cm
Step 4: Calculate the magnification. The magnification Mis given by:
M=−q
p
4
Substitute q=−12 cm and p= 30 cm into the equation:
M=−
−12
30
M= 0.4
Therefore, the image distance is -12 cm and the magnification is 0.4.
Question 5
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Calculate the image distance and magnification.
Solution
Step 1: Recall the mirror equation for concave mirrors, which relates the object
distance (do), image distance (di), and focal length (f):
1
f=1
do
+1
di
Step 2: Substitute the given values into the mirror equation:
1
10 =1
20 +1
di
Step 3: Solve for di:
1
10 −1
20 =1
di
=⇒1
di
=1
10 −1
20
Step 4: Simplify the equation:
1
di
=2−1
20 =⇒1
di
=1
20
Step 5: Therefore, the image distance is di= 20 cm.
Step 6: To find the magnification (M), we use the formula:
M=−di
do
Step 7: Substitute the values for diand do:
M=−20
20 =−1
Step 8: The magnification is −1, indicating that the image is the same size
as the object but inverted.
5
Question 6
Question
An object is placed 20 cm in front of a convex lens of focal length 10 cm.
Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values and the information provided by the problem.
Object distance (do) = -20 cm (negative because it is in front of the lens)
Focal length (f) = 10 cm
Step 2: Apply the lens equation to find the image distance (di). The lens
equation is given by: 1
f=1
di
+1
do
Plugging in the values: 1
10 =1
di
+1
−20
Solving for digives:
1
10 +1
20 =1
di
=⇒3
20 =1
di
=⇒di=20
3cm ≈6.67 cm
Step 3: Calculate the magnification (M) produced by the lens. The magni-
fication is given by:
M=−di
do
Substitute the values in:
M=−
20
3
−20 =20
3×20 =1
3
Therefore, the image distance is approximately 6.67 cm and the magnifica-
tion produced by the lens is 1
3.
Question 7
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Determine the image distance and the magnification produced by the lens.
6
Solution
Step 1: Identify the given values and the lens/mirror equation.
Object distance, do=−20 cm (negative because the object is in front of
the lens)
Focal length, f= 15 cm
Lens equation: 1
f=1
do+1
di
Step 2: Substitute the given values into the lens equation to find the image
distance, di.
1
15 =1
−20 +1
di
1
15 =−1
20 +1
di
1
15 +1
20 =1
di
4
60 +3
60 =1
di
7
60 =1
di
di=60
7cm
Step 3: Calculate the magnification produced by the lens using the formula
m=−di
do.
m=−
60
7
−20
m=−60
7×
−1
20
m=3
7≈0.43
Therefore, the image distance is 60
7cm and the magnification produced by
the lens is approximately 0.43.
Question 8
Question
An object is placed 30 cm in front of a concave mirror with a focal length of
20 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
7
Solution
Step 1: Identify the given values and the mirror formula. Given: - Object
distance (u) = -30 cm (negative sign indicates object is in front of mirror) -
Focal length (f) = -20 cm (negative sign for concave mirror) Mirror formula:
1
f=1
v+1
u
Step 2: Solve for the image distance (v). Substitute the given values into
the mirror formula: 1
−20 =1
v+1
−30
Solve for v: 1
v=−1
20 +1
30 =−3
60 +2
60 =−1
60
v=−60 cm
Step 3: Calculate the magnification (m). Magnification formula:
m=−v
u
Substitute the values of v and u:
m=−60
−30 = 2
Step 4: Determine if the image is real or virtual. For concave mirrors, the
image is real if the image distance (v) is positive. Otherwise, the image is
virtual. In this case, v is -60 cm, so the image is virtual.
Therefore, the image distance is -60 cm, the magnification is 2, and the
image is virtual.
Question 9
Question
A concave mirror with a focal length of 10 cm forms an image that is 2/3 the
size of the object and located 15 cm from the mirror. Determine the object
distance from the mirror.
Solution
Let the object distance be denoted by p, the image distance by q, the focal
length by f, the magnification by M, the object height by ho, and the image
height by hi. We are given that f=−10 cm, hi=−2
3ho, and q=−15 cm.
Step 1: Use the magnification formula to relate the object distance to the
image distance:
−hi
ho
=q
p
8
−2
3=−15
p
Solving for p:
p=15
2×3 = 22.5 cm
Therefore, the object distance from the mirror is 22.5 cm.
Question 10
Question
A concave mirror with a focal length of 10 cm forms an image that is 3 times the
size of the object. If the object is placed 20 cm in front of the mirror, determine
the position of the image relative to the mirror.
Solution
Step 1: Given that the concave mirror has a focal length of -10 cm and the object
distance do=−20 cm. Step 2: We can use the mirror equation 1
f=1
do+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Step 3: Substituting the values into the mirror equation, we have 1
−10 =1
−20 +1
di.
Step 4: Simplifying the equation gives 1
−10 =1
−20 +1
di. Step 5: Solving for di,
we have 1
di=1
−10 −1
−20 . Step 6: Hence, 1
di=2+1
20 =3
20 . Step 7: Therefore,
di=20
3cm. Step 8: Since the image is formed on the same side as the object
for a concave mirror, the position of the image is negative. Step 9: Thus, the
position of the image relative to the mirror is di=−20
3cm.
Question 11
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
15 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
Solution
Step 1: Identify the given values and mirror equation.
Object distance, p=−20 cm (negative sign indicates object is in front of
mirror)
Focal length, f= 15 cm
Mirror equation: 1
f=1
p+1
q
9
Step 2: Substitute the given values into the mirror equation to find the image
distance, q.
1
15 =1
−20 +1
q
1
15 =−1
20 +1
q
1
15 +1
20 =1
q
q=1
1
15 +1
20
q=60
7≈8.57 cm
Step 3: Calculate the magnification, m.
m=−q
p
m=−8.57
−20
m≈0.43
Step 4: Determine if the image is real or virtual by the sign of q. Since qis
positive, the image is virtual.
Therefore, the image distance is approximately 8.57 cm, the magnification
is approximately 0.43, and the image formed is virtual.
Question 12
Question
A concave mirror with a focal length of 15 cm forms an image of an object
located 30 cm from the mirror. Calculate the position and nature of the image
using the mirror equation.
Solution
Step 1: Identify the given values and mirror equation.
Given: f=−15 cm, do=−30 cm (negative values indicate distances on the
same side as the object), and using the mirror equation:
1
f=1
do
+1
di
Step 2: Substitute the values into the mirror equation.
1
−15 =1
−30 +1
di
10
Step 3: Solve for the image distance (di).
−1
15 =−1
30 +1
di
−1
15 +1
30 =1
di
2
30 −1
30 =1
di
1
30 =1
di
di= 30 cm
Step 4: Calculate the position and nature of the image.
The image distance di= 30 cm, which is positive, indicating that the image is
formed on the opposite side of the mirror (real image). Therefore, the position
of the image is 30 cm from the mirror, and the nature of the image is real.
Question 13
Question
An object is placed in front of a concave mirror with a focal length of 15 cm.
The image is formed 25 cm in front of the mirror. Determine the magnification
of the image.
Solution
Let’s denote the object distance as do, the image distance as di, and the focal
length as f. The magnification, denoted by M, is given by the formula:
M=−di
do
Given that f= 15 cm and di= 25 cm, we need to find dofirst.
Step 1: Find the object distance From the mirror equation:
1
f=1
do
+1
di
Substitute f= 15 cm and di= 25 cm:
1
15 =1
do
+1
25
1
do
=1
15 −1
25
11
1
do
=5−3
75
1
do
=2
75
do=75
2
do= 37.5 cm
So, do= 37.5 cm.
Step 2: Find the magnification Now, we can substitute do= 37.5 cm
and di= 25 cm into the formula for magnification:
M=−25
37.5
M=−2
3
Thus, the magnification of the image is −2
3.
Question 14
Question
An object is placed 8 cm in front of a concave mirror with a focal length of 12
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the mirror equation. Given: - Object
distance, do=−8 cm (since it is in front of the mirror), - Focal length, f= 12
cm.
The mirror equation is given by:
1
f=1
do
+1
di
Step 2: Plug in the given values to find the image distance. Substitute
f= 12 cm and do=−8 cm into the mirror equation:
1
12 =1
−8+1
di
Step 3: Solve for image distance.
1
di
=1
12 −1
−8
12
1
di
=2−3
24
1
di
=−1
24
di=−24 cm
Therefore, the image distance is −24 cm.
Step 4: Calculate the magnification using the formula m=−di
do. Substitute
di=−24 cm and do=−8 cm into the magnification formula:
m=−
−24
−8=−3
Therefore, the magnification is −3.
Question 15
Question
An object is placed 20 cm to the left of a converging lens with a focal length of
15 cm. Determine the position and magnification of the image formed by the
lens.
Solution
Step 1: Identify the given quantities and the lens equation. Given: - Object
distance (u) = -20 cm (negative because it is to the left of the lens) - Focal length
(f) = 15 cm The lens equation relates the object distance, image distance, and
focal length: 1
f=1
u+1
v
Step 2: Substitute the known values into the lens equation and solve for the
image distance (v). Plugging in the values:
1
15 cm =1
−20 cm +1
v
Solving for v:
v=1
1
15 −1
20
Step 3: Calculate the image distance (v). Simplify the expression:
v=1
4
60 −3
60
=60
1= 60 cm
13
Step 4: Determine the position and nature of the image formed. Since the
object distance is negative, the image distance is positive, indicating that the
image is formed on the opposite side where the object is located. Therefore, the
image is real and located 60 cm to the right of the lens.
Step 5: Calculate the magnification. The magnification can be calculated
using the formula:
M=−v
u
Substitute the values:
M=−60 cm
−20 cm = 3
Step 6: Final answer The image is real and located 60 cm to the right of the
lens, with a magnification of 3.
Question 16
Question
An object is placed 18 cm in front of a concave mirror with a focal length of
12 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
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 (distance
of the object from the mirror), - diis the image distance (distance of the image
from the mirror).
Given f=−12 cm, do=−18 cm, and di=?.
Step 2: Substitute the values into the mirror equation and solve for di:
1
−12 =1
−18 +1
di
−1
12 =−1
18 +1
di
−1
12 +1
18 =1
di
3
36 −2
36 =1
di
1
36 =1
di
14
di= 36 cm
Step 3: Calculate the magnification Musing the formula:
M=−di
do
M=−36
−18
M= 2
Step 4: Determine if the image is real or virtual based on the sign of di. A
negative diindicates a virtual image, while a positive diindicates a real image.
Since diis positive, the image is real.
Therefore, the image distance is 36 cm, the magnification is 2, and the image
formed is real.
Question 17
Question
A convex lens has a focal length of 10 cm and is placed 20 cm from an object.
If the image formed is virtual and upright, find the distance of the image from
the lens.
Solution
Step 1: Given that the focal length of the convex lens, f= 10 cm, the object
distance, u=−20 cm (negative sign indicates the object is placed to the left of
the lens), and the image distance, vis to be determined.
Step 2: Using the lens formula 1
f=1
v−1
u, we can substitute f= 10 cm and
u=−20 cm to solve for v.1
10 =1
v−1
−20
Step 3: Simplifying the above equation, we get
1
10 =1
v+1
20
Step 4: Combining the fractions on the right side, we have
1
10 =2 + v
20v
Step 5: Cross multiplying, we get
20 + 2v= 10v
15
Step 6: Simplifying the above equation gives
20 = 8v
Step 7: Dividing by 8 on both sides, we find
v=20
8= 2.5 cm
Step 8: Therefore, the distance of the virtual upright image from the lens is
2.5 cm.
Question 18
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Calculate the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values and the type of lens.
Object distance, do=−20 cm (negative as the object is placed to the left
of the lens)
Focal length, f= 15 cm (positive for a converging lens)
Step 2: Use the lens formula to find the image distance. The lens formula is
given by: 1
f=1
do
+1
di
Substitute the given values into the lens formula:
1
15 =1
−20 +1
di
Step 3: Solve for the image distance, di.
1
di
=1
15 −1
−20
1
di
=20 −15
300
di=300
5= 60 cm
Step 4: Calculate the magnification, m. The magnification is given by:
m=−di
do
16
Substitute the values of diand dointo the formula:
m=−60
−20
m= 3
Therefore, the image distance is 60 cm and the magnification of the image
formed is 3.
Question 19
Question
A convex lens with a focal length of 20 cm is placed 40 cm to the left of a
concave mirror with a focal length of 15 cm. Determine the final image position
and magnification when an object is placed 10 cm to the left of the lens.
Solution
Step 1: Calculate the image position for the lens using the lens equation:
Given: - flens = 20 cm - dobject,lens =−10 cm
The lens equation is given by:
1
flens
=1
dobject,lens
+1
dimage,lens
Substitute in the values:
1
20 =1
−10 +1
dimage,lens
Solve for dimage,lens:
1
dimage,lens
=1
20 +1
10 =3
20
dimage,lens =20
3cm
Step 2: Calculate the object position for the mirror:
Using the mirror equation:
1
fmirror
=1
dobject,mirror
+1
dimage,mirror
Given: - fmirror =−15 cm - dobject,mirror = 40 cm - dimage,mirror =−20/3 cm
Substitute in the values:
1
−15 =1
40 +1
−20/3
17
−3
20 =1
40 −3
20
−3
20 =−2
40 −3
20
From this equation we find dobject,mirror = 40 cm.
Step 3: Calculate the image position for the mirror:
Use the mirror equation again with the newly found object position dobject,mirror =
40 cm: 1
−15 =1
40 +1
dimage,mirror
−3
20 =1
40 +1
dimage,mirror
−3
20 −1
40 =1
dimage,mirror
Solving for dimage,mirror gives:
dimage,mirror =−60 cm
Step 4: Calculate the final image position:
Since the mirror forms an image to the left, the final image position is the
sum of the object position for the mirror and the distance from the mirror to
the final image:
dfinal image =dobject,mirror +dimage,mirror
dfinal image = 40 + (−60) = −20 cm
Step 5: Calculate the magnification:
The magnification is given by:
M=himage
hobject
=−dimage,mirror
dobject,mirror
Substitute the values:
M=−
−60
40 =3
2
Therefore, the final image position is −20 cm to the left of the mirror and
the magnification is 3
2.
Question 20
Question
An object is placed 30 cm from a concave mirror with a focal length of 10 cm.
Find the image distance and magnification.
18
Solution
Step 1: Identify the given values.
Object distance, do=−30 cm (negative because it is in front of the mirror)
Focal length, f=−10 cm (negative for a concave mirror)
Step 2: Use the mirror equation to find the image distance, di.
1
f=1
do
+1
di
Substitute do=−30 cm and f=−10 cm into the mirror equation:
1
−10 =1
−30 +1
di
Step 3: Solve for di.
−1
30 +1
di
=−1
10
1
di
=−1
10 +1
30
1
di
=−3
30 +1
30
1
di
=−2
30
di=−15 cm
The image distance is −15 cm, which means the image is formed on the same
side as the object (virtual image).
Step 4: Calculate the magnification, M.
M=−di
do
M=−
−15
−30
M=1
2
The magnification is 1
2, indicating that the image is half the size of the
object.
Question 21
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
10 cm. If the image is virtual, upright, and magnified, determine the image
distance.
19
Solution
Step 1: Identify the given values and the mirror equation. Given: Object
distance, do=−20 cm (negative because it is in front of the mirror) Focal
length, f=−10 cm (negative because the mirror is concave) Image distance,
di=?
The mirror equation is: 1
f=1
di
+1
do
Step 2: Substitute the given values into the mirror equation.
1
−10 =1
di
+1
−20
Step 3: Solve the equation for the image distance.
−1
10 =1
di
−1
20
−1
10 +1
20 =1
di
1
20 =1
di
di= 20 cm
Step 4: Verify the sign of the image distance. The positive value of the
image distance indicates that the image is virtual and formed on the same side
as the object, consistent with the given information.
Therefore, the image distance is 20 cm.
Question 22
Question
An object is placed 15 cm in front of a concave mirror of focal length 10 cm.
Determine the magnification of the image and the distance of the image from
the mirror.
Solution
Step 1: Identify the given variables.
The given variables are:
u=−15 cm (object distance)
f=−10 cm (focal length)
20
Step 2: Calculate the image distance using the mirror equation. The mirror
equation is given by:
−1
f=1
v+1
u
Substitute f=−10 cm and u=−15 cm into the equation:
−1
−10 =1
v+1
−15
0.1 = 1
v−1
15
1
v= 0.1 + 1
15
1
v=1
10 +1
15
1
v=3+2
30
1
v=5
30
v=30
5= 6 cm
Therefore, the image distance vis 6 cm.
Step 3: Calculate the magnification of the image. The magnification mis
given by:
m=−v
u
Substitute v= 6 cm and u=−15 cm into the equation:
m=−6
−15 = 0.4
Therefore, the magnification of the image is 0.4.
Question 23
Question
An object is placed 15 cm in front of a convex lens with a focal length of 10 cm.
Determine the image distance and magnification produced by the lens.
21
Solution
Step 1: Identify the given values and the unknowns.
Given:
Object distance, do=−15 cm (negative since the object is in front of the
lens)
Focal length, f= 10 cm
Unknowns:
Image distance, di
Magnification, m
Step 2: Apply the lens equation to calculate the image distance.
The lens equation is given by:
1
f=1
di
+1
do
Substitute f= 10 cm and do=−15 cm:
1
10 =1
di
+1
−15
Solve for di:1
di
=1
10 +1
15 =3+2
30 =5
30 =1
6
di= 6 cm
Step 3: Calculate the magnification using the magnification formula.
The magnification mis given by:
m=−di
do
Substitute di= 6 cm and do=−15 cm:
m=−6
−15 =2
5
Therefore, the image distance is 6 cm and the magnification is 2
5.
Question 24
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
10 cm. Determine the position and magnification of the image formed by the
mirror.
22
Solution
Step 1: Identify the given values and variables. Given: - Object distance (do)
= -20 cm (negative since the object is in front of the mirror) - Focal length (f)
= 10 cm - Since the mirror is concave, the focal length is negative. - Using the
sign convention: object distance (do) is negative for real objects placed in front
of the mirror.
Step 2: Apply the mirror formula to find the image distance. The mirror
equation is given by:
−1
f=1
di
+1
do
Substitute the given values into the equation:
−1
10 =1
di
+1
−20
Solve for di:
−1
10 =1
di
−1
20
−1
10 +1
20 =1
di
1
20 =1
di
di= 20 cm
Step 3: Calculate the magnification of the image. The magnification (m) is
given by:
m=−di
do
Substitute the values of diand do:
m=−20
−20
m=−1
Step 4: Analyze the results. The image is located 20 cm behind the mirror,
which indicates that it is a real image. The negative value of magnification
indicates that the image is inverted.
Therefore, the image is formed 20 cm behind the mirror with a magnification
of -1.
Question 25
Question
A concave mirror with a focal length of 20 cm forms an image of an object
located 30 cm from the mirror. Determine the position and nature of the image
formed.
23
Solution
Step 1: Given that the focal length of the concave mirror is 20 cm and the
object distance is 30 cm. Step 2: Using the mirror equation 1
f=1
do+1
di, where
fis the focal length, dois the object distance, and diis the image distance.
Step 3: Substitute the given values into the mirror equation to solve for di:
1
20 =1
30 +1
di. Step 4: Simplifying the equation, we get 1
di=1
20 −1
30 . Step 5:
Find a common denominator: 1
di=3
60 −2
60 . Step 6: Therefore, 1
di=1
60 . Step 7:
Solving for di, we get di= 60 cm. Step 8: Since the image distance is positive,
the image is formed on the same side as the object (real image). Step 9: The
position of the image is 60 cm from the mirror surface. Step 10: The nature of
the image is real and inverted.
Therefore, the image formed by the concave mirror is real, inverted, and
located 60 cm from the mirror surface.
Question 26
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Determine the image distance and magnification using the mirror equation.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
Object distance do=−20 cm (negative because it is in front of the mirror)
Focal length f= 15 cm
Mirror equation: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation to find the
image distance.
Substitute f= 15 cm and do=−20 cm into the mirror equation:
1
15 =1
−20 +1
di
Step 3: Solve for the image distance.
1
15 =−1
20 +1
di
1
di
=1
15 +1
20
1
di
=4
60 +3
60
24
1
di
=7
60
di=60
7≈8.57 cm
Therefore, the image distance is approximately 8.57 cm.
Step 4: Calculate the magnification using the magnification equation.
Magnification equation: m=−di
do
Step 5: Substitute the image distance and object distance to find the mag-
nification.
Substitute di= 8.57 cm and do=−20 cm into the magnification equation:
m=−8.57
−20
m= 0.4285
Therefore, the magnification is 0.4285.
Question 27
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
15 cm. Calculate the image distance and verify whether the image is real or
virtual.
Solution
Step 1: Identify the given values. The object distance ois 20 cm, and the focal
length fis 15 cm. The focal length is negative for a concave mirror.
Step 2: Apply the mirror equation. The mirror equation is given by 1
f=
1
o+1
i, where fis the focal length, ois the object distance, and iis the image
distance.
Step 3: Substitute the given values into the mirror equation. Substitute
f=−15 cm and o= 20 cm:
−1
15 =1
20 +1
i
Step 4: Solve for i.
−1
15 =1
20 +1
i
−1
15 −1
20 =1
i
−4
60 −3
60 =1
i
25
−7
60 =1
i
Solving for igives:
i=−60
7cm ≈ −8.57 cm
Step 5: Determine if the image is real or virtual. For concave mirrors, the
image is real if iis negative (in front of the mirror) and virtual if iis positive
(behind the mirror). Since iis negative in this case, the image is real.
Therefore, the image distance is approximately -8.57 cm, and the image
formed is real.
Question 28
Question
An object is placed 20 cm in front of a converging lens with a focal length of 10
cm. Determine the image distance and height of the image formed by the lens.
Solution
Step 1: Identify the given values and the lens equation. Given: f= 10 cm (focal
length), o=−20 cm (object distance).
The lens equation is given by:
1
f=1
o+1
i
Step 2: Substitute the given values into the lens equation. Substitute f= 10
cm and o=−20 cm into the lens equation to get:
1
10 =1
−20 +1
i
Step 3: Solve for i. Solving the equation above for 1/i, we get:
1
i=1
10 −1
20 =2−1
20 =1
20
Thus, i= 20 cm.
Step 4: Calculate the magnification. The magnification (M) is given by:
M=−i
o
Substitute i= 20 cm and o=−20 cm into the magnification equation:
M=−20
−20 = 1
26
Step 5: Calculate the image height. The height of the image can be deter-
mined using the magnification. Since the magnification is 1, the image is the
same size as the object. Therefore, the height of the image is the same as the
height of the object.
Step 6: Write the final answer. The image distance is 20 cm and the height
of the image is the same as the height of the object.
Question 29
Question
A converging lens forms an image of an object located 20 cm from the lens. The
image is real, inverted, and four times the size of the object. Find the focal
length of the lens.
Solution
Step 1: Identify the given values and the sign conventions. Step 2: Use the lens
equation to find the focal length of the lens.
Step 1: Given values and sign conventions: The object distance p=−20
cm (since the object is located on the opposite side of the incoming light).
The image distance qis positive for a real image. The magnification M=−4
(negative sign indicates an inverted image).
Step 2: Applying the lens equation: The lens equation is given by:
1
f=1
p+1
q
Substitute the given values into the lens equation:
1
f=1
−20 +1
q
Since M=−q
p, we have q=−4p. Substitute q=−4pinto the equation:
1
f=1
−20 +1
−4p
Simplify the equation:
1
f=−4−20
(−20)(−4) =−24
80 =−3
10
Therefore, the focal length of the lens is f=−10
3cm.
Question 30
Question
A convex lens has a focal length of 15 cm. An object is placed 25 cm in front of
the lens. Calculate the image distance and magnification produced by the lens.
27
Solution
Step 1: Identify the given values: The focal length of the convex lens, f= 15
cm, The object distance, u=−25 cm (negative because the object is located to
the left of the lens), The image distance, v=?, The magnification, m=?.
Step 2: Apply the lens equation to find the image distance: The lens equation
is given by: 1
f=1
v+1
u
Substitute f= 15 cm and u=−25 cm:
1
15 =1
v+1
−25
Solving for v:1
v=1
15 +1
25 =5+3
75 =8
75
v=75
8= 9.375 cm
Step 3: Calculate the magnification: The magnification can be calculated
using the formula:
m=−v
u
Substitute u=−25 cm and v= 9.375 cm:
m=−9.375
−25 = 0.375
Step 4: Interpret the results: The image distance is 9.375 cm in front of
the lens, and the magnification is 0.375. The positive value of magnification
indicates the image is upright relative to the object and diminished in size.
Question 31
Question
A converging lens has a focal length of 15 cm. An object is placed 30 cm in
front of the lens. Calculate the position and nature (real or virtual) of the image
formed by the lens.
Solution
Step 1: Given data The focal length of the lens, f= 15 cm
The object distance, u=−30 cm (negative because the object is placed in front
of the lens)
28
Step 2: Using the lens equation The lens equation relates the object distance
(u), the image distance (v), and the focal length (f) of the lens:
1
f=1
v+1
u
Step 3: Substitute the given values into the lens equation
1
15 =1
v+1
−30
Step 4: Solve for v1
v=1
15 −1
30
1
v=2
30 −1
30
1
v=1
30
v= 30 cm
Step 5: Interpret the result The positive value of vindicates that the image
is formed on the opposite side of the lens from the object (real image).
Therefore, the image is formed 30 cm in front of the lens and the nature of
the image is real.
Question 32
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.
Solution
Step 1: Identify the given values and the known lens equation. Given: - Object
distance, p=−20 cm (negative sign indicates that the object is located to the
left of the lens). - Focal length, f= 10 cm. The lens equation is:
1
f=1
p+1
q
Step 2: Substitute the known values into the lens equation.
1
10 =1
−20 +1
q
Step 3: Solve for the image distance, q.
1
10 =−1
20 +1
q
29
1
10 +1
20 =1
q
3
20 =1
q
q=20
3cm
Therefore, the image distance is q=20
3cm.
Step 4: Calculate the magnification using the magnification formula. The
magnification, M, is given by:
M=−q
p
Substitute the values of qand pto find the magnification.
M=−
20
3
−20
M=20
3×20
M=1
3
So, the magnification produced by the lens is 1
3.
Question 33
Question
A concave mirror with a focal length of 20 cm forms a real image that is mag-
nified 3 times. If the object is located 30 cm from the mirror, what is the
magnification produced by the mirror?
Solution
Step 1: First, we need to determine the image distance using the mirror equa-
tion: 1
f=1
do
+1
di
Given that f=−20 cm (since it is a concave mirror) and do=−30 cm (since
the object is placed in front of the mirror), we can rearrange the mirror equation
to solve for di:1
−20 =1
−30 +1
di
−3
20 =−3
30 +1
di
−3
20 +1
3=1
di
30
−9
60 +20
60 =1
di
11
60 =1
di
di=60
11
di≈5.45 cm
Step 2: Next, we can calculate the magnification produced by the mirror
using the magnification equation:
M=−di
do
Substitute the values of diand dointo the formula:
M=−5.45
−30
M=5.45
30
M= 0.183
Therefore, the magnification produced by the mirror is 0.183.
Question 34
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Calculate the image distance and determine if the image is real or virtual.
Solution
Step 1: Use the mirror equation to find the image distance.
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: Substitute the given values into the equation.
1
10 cm =1
20 cm +1
di
Step 3: Solve for di.1
di
=1
10 cm −1
20 cm
1
di
=2
20 cm −1
20 cm =1
20 cm
di= 20 cm
Step 4: Since the image distance is positive, the image is real.
31
Question 35
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the position and nature of the image formed.
Solution
Step 1: Given that the object distance (do) is -20 cm (since the object is in front
of the mirror) and the focal length (f) is -10 cm (since the mirror is concave),
we can use the mirror equation:
1
f=1
do
+1
di
Substitute the known values:
1
−10 =1
−20 +1
di
Step 2: Solve for the image distance (di):
−1
10 =−1
20 +1
di
−1
10 +1
20 =1
di
1
di
=2−1
20 =1
20
Thus, di= 20 cm.
Step 3: To determine the nature of the image, we can use the magnification
formula:
m=−di
do
Substitute the values:
m=−20
−20 = 1
Since the magnification is positive, the image is upright.
Step 4: Since the image distance is positive, the image is formed on the
same side as the object (in front of the mirror). Therefore, the image formed is
virtual, upright, and magnified.
32
