PHYS 101 - ELEMENTS OF PHYSICS
- Lens and mirror equations
Question Bank - Set 2
Liberty University
Question 1
Question
A converging lens with a focal length of 15 cm is used to form an image of an
object placed 30 cm from the lens. Calculate the image distance, magnification,
and height of the image if the object height is 2 cm.
Solution
Step 1: Given that the focal length of the converging lens, f= 15 cm, the object
distance from the lens, do=−30 cm, and the object height, ho= 2 cm.
Step 2: We can use the lens equation to find the image distance:
1
f=1
do
+1
di
Step 3: Substituting the given values into the lens equation, we have:
1
15 =1
−30 +1
di
Step 4: Solving for the image distance, di:
1
di
=1
15 +1
30 =2
30 +1
30 =3
30 =1
10
Step 5: Therefore, the image distance, di= 10 cm.
Step 6: To find the magnification, we use the formula:
M=−di
do
Step 7: Substituting the values into the magnification formula:
M=−10
−30 =1
3
Step 8: Thus, the magnification is M=1
3.
Step 9: Finally, we can find the height of the image using the magnification
formula:
hi=M·ho
Step 10: Substituting the values, we get:
hi=1
3·2 = 2
3
Step 11: Therefore, the height of the image is hi=2
3cm.
Question 2
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, magnification, and describe
the nature of the image.
Solution
Step 1: Given that the focal length of the mirror (f) = -15 cm (concave mirror)
Step 2: Given that the object distance (do) = −30cm(sinceitisinf rontofthemirror, theobjectdistanceisnegative)Step3 :
Usingthemirrorequation1f=1
do+1
di
, where fis the focal length, dois the object
distance, and diis the image distance. Step 4: Plug in the values: 1
−15 =1
−30 +1
di
Step 5: Simplify the equation to solve for di:−2
30 =1
diStep 6: Solve for
di:di=−15 cm Step 7: Calculate the magnification (m) using the formula
m=−di
doStep 8: Substitute the values: m=−−15
−30 = 0.5 Step 9: Since the
magnification is positive, the image is upright. Step 10: Since diis negative, the
image is formed behind the mirror, hence the image is real. Step 11: Therefore,
the image distance is −15 cm, the magnification is 0.5, and the image is real
and upright.
Question 3
Question
A concave mirror with a radius of curvature of 20 cm forms an image of an
object 10 cm in front of the mirror. Find the position and nature of the image.
2
Solution
Step 1: Given values: The radius of curvature R=−20 cm (negative for concave
mirror) and the object distance u=−10 cm (negative for object in front of the
mirror).
Step 2: Find the mirror equation: The mirror equation relates the object
distance (u), image distance (v), and focal length (f) for mirrors:
1
f=1
v+1
u
Since f=R
2for a concave mirror, we have f=−20
2=−10 cm.
Plugging in fand uinto the mirror equation, we get:
1
−10 =1
v+1
−10
Step 3: Solve for v: Solving the equation, we get:
1
v=1
−10 −1
−10 =2
−10
v=−10
2=−5 cm
Step 4: Determine the position and nature of the image: Since the image
distance vis negative, the image is formed on the same side as the object (in
this case, in front of the mirror). Since vis negative, the image is real.
Therefore, the image is formed 5 cm in front of the mirror (on the same side
as the object) and it is a real image.
Question 4
Question
An object is placed 20 cm from a concave mirror of focal length 15 cm. Deter-
mine the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given: - Object
distance (o) = -20 cm (negative sign indicates that the object is located to the
left of the mirror) - Focal length (f) = -15 cm (negative sign indicates that the
mirror is concave) The mirror equation is given by:
1
f=1
o+1
i
Step 2: Substitute the given values into the mirror equation.
1
−15 =1
−20 +1
i
3
Step 3: Solve for the image distance (i).
−1
15 =−1
20 +1
i
−1
15 +1
20 =1
i
−4+3
60 =1
i
−1
60 =1
i
i=−60 cm
Therefore, the image distance is -60 cm, which means the image is formed
on the same side as the object (virtual image).
Step 4: Calculate the magnification. The magnification (m) is given by:
m=−i
o
m=−−60
−20
m= 3
Therefore, the magnification produced by the mirror is 3.
Question 5
Question
An object is placed 20 cm from 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 known values and the mirror formula.
Given: Object distance, u=−20 cm (since the object is in front of the
mirror) Focal length, f=−15 cm (since the mirror is concave)
The mirror equation is given by:
1
f=1
u+1
v
Step 2: Substitute the values into the mirror equation and solve for the
image distance.
1
−15 =1
−20 +1
v
4
−1
15 =−1
20 +1
v
1
v=−1
15 +1
20
1
v=−4
60 +3
60
1
v=−1
60
v=−60 cm
So, the image is formed at a distance of 60 cm behind the mirror.
Step 3: Determine the nature of the image.
Since the object distance is negative and the image distance is also negative,
the image is formed on the same side as the object. Therefore, the image formed
is virtual and erect.
Step 4: State the final answer.
The image is formed 60 cm behind the concave mirror and is virtual and
erect.
Question 6
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 30
cm. Determine the image distance and the magnification of the image.
Solution
Step 1: Identify the given values and the mirror formula.
Object distance, p=−20 cm (negative because it is in front of the mirror).
Focal length, f=−30 cm (negative because it is a concave mirror).
The mirror formula is given by:
1
f=1
p+1
q
where pis the object distance and qis the image distance.
Step 2: Plug in the given values into the mirror formula.
1
−30 =1
−20 +1
q
5
Step 3: Solve for the image distance, q.
1
−30 =1
−20 +1
q
1
q=1
−30 −1
−20
1
q=−1
60
q=−60 cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−q
p
Step 5: Plug in the values of pand qto find the magnification, M.
M=−−60
−20
M= 3
Therefore, the image distance is −60 cm and the magnification of the image
is 3.
Question 7
Question
A concave mirror with a focal length of 15 cm forms an image that is three
times the size of the object. If the object is placed 20 cm in front of the mirror,
find the image distance and magnification.
Solution
Step 1: Given the object distance and focal length, we can use the mirror
equation to find the image distance.
1
f=1
do
+1
di
Step 2: Substitute f=−15 cm (since it is a concave mirror), do=−20 cm,
and let di=−x(to indicate that it is a real image).
1
−15 =1
−20 +1
−x
Step 3: Solve for di:
−1
15 =−1
20 −1
x
6
1
x=1
20 −1
15
Step 4: Find a common denominator:
1
x=3
60 −4
60
1
x=−1
60
x=−60 cm
Step 5: The image distance is −60 cm, indicating the image is real and
located 60 cm away from the mirror.
Step 6: Next, we can find the magnification using the magnification equation:
magnification =image height
object height =−di
do
Step 7: Substitute di=−60 cm and do=−20 cm:
magnification =−−60
−20 = 3
Step 8: Therefore, the image distance is −60 cm and the magnification is 3.
Question 8
Question
A convex lens has a focal length of 15 cm. An object is placed 30 cm in front
of the lens. Determine the image distance and magnification produced by the
lens.
Solution
Step 1: Identify the given values and the lens equation.
Given: f= 15 cm, do=−30 cm
Lens equation: 1
f=1
di+1
do
7
Step 2: Plug in the values and solve for di.
1
15 =1
di
+1
−30
1
15 =1
di
−1
30
1
di
=1
15 +1
30
1
di
=2
30 +1
30
1
di
=3
30
di=30
3
di= 10 cm
Step 3: Calculate the magnification.
M=−di
do
M=−10
−30
M=1
3
Step 4: Write the final answer. The image distance of the lens is 10 cm and
the magnification is 1
3.
Question 9
Question
A converging lens of focal length 15 cm is placed 25 cm from an object. If
the image formed is virtual and 3 times the size of the object, find the image
distance and magnification.
Solution
Step 1: Given that the focal length of the converging lens f= 15 cm, the object
distance u=−25 cm, and the magnification m=−3.
Step 2: Use the lens formula 1
f=1
v+1
u, where fis the focal length, vis the
image distance, and uis the object distance.
Step 3: Substituting the given values into the lens formula, we have 1
15 =
1
v+1
−25 .
Step 4: Solve for v:1
15 =1
v−1
25
8
1
v=1
15 +1
25
1
v=8
75
v=75
8
v= 9.375 cm
Step 5: Now, let’s find the magnification using the formula m=−v
u.
Step 6: Substituting the values of vand u, we get m=−9.375
−25 .
Step 7: Calculate the magnification:
m=9.375
25
m=−0.375
Step 8: Therefore, the image distance is 9.375 cm and the magnification is
-0.375.
Question 10
Question
An object is placed 15 cm in front of a convex lens with a focal length of 10 cm.
Calculate the position and size of the image formed.
Solution
Step 1: Identify the given values and the formula to use. Given: - Object
distance, u=−15 cm (negative because it is in front of the lens) - Focal length,
f= 10 cm
We will use the lens equation:
1
f=1
v+1
u
Step 2: Substitute the values into the lens equation.
1
10 =1
v+1
−15
Step 3: Solve for the image distance, v.
1
v=1
10 +1
15 =3+2
30 =5
30 =1
6
So, v= 6 cm.
9
Step 4: Determine the image size using magnification equation. The magni-
fication, m, is given by:
m=−v
u
Substitute u=−15 cm and v= 6 cm:
m=−6
−15 =2
5
Step 5: Analyze the magnification for the size of the image. Since the
magnitude of mis less than 1, the image is smaller than the object.
Step 6: Summarize the results. The image is formed at a distance of 6 cm
from the lens. It is real and inverted, smaller than the object, and appears on
the same side as the object (since the object is in front of the lens).
Question 11
Question
A concave mirror has a focal length of 20 cm. An object is placed 30 cm in front
of the mirror. Determine the image distance, magnification, and nature of the
image formed.
Solution
Step 1: Identify the given values and the mirror formula. Given: f=−20 cm,
do=−30 cm.
The mirror formula is: 1
f=1
di
+1
do
Step 2: Calculate the image distance. Substitute the given values into the
mirror formula: 1
−20 =1
di
+1
−30
−1
20 =1
di
−1
30
1
di
=1
30 −1
20
1
di
=2−3
60
1
di
=−1
60
di=−60 cm
10
Therefore, the image distance is di=−60 cm.
Step 3: Calculate the magnification. The magnification formula is:
m=−di
do
Substitute the calculated values into the magnification formula:
m=−−60
−30
m=−2
Therefore, the magnification is m=−2.
Step 4: Determine the nature of the image. Since the magnification is neg-
ative, the image is inverted. Since the magnification’s absolute value is greater
than 1, the image is larger than the object. Therefore, the image formed is a
virtual and magnified image.
In conclusion, the image distance is −60 cm, the magnification is −2, and
the image formed is virtual and magnified.
Question 12
Question
A converging lens with a focal length of 15 cm and a concave mirror with a
focal length of -10 cm are placed 25 cm apart along the same optical axis. An
object is placed 40 cm in front of the lens. Determine the final image distance
from the lens when viewed from the mirror.
Solution
Step 1: Calculate the image distance from the lens using the lens equation.
1
flens
=1
do
+1
di,lens
Given that flens = 15 cm, do=−40 cm (as the object is 40 cm in front of
the lens), and the lens equation is solved for di,lens.
1
15 =1
−40 +1
di,lens
1
di,lens
=1
15 −1
−40
1
di,lens
=8
120
11
di,lens =120
8
di,lens = 15 cm
Therefore, the image distance from the lens is 15 cm.
Step 2: Calculate the object distance from the mirror using the mirror equa-
tion. 1
fmirror
=1
di,lens
+1
do,mirror
Given that fmirror =−10 cm and di,lens = 15 cm, the mirror equation is
solved for do,mirror.
1
−10 =1
15 +1
do,mirror
1
do,mirror
=−1
10 −1
15
1
do,mirror
=−3
30 −2
30
1
do,mirror
=−5
30
do,mirror =30
−5
do,mirror =−6 cm
Therefore, the object distance from the mirror is -6 cm.
Question 13
Question
A concave mirror with a focal length of 15 cm is placed 25 cm from an object.
What is the magnification of the image formed by the mirror?
Solution
Step 1: Write down the mirror equation, which relates the object distance (p),
image distance (q), and focal length (f) of a mirror:
1
f=1
p+1
q
12
Step 2: Substitute the given values into the mirror equation:
1
15 cm =1
25 cm +1
q
Step 3: Solve for q:1
q=1
15 cm −1
25 cm
1
q=1
75 cm
q= 75 cm
Step 4: Calculate the magnification (M) using the formula:
M=−q
p
Step 5: Substitute the known values into the magnification formula:
M=−75 cm
25 cm
M=−3
Answer: The magnification of the image formed by the concave mirror is
−3.
Question 14
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 known parameters. Given: Object dis-
tance, do=−20 cm Focal length, f= 15 cm
Step 2: Use the mirror equation to find the image distance. The mirror
equation relates the object distance, image distance, and focal length:
1
f=1
do
+1
di
Substitute the known values into the equation:
1
15 =1
−20 +1
di
13
Solve for di:
di=1
1
15 −1
−20
Step 3: Calculate the image distance.
di=1
1
15 +1
20
=−60 cm
Step 4: Calculate the magnification. The magnification is given by the
formula:
m=−di
do
Substitute the values:
m=−−60
−20 = 3
Step 5: Determine if the image is real or virtual. The image distance is
negative, indicating that the image is formed on the same side as the object.
Therefore, the image is virtual.
Therefore, the image distance is -60 cm, the magnification is 3, and the
image is virtual.
Question 15
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: Identify the given values and the mirror equation. Given: - Object
distance (u) = -20 cm (negative because the object is in front of the mirror) -
Focal length (f) = 10 cm The mirror equation relates the object distance (u),
image distance (v), and focal length (f) as follows:
1
f=1
v+1
u
Step 2: Substitute the given values into the mirror equation.
1
10 =1
v+1
−20
Step 3: Solve for the image distance (v).
1
v=1
10 −1
−20
14
1
v=2
20 +1
20 =3
20
v=20
3cm = 6.67cm
Therefore, the image distance is 6.67 cm.
Step 4: Calculate the magnification. The magnification (M) is given by the
formula:
M=−v
u
Step 5: Substitute the calculated values into the magnification formula.
M=−6.67
−20 = 0.333
Therefore, the magnification is 0.333.
Question 16
Question
A concave mirror creates a real image of a tree that is 2.5 meters tall. The image
is located 7.5 meters from the mirror, and the magnification is -0.6. Determine
the focal length of the mirror.
Solution
Step 1: Identify the given values and the magnification equation. Given: -
Object height, ho= 2.5 m - Image distance, di=−7.5 m (since the image is
real, the distance is negative) - Magnification, M=−0.6
The magnification equation for mirrors is given by:
M=−di
do
Step 2: Find the object distance. Let’s first find the height of the image
using the magnification formula for mirrors:
M=hi
ho
=−di
do
Substitute the known values:
−0.6 = −7.5
do
Solve for do:
do=7.5
0.6= 12.5 m
15
Step 3: Use the mirror equation to calculate the focal length. The mirror
equation for mirrors is: 1
f=1
do
+1
di
Substitute the known values:
1
f=1
12.5+1
−7.5
1
f=1
12.5−1
7.5
1
f=1
75
f= 75 m
Therefore, the focal length of the concave mirror is 75 meters.
Question 17
Question
An object is placed 20 cm to the left of a converging 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 lens equation.
Object distance(u) = −20 cm
Focal length(f) = 10 cm
Lens equation: 1
f=1
u+1
v
Step 2: Solve for the image distance using the lens equation.
1
f=1
u+1
v
1
10 =1
−20 +1
v
1
v=1
10 +1
20
1
v=1
10 +1
10
1
v=2
10
v=10
2
v= 5 cm
16
The image distance is 5 cm and is positive, indicating the image is formed
on the right side of the lens.
Step 3: Calculate the magnification using the magnification formula.
Magnification (m) = −v
u
=−5
−20
=1
4
Therefore, the image distance produced by the lens is 5 cm and the magni-
fication is 1
4.
Question 18
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 object.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance (p) = −15 cm (since the object is in front of the mirror, the distance is negative)
Focal length (f) = −10 cm
Mirror equation: 1
f=1
p+1
q
Step 2: Substitute the known values into the mirror equation to solve for
the image distance (q).
1
−10 =1
−15 +1
q
−1
10 =−1
15 +1
q
1
q=−1
10 +1
15
1
q=3
30 −2
30
1
q=1
30
q= 30 cm
17
Step 3: Calculate the magnification of the object using the magnification
equation.
Magnification (m) = −q
p
m=−30
−15
m= 2
Therefore, the image distance is 30 cm and the magnification of the object
is 2.
Question 19
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 the object is in front of the lens) - Focal length,
f= 20 cm - Lens formula: 1
f=1
v+1
u
Step 2: Substitute the known values into the lens formula and solve for v.
1
20 =1
v+1
−30
1
20 =1
v−1
30
1
v=1
20 +1
30 =3+2
60 =5
60 =1
12
So, v= 12 cm.
Step 3: Calculate the magnification using the formula m=−v
u.
m=−12
−30 =2
5
Step 4: State the results. The image distance is 12 cm to the right of the
lens. The magnification produced by the lens is 2
5.
Question 20
Question
A 4 cm tall object is placed 30 cm in front of a concave mirror with a focal
length of 20 cm. Determine the position and height of the image.
18
Solution
Step 1: Identify the given values. The object height (ho) is 4 cm, the object
distance (do) is -30 cm (since it is in front of the mirror and the mirror formula
considers distances measured from the mirror to the left as negative), and the
focal length (f) is -20 cm (since it is a concave mirror).
Step 2: Calculate the magnification of the mirror. The magnification (M)
of a mirror is given by:
M=−f
do
M=−−20
−30
M=−2
3
Step 3: Use the magnification equation to find the image height. The image
height (hi) can be found using the magnification:
M=hi
ho
−2
3=hi
4
hi=−8
3cm
Step 4: Calculate the image distance. The image distance (di) can be found
using the mirror equation: 1
f=1
do
+1
di
1
−20 =1
−30 +1
di
Solving for digives:
di=−60 cm
Therefore, the image is located 60 cm in front of the mirror, and its height
is 8
3cm.
Question 21
Question
A converging lens with a focal length of 15 cm is placed 20 cm away from an
object. Calculate the position of the image formed by the lens.
19
Solution
Step 1: Given that the focal length of the lens is 15 cm and the object distance
is 20 cm, we can use the lens equation to find the image distance. The lens
equation is given by: 1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Step 2: Substitute f= 15 cm and do=−20 cm (since the object is in front
of the lens) into the lens equation to solve for di:
1
15 =1
−20 +1
di
Step 3: Simplifying the equation:
1
di
=1
15 −1
−20
Step 4: Finding a common denominator:
1
di
=4
60 −3
60 =1
60
Step 5: Solving for di:
di= 60 cm
Step 6: Therefore, the image is formed 60 cm away from the lens. Since the
image distance is positive, the image is formed on the opposite side of the lens
from the object and is real.
Question 22
Question
An object is placed 20 cm from a diverging lens of focal length -10 cm. Determine
the image distance and magnification.
Solution
Step 1: Identify the given values The given values are: Object distance (u) =
-20 cm Focal length of the lens (f) = -10 cm
Step 2: Apply the lens equation The lens equation is given by:
1
f=1
v+1
u
where f = focal length of the lens, v = image distance, u = object distance.
20
Substitute the given values:
1
−10 =1
v+1
−20
Step 3: Solve for image distance (v)
1
−10 =1
v−1
20
1
v=1
−10 +1
20
1
v=2−1
20
1
v=1
20
v= 20 cm
Thus, the image distance is 20 cm.
Step 4: Calculate magnification The magnification (M) is given by:
M=−v
u
Substitute the values of v and u:
M=−20
−20
M= 1
Therefore, the magnification is 1.
Question 23
Question
A converging lens has a focal length of 15 cm. An object is placed 20 cm in
front of the lens. Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the focal length. Given:
f= 15 cm, do=−20 cm
where fis the focal length of the lens and dois the object distance.
Step 2: Apply the lens equation to find the image distance. The lens equation
is given by: 1
f=1
do
+1
di
21
Substitute the given values:
1
15 =1
−20 +1
di
Solve for di:1
di
=1
15 −1
−20
1
di
=4
60 +3
60
1
di
=7
60
di=60
7cm
Step 3: Calculate the magnification using the magnification formula. The
magnification is given by:
M=−di
do
Substitute the values:
M=−60/7
−20
M=60
7×20
M=3
7
Therefore, the image distance is 60
7cm and the magnification is 3
7.
Question 24
Question
A concave mirror with a focal length of 15 cm forms an image that is 1/3 the
size of the object. If the object is located 30 cm from the mirror, what is the
image distance?
Solution
Step 1: Given the mirror forms an image that is 1/3 the size of the object, we
can determine the magnification using the magnification formula:
m=−v
u
where mis the magnification, vis the image distance, and uis the object
distance.
22
Step 2: Since the image is 1/3 the size of the object, we have m=−1
3.
Step 3: Plugging in the given values into the magnification formula:
−1
3=−v
30
Step 4: Solve for vto find the image distance:
v= 10 cm
Therefore, the image distance is 10 cm.
Question 25
Question
A concave mirror with a focal length of 15 cm forms an image of an object located
30 cm from the mirror. If the magnification of the image is -2, determine the
image distance and characterize the image in terms of size and orientation.
Solution
Step 1: Identify the given quantities and sign conventions.
f=−15 cm (focal length of the concave mirror)
do=−30 cm (object distance, negative since the object is located in front
of the mirror)
m=−2 (magnification, negative indicating an inverted image)
di(image distance, to be determined)
Step 2: Use the mirror equation 1
f=1
di+1
doto find di.
1
f=1
di
+1
do
1
−15 =1
di
+1
−30
−1
15 =1
di
−1
30
−2
30 =1
di
di=−15 cm
Step 3: Analyze the characteristics of the image based on the sign and
magnitude of the image distance and magnification.
Since di=−15 cm (<0), the image is formed in front of the mirror.
23
The negative magnification value m=−2 indicates an inverted image.
The magnitude of the magnification (|m|= 2) greater than 1, suggests
that the image is enlarged.
Step 4: Summarize the characteristics of the image.
The image is located 15 cm in front of the mirror.
The image is inverted.
The image is enlarged.
Question 26
Question
A concave mirror with a focal length of 20 cm forms an image that is 1/3 the
size of the object and located 30 cm in front of the mirror. Determine the object
distance (do) and the image distance (di).
Solution
Step 1: Given that the focal length (f) of the concave mirror is 20 cm, the
magnification (M) is 1/3, and the object distance (do) is 30 cm.
Step 2: Recall that the magnification for a mirror is given by the equation
M=−di
do, where diis the image distance.
Step 3: Substituting the given values into the magnification formula, we
have 1
3=−di
30 .
Step 4: Solving for di, we find di=−10 cm.
Step 5: Now, we can use the mirror equation 1
f=1
do+1
dito find do.
Step 6: Substituting the known values into the mirror equation, we get
1
20 =1
do+1
−10 .
Step 7: Solving for do, we find do=−15 cm.
Step 8: Finally, we note that the negative sign for doand diindicates that
they are on the same side as the object, as per the sign convention for mirror
equations. Thus, the object distance is 15 cm and the image distance is 10 cm.
Question 27
Question
A converging lens with a focal length of 10 cm is placed 20 cm from an object.
Calculate the image distance and magnification produced by the lens.
24
Solution
Step 1: Given that the object distance do=−20 cm and the focal length f= 10
cm, we can use the lens equation 1
f=1
do+1
dito find the image distance di.
Step 2: Substitute the known values into the lens equation:
1
10 =1
−20 +1
di
Step 3: Simplify the equation:
1
10 =−1
20 +1
di
Step 4: Solve for 1
di:
1
di
=1
10 +1
20 =3
20
Step 5: Find the value of diby taking the reciprocal:
di=20
3cm
Step 6: Now we can calculate the magnification Musing the magnification
formula M=−di
do.
Step 7: Substitute the known values into the magnification formula:
M=−
20
3
−20 =1
3
Step 8: Therefore, the image distance is 20
3cm and the magnification is 1
3.
Question 28
Question
An object is placed 25 cm in front of a concave mirror of focal length 15 cm.
Determine the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given quantities and the mirror equation. The given quan-
tities are: - Object distance, do=−25 cm (since the object is placed in front
of the mirror) - Focal length, f=−15 cm (for a concave mirror, focal length is
negative) The mirror equation is given by:
−1
f=1
di
+1
do
where diis the image distance.
25
Step 2: Substitute the given values into the mirror equation.
−1
−15 =1
di
+1
−25
Step 3: Solve for di. Simplifying the equation, we get:
1
15 =1
di
−1
25
1
di
=1
15 +1
25
1
di
=5+3
75
1
di
=8
75
di=75
8
di= 9.375 cm
Therefore, the image distance is 9.375 cm.
Step 4: Calculate the magnification. The magnification, M, is given by:
M=−di
do
Substitute the values of diand do:
M=−9.375
−25
M= 0.375
Therefore, the magnification produced by the mirror is 0.375.
Question 29
Question
An object is placed 30 cm in front of a thin lens with a focal length of 20 cm.
Determine the position and magnification of the image formed by the lens.
26
Solution
Step 1: Given the object distance (do=−30 cm) and the focal length of the
lens (f= 20 cm), we can use the lens equation to find the image distance:
1
f=1
do
+1
di
Step 2: Plug in the values:
1
20 =1
−30 +1
di
Step 3: Solve for the image distance, di:
1
di
=1
20 −1
30 =3−2
60 =1
60
di= 60 cm
Step 4: Calculate the magnification (M) using the magnification equation:
M=−di
do
Step 5: Plug in the values:
M=−60
−30 = 2
Step 6: Therefore, the image is formed 60 cm behind the lens and the image
is magnified by a factor of 2.
Question 30
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 produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given: - Object
distance (do) = -30 cm (negative since the object is in front of the mirror) -
Focal length (f) = -20 cm (negative since the concave mirror has a focus in
front of it) The mirror equation is:
1
f=1
do
+1
di
27
Step 2: Substitute the known values into the mirror equation and solve for
the image distance (di). 1
−20 =1
−30 +1
di
−3
60 =−2
60 +1
di
−1
60 =1
di
di=−60cm
Step 3: Calculate the magnification using the magnification equation:
m=−di
do
m=−−60
−30 = 2
Therefore, the image distance is 60 cm in front of the mirror and the mag-
nification produced by the mirror is 2.
Question 31
Question
A converging lens of focal length 15 cm is placed 25 cm to the left of a diverging
lens of focal length -20 cm. An object is placed 40 cm to the left of the converging
lens. Calculate the position of the final image formed.
Solution
Step 1: Calculate the position of the image formed by the converging lens. Step
2: Use this position as the object distance for the diverging lens to find the final
image position.
Step 1: Given: f1= 15 cm (focal length of converging lens) u1=−40 cm
(object distance for converging lens)
Using the lens formula: 1
f1
=1
v1
+1
u1
1
15 =1
v1
+1
−40
1
v1
=1
15 +1
40
1
v1
=40 + 15
600
28
v1=600
55 ≈10.91 cm
So, the image formed by the converging lens is approximately 10.91 cm to
the right of the lens.
Step 2: Now, we use v1as the object distance for the diverging lens.
Given: f2=−20 cm (focal length of diverging lens) u2=v1= 10.91 cm
Using the lens formula for the diverging lens:
1
f2
=1
v2
+1
u2
1
−20 =1
v2
+1
10.91
1
v2
=−1
20 −1
10.91
1
v2
=−10.91 + 20
218
v2=218
9.09 ≈23.98 cm
Therefore, the final image is formed approximately 23.98 cm to the right of
the diverging lens.
Question 32
Question
An object located 40 cm in front of a concave mirror produces a real image
three times the size of the object. If the image is located at a distance of 30 cm
from the mirror, determine the focal length of the mirror.
Solution
Step 1: Given that the object distance is p=−40 cm and the image distance is
q=−30 cm. The magnification mis given by
m=−3
since the image is three times the size of the object.
Step 2: The magnification formula for mirrors is given by
m=−q
p.
Step 3: Substituting the given values into the magnification formula, we
have
−3 = −−30
−40.
29
Step 4: Solving the equation above, we get the object distance as p=−40 cm
and the image distance as q=−30 cm.
Step 5: The mirror equation is given by
1
f=1
p+1
q.
Step 6: Substituting the values of pand q, we have
1
f=1
−40 +1
−30.
Step 7: Simplifying the equation above gives
1
f=−3
120 −4
120 =−7
120.
Step 8: Solving for f, we find
f=−120
7cm ≈ −17.14 cm.
Step 9: Therefore, the focal length of the concave mirror is approximately
−17.14 cm.
Question 33
Question
An object is placed 25 cm from a concave mirror with a focal length of 15 cm.
Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the mirror formula.
Given values:
f=−15 cm
do= 25 cm
Mirror formula: 1
f=1
do
+1
di
Step 2: Calculate the image distance using the mirror formula.
Substitute the given values into the mirror formula:
1
−15 =1
25 +1
di
Solve for di:
−1
15 =1
25 +1
di
30
1
di
=−1
15 −1
25
1
di
=−8
75
di=−75
8
di=−9.375 cm
Step 3: Calculate the magnification.
The magnification is given by:
M=−di
do
Substitute the calculated values:
M=−−9.375
25
M= 0.375
Therefore, the image distance is -9.375 cm and the magnification is 0.375.
Question 34
Question
A concave mirror with a focal length of 20 cm forms an inverted image that is
three times the size of the object. If the object is placed 30 cm from the mirror,
what is the object distance, image distance, and the magnification of the image?
Solution
Step 1: Given data: The focal length of the mirror, f=−20 cm (concave mirror
has negative focal length) The magnification, m=−3 (negative sign indicates
an inverted image) The object distance, u=−30 cm
Step 2: Lens and mirror equations: The mirror equation relates the object
distance, image distance, and focal length:
1
f=1
u+1
v
The magnification equation is:
m=−v
u
Step 3: Calculate the image distance: Substitute the given values into the
mirror equation to solve for v:
1
−20 =1
−30 +1
v
31
1
v=1
−20 −1
−30
1
v=3
60
v=−60 cm
Step 4: Calculate the magnification of the image: Substitute the calculated
image distance into the magnification equation:
m=−(−60)
−30
m= 2
Step 5: Calculate the object distance: Use the mirror equation with the
calculated image distance to find the object distance:
1
−20 =1
−30 +1
−60
1
u=1
−20 −1
−30
1
u=3
60
u=−20 cm
Therefore, the object distance is -20 cm, the image distance is -60 cm, and
the magnification of the image is 2.
Question 35
Question
An object is placed 30 cm in front of a concave mirror of focal length 20 cm.
Determine the image distance, magnification, and whether the image is real or
virtual.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
Object distance, p=−30 cm (since the object is in front of the mirror)
Focal length, f=−20 cm (negative for a concave mirror)
32
Step 7: Substituting the values into the magnification formula:
M=−10
−30 =1
3
Step 8: Thus, the magnification is M=1
3.
Step 9: Finally, we can find the height of the image using the magnification
formula:
hi=M·ho
Step 10: Substituting the values, we get:
hi=1
3·2 = 2
3
Step 11: Therefore, the height of the image is hi=2
3cm.
Question 2
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, magnification, and describe
the nature of the image.
Solution
Step 1: Given that the focal length of the mirror (f) = -15 cm (concave mirror)
Step 2: Given that the object distance (do) = −30cm(sinceitisinf rontof themirror, theobjectdistanceisnegative)Step3 :
Usingthemirrorequation1f=1
do+1
di
, where fis the focal length, dois the object
distance, and diis the image distance. Step 4: Plug in the values: 1
−15 =1
−30 +1
di
Step 5: Simplify the equation to solve for di:−2
30 =1
diStep 6: Solve for
di:di=−15 cm Step 7: Calculate the magnification (m) using the formula
m=−di
doStep 8: Substitute the values: m=−−15
−30 = 0.5 Step 9: Since the
magnification is positive, the image is upright. Step 10: Since diis negative, the
image is formed behind the mirror, hence the image is real. Step 11: Therefore,
the image distance is −15 cm, the magnification is 0.5, and the image is real
and upright.
Question 3
Question
A concave mirror with a radius of curvature of 20 cm forms an image of an
object 10 cm in front of the mirror. Find the position and nature of the image.
2
Solution
Step 1: Given values: The radius of curvature R=−20 cm (negative for concave
mirror) and the object distance u=−10 cm (negative for object in front of the
mirror).
Step 2: Find the mirror equation: The mirror equation relates the object
distance (u), image distance (v), and focal length (f) for mirrors:
1
f=1
v+1
u
Since f=R
2for a concave mirror, we have f=−20
2=−10 cm.
Plugging in fand uinto the mirror equation, we get:
1
−10 =1
v+1
−10
Step 3: Solve for v: Solving the equation, we get:
1
v=1
−10 −1
−10 =2
−10
v=−10
2=−5 cm
Step 4: Determine the position and nature of the image: Since the image
distance vis negative, the image is formed on the same side as the object (in
this case, in front of the mirror). Since vis negative, the image is real.
Therefore, the image is formed 5 cm in front of the mirror (on the same side
as the object) and it is a real image.
Question 4
Question
An object is placed 20 cm from a concave mirror of focal length 15 cm. Deter-
mine the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given: - Object
distance (o) = -20 cm (negative sign indicates that the object is located to the
left of the mirror) - Focal length (f) = -15 cm (negative sign indicates that the
mirror is concave) The mirror equation is given by:
1
f=1
o+1
i
Step 2: Substitute the given values into the mirror equation.
1
−15 =1
−20 +1
i
3
Step 3: Solve for the image distance (i).
−1
15 =−1
20 +1
i
−1
15 +1
20 =1
i
−4+3
60 =1
i
−1
60 =1
i
i=−60 cm
Therefore, the image distance is -60 cm, which means the image is formed
on the same side as the object (virtual image).
Step 4: Calculate the magnification. The magnification (m) is given by:
m=−i
o
m=−−60
−20
m= 3
Therefore, the magnification produced by the mirror is 3.
Question 5
Question
An object is placed 20 cm from 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 known values and the mirror formula.
Given: Object distance, u=−20 cm (since the object is in front of the
mirror) Focal length, f=−15 cm (since the mirror is concave)
The mirror equation is given by:
1
f=1
u+1
v
Step 2: Substitute the values into the mirror equation and solve for the
image distance.
1
−15 =1
−20 +1
v
4
−1
15 =−1
20 +1
v
1
v=−1
15 +1
20
1
v=−4
60 +3
60
1
v=−1
60
v=−60 cm
So, the image is formed at a distance of 60 cm behind the mirror.
Step 3: Determine the nature of the image.
Since the object distance is negative and the image distance is also negative,
the image is formed on the same side as the object. Therefore, the image formed
is virtual and erect.
Step 4: State the final answer.
The image is formed 60 cm behind the concave mirror and is virtual and
erect.
Question 6
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 30
cm. Determine the image distance and the magnification of the image.
Solution
Step 1: Identify the given values and the mirror formula.
Object distance, p=−20 cm (negative because it is in front of the mirror).
Focal length, f=−30 cm (negative because it is a concave mirror).
The mirror formula is given by:
1
f=1
p+1
q
where pis the object distance and qis the image distance.
Step 2: Plug in the given values into the mirror formula.
1
−30 =1
−20 +1
q
5
Step 3: Solve for the image distance, q.
1
−30 =1
−20 +1
q
1
q=1
−30 −1
−20
1
q=−1
60
q=−60 cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−q
p
Step 5: Plug in the values of pand qto find the magnification, M.
M=−−60
−20
M= 3
Therefore, the image distance is −60 cm and the magnification of the image
is 3.
Question 7
Question
A concave mirror with a focal length of 15 cm forms an image that is three
times the size of the object. If the object is placed 20 cm in front of the mirror,
find the image distance and magnification.
Solution
Step 1: Given the object distance and focal length, we can use the mirror
equation to find the image distance.
1
f=1
do
+1
di
Step 2: Substitute f=−15 cm (since it is a concave mirror), do=−20 cm,
and let di=−x(to indicate that it is a real image).
1
−15 =1
−20 +1
−x
Step 3: Solve for di:
−1
15 =−1
20 −1
x
6
1
x=1
20 −1
15
Step 4: Find a common denominator:
1
x=3
60 −4
60
1
x=−1
60
x=−60 cm
Step 5: The image distance is −60 cm, indicating the image is real and
located 60 cm away from the mirror.
Step 6: Next, we can find the magnification using the magnification equation:
magnification =image height
object height =−di
do
Step 7: Substitute di=−60 cm and do=−20 cm:
magnification =−−60
−20 = 3
Step 8: Therefore, the image distance is −60 cm and the magnification is 3.
Question 8
Question
A convex lens has a focal length of 15 cm. An object is placed 30 cm in front
of the lens. Determine the image distance and magnification produced by the
lens.
Solution
Step 1: Identify the given values and the lens equation.
Given: f= 15 cm, do=−30 cm
Lens equation: 1
f=1
di+1
do
7
Step 2: Plug in the values and solve for di.
1
15 =1
di
+1
−30
1
15 =1
di
−1
30
1
di
=1
15 +1
30
1
di
=2
30 +1
30
1
di
=3
30
di=30
3
di= 10 cm
Step 3: Calculate the magnification.
M=−di
do
M=−10
−30
M=1
3
Step 4: Write the final answer. The image distance of the lens is 10 cm and
the magnification is 1
3.
Question 9
Question
A converging lens of focal length 15 cm is placed 25 cm from an object. If
the image formed is virtual and 3 times the size of the object, find the image
distance and magnification.
Solution
Step 1: Given that the focal length of the converging lens f= 15 cm, the object
distance u=−25 cm, and the magnification m=−3.
Step 2: Use the lens formula 1
f=1
v+1
u, where fis the focal length, vis the
image distance, and uis the object distance.
Step 3: Substituting the given values into the lens formula, we have 1
15 =
1
v+1
−25 .
Step 4: Solve for v:1
15 =1
v−1
25
8
1
v=1
15 +1
25
1
v=8
75
v=75
8
v= 9.375 cm
Step 5: Now, let’s find the magnification using the formula m=−v
u.
Step 6: Substituting the values of vand u, we get m=−9.375
−25 .
Step 7: Calculate the magnification:
m=9.375
25
m=−0.375
Step 8: Therefore, the image distance is 9.375 cm and the magnification is
-0.375.
Question 10
Question
An object is placed 15 cm in front of a convex lens with a focal length of 10 cm.
Calculate the position and size of the image formed.
Solution
Step 1: Identify the given values and the formula to use. Given: - Object
distance, u=−15 cm (negative because it is in front of the lens) - Focal length,
f= 10 cm
We will use the lens equation:
1
f=1
v+1
u
Step 2: Substitute the values into the lens equation.
1
10 =1
v+1
−15
Step 3: Solve for the image distance, v.
1
v=1
10 +1
15 =3+2
30 =5
30 =1
6
So, v= 6 cm.
9
Step 4: Determine the image size using magnification equation. The magni-
fication, m, is given by:
m=−v
u
Substitute u=−15 cm and v= 6 cm:
m=−6
−15 =2
5
Step 5: Analyze the magnification for the size of the image. Since the
magnitude of mis less than 1, the image is smaller than the object.
Step 6: Summarize the results. The image is formed at a distance of 6 cm
from the lens. It is real and inverted, smaller than the object, and appears on
the same side as the object (since the object is in front of the lens).
Question 11
Question
A concave mirror has a focal length of 20 cm. An object is placed 30 cm in front
of the mirror. Determine the image distance, magnification, and nature of the
image formed.
Solution
Step 1: Identify the given values and the mirror formula. Given: f=−20 cm,
do=−30 cm.
The mirror formula is: 1
f=1
di
+1
do
Step 2: Calculate the image distance. Substitute the given values into the
mirror formula: 1
−20 =1
di
+1
−30
−1
20 =1
di
−1
30
1
di
=1
30 −1
20
1
di
=2−3
60
1
di
=−1
60
di=−60 cm
10
Therefore, the image distance is di=−60 cm.
Step 3: Calculate the magnification. The magnification formula is:
m=−di
do
Substitute the calculated values into the magnification formula:
m=−−60
−30
m=−2
Therefore, the magnification is m=−2.
Step 4: Determine the nature of the image. Since the magnification is neg-
ative, the image is inverted. Since the magnification’s absolute value is greater
than 1, the image is larger than the object. Therefore, the image formed is a
virtual and magnified image.
In conclusion, the image distance is −60 cm, the magnification is −2, and
the image formed is virtual and magnified.
Question 12
Question
A converging lens with a focal length of 15 cm and a concave mirror with a
focal length of -10 cm are placed 25 cm apart along the same optical axis. An
object is placed 40 cm in front of the lens. Determine the final image distance
from the lens when viewed from the mirror.
Solution
Step 1: Calculate the image distance from the lens using the lens equation.
1
flens
=1
do
+1
di,lens
Given that flens = 15 cm, do=−40 cm (as the object is 40 cm in front of
the lens), and the lens equation is solved for di,lens.
1
15 =1
−40 +1
di,lens
1
di,lens
=1
15 −1
−40
1
di,lens
=8
120
11
di,lens =120
8
di,lens = 15 cm
Therefore, the image distance from the lens is 15 cm.
Step 2: Calculate the object distance from the mirror using the mirror equa-
tion. 1
fmirror
=1
di,lens
+1
do,mirror
Given that fmirror =−10 cm and di,lens = 15 cm, the mirror equation is
solved for do,mirror.
1
−10 =1
15 +1
do,mirror
1
do,mirror
=−1
10 −1
15
1
do,mirror
=−3
30 −2
30
1
do,mirror
=−5
30
do,mirror =30
−5
do,mirror =−6 cm
Therefore, the object distance from the mirror is -6 cm.
Question 13
Question
A concave mirror with a focal length of 15 cm is placed 25 cm from an object.
What is the magnification of the image formed by the mirror?
Solution
Step 1: Write down the mirror equation, which relates the object distance (p),
image distance (q), and focal length (f) of a mirror:
1
f=1
p+1
q
12
Step 2: Substitute the given values into the mirror equation:
1
15 cm =1
25 cm +1
q
Step 3: Solve for q:1
q=1
15 cm −1
25 cm
1
q=1
75 cm
q= 75 cm
Step 4: Calculate the magnification (M) using the formula:
M=−q
p
Step 5: Substitute the known values into the magnification formula:
M=−75 cm
25 cm
M=−3
Answer: The magnification of the image formed by the concave mirror is
−3.
Question 14
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 known parameters. Given: Object dis-
tance, do=−20 cm Focal length, f= 15 cm
Step 2: Use the mirror equation to find the image distance. The mirror
equation relates the object distance, image distance, and focal length:
1
f=1
do
+1
di
Substitute the known values into the equation:
1
15 =1
−20 +1
di
13
Solve for di:
di=1
1
15 −1
−20
Step 3: Calculate the image distance.
di=1
1
15 +1
20
=−60 cm
Step 4: Calculate the magnification. The magnification is given by the
formula:
m=−di
do
Substitute the values:
m=−−60
−20 = 3
Step 5: Determine if the image is real or virtual. The image distance is
negative, indicating that the image is formed on the same side as the object.
Therefore, the image is virtual.
Therefore, the image distance is -60 cm, the magnification is 3, and the
image is virtual.
Question 15
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: Identify the given values and the mirror equation. Given: - Object
distance (u) = -20 cm (negative because the object is in front of the mirror) -
Focal length (f) = 10 cm The mirror equation relates the object distance (u),
image distance (v), and focal length (f) as follows:
1
f=1
v+1
u
Step 2: Substitute the given values into the mirror equation.
1
10 =1
v+1
−20
Step 3: Solve for the image distance (v).
1
v=1
10 −1
−20
14
1
v=2
20 +1
20 =3
20
v=20
3cm = 6.67cm
Therefore, the image distance is 6.67 cm.
Step 4: Calculate the magnification. The magnification (M) is given by the
formula:
M=−v
u
Step 5: Substitute the calculated values into the magnification formula.
M=−6.67
−20 = 0.333
Therefore, the magnification is 0.333.
Question 16
Question
A concave mirror creates a real image of a tree that is 2.5 meters tall. The image
is located 7.5 meters from the mirror, and the magnification is -0.6. Determine
the focal length of the mirror.
Solution
Step 1: Identify the given values and the magnification equation. Given: -
Object height, ho= 2.5 m - Image distance, di=−7.5 m (since the image is
real, the distance is negative) - Magnification, M=−0.6
The magnification equation for mirrors is given by:
M=−di
do
Step 2: Find the object distance. Let’s first find the height of the image
using the magnification formula for mirrors:
M=hi
ho
=−di
do
Substitute the known values:
−0.6 = −7.5
do
Solve for do:
do=7.5
0.6= 12.5 m
15
Step 3: Use the mirror equation to calculate the focal length. The mirror
equation for mirrors is: 1
f=1
do
+1
di
Substitute the known values:
1
f=1
12.5+1
−7.5
1
f=1
12.5−1
7.5
1
f=1
75
f= 75 m
Therefore, the focal length of the concave mirror is 75 meters.
Question 17
Question
An object is placed 20 cm to the left of a converging 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 lens equation.
Object distance(u) = −20 cm
Focal length(f) = 10 cm
Lens equation: 1
f=1
u+1
v
Step 2: Solve for the image distance using the lens equation.
1
f=1
u+1
v
1
10 =1
−20 +1
v
1
v=1
10 +1
20
1
v=1
10 +1
10
1
v=2
10
v=10
2
v= 5 cm
16
The image distance is 5 cm and is positive, indicating the image is formed
on the right side of the lens.
Step 3: Calculate the magnification using the magnification formula.
Magnification (m) = −v
u
=−5
−20
=1
4
Therefore, the image distance produced by the lens is 5 cm and the magni-
fication is 1
4.
Question 18
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 object.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance (p) = −15 cm (since the object is in front of the mirror, the distance is negative)
Focal length (f) = −10 cm
Mirror equation: 1
f=1
p+1
q
Step 2: Substitute the known values into the mirror equation to solve for
the image distance (q).
1
−10 =1
−15 +1
q
−1
10 =−1
15 +1
q
1
q=−1
10 +1
15
1
q=3
30 −2
30
1
q=1
30
q= 30 cm
17
Step 3: Calculate the magnification of the object using the magnification
equation.
Magnification (m) = −q
p
m=−30
−15
m= 2
Therefore, the image distance is 30 cm and the magnification of the object
is 2.
Question 19
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 the object is in front of the lens) - Focal length,
f= 20 cm - Lens formula: 1
f=1
v+1
u
Step 2: Substitute the known values into the lens formula and solve for v.
1
20 =1
v+1
−30
1
20 =1
v−1
30
1
v=1
20 +1
30 =3+2
60 =5
60 =1
12
So, v= 12 cm.
Step 3: Calculate the magnification using the formula m=−v
u.
m=−12
−30 =2
5
Step 4: State the results. The image distance is 12 cm to the right of the
lens. The magnification produced by the lens is 2
5.
Question 20
Question
A 4 cm tall object is placed 30 cm in front of a concave mirror with a focal
length of 20 cm. Determine the position and height of the image.
18
Solution
Step 1: Identify the given values. The object height (ho) is 4 cm, the object
distance (do) is -30 cm (since it is in front of the mirror and the mirror formula
considers distances measured from the mirror to the left as negative), and the
focal length (f) is -20 cm (since it is a concave mirror).
Step 2: Calculate the magnification of the mirror. The magnification (M)
of a mirror is given by:
M=−f
do
M=−−20
−30
M=−2
3
Step 3: Use the magnification equation to find the image height. The image
height (hi) can be found using the magnification:
M=hi
ho
−2
3=hi
4
hi=−8
3cm
Step 4: Calculate the image distance. The image distance (di) can be found
using the mirror equation: 1
f=1
do
+1
di
1
−20 =1
−30 +1
di
Solving for digives:
di=−60 cm
Therefore, the image is located 60 cm in front of the mirror, and its height
is 8
3cm.
Question 21
Question
A converging lens with a focal length of 15 cm is placed 20 cm away from an
object. Calculate the position of the image formed by the lens.
19
Solution
Step 1: Given that the focal length of the lens is 15 cm and the object distance
is 20 cm, we can use the lens equation to find the image distance. The lens
equation is given by: 1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Step 2: Substitute f= 15 cm and do=−20 cm (since the object is in front
of the lens) into the lens equation to solve for di:
1
15 =1
−20 +1
di
Step 3: Simplifying the equation:
1
di
=1
15 −1
−20
Step 4: Finding a common denominator:
1
di
=4
60 −3
60 =1
60
Step 5: Solving for di:
di= 60 cm
Step 6: Therefore, the image is formed 60 cm away from the lens. Since the
image distance is positive, the image is formed on the opposite side of the lens
from the object and is real.
Question 22
Question
An object is placed 20 cm from a diverging lens of focal length -10 cm. Determine
the image distance and magnification.
Solution
Step 1: Identify the given values The given values are: Object distance (u) =
-20 cm Focal length of the lens (f) = -10 cm
Step 2: Apply the lens equation The lens equation is given by:
1
f=1
v+1
u
where f = focal length of the lens, v = image distance, u = object distance.
20
Substitute the given values:
1
−10 =1
v+1
−20
Step 3: Solve for image distance (v)
1
−10 =1
v−1
20
1
v=1
−10 +1
20
1
v=2−1
20
1
v=1
20
v= 20 cm
Thus, the image distance is 20 cm.
Step 4: Calculate magnification The magnification (M) is given by:
M=−v
u
Substitute the values of v and u:
M=−20
−20
M= 1
Therefore, the magnification is 1.
Question 23
Question
A converging lens has a focal length of 15 cm. An object is placed 20 cm in
front of the lens. Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the focal length. Given:
f= 15 cm, do=−20 cm
where fis the focal length of the lens and dois the object distance.
Step 2: Apply the lens equation to find the image distance. The lens equation
is given by: 1
f=1
do
+1
di
21
Substitute the given values:
1
15 =1
−20 +1
di
Solve for di:1
di
=1
15 −1
−20
1
di
=4
60 +3
60
1
di
=7
60
di=60
7cm
Step 3: Calculate the magnification using the magnification formula. The
magnification is given by:
M=−di
do
Substitute the values:
M=−60/7
−20
M=60
7×20
M=3
7
Therefore, the image distance is 60
7cm and the magnification is 3
7.
Question 24
Question
A concave mirror with a focal length of 15 cm forms an image that is 1/3 the
size of the object. If the object is located 30 cm from the mirror, what is the
image distance?
Solution
Step 1: Given the mirror forms an image that is 1/3 the size of the object, we
can determine the magnification using the magnification formula:
m=−v
u
where mis the magnification, vis the image distance, and uis the object
distance.
22
Step 2: Since the image is 1/3 the size of the object, we have m=−1
3.
Step 3: Plugging in the given values into the magnification formula:
−1
3=−v
30
Step 4: Solve for vto find the image distance:
v= 10 cm
Therefore, the image distance is 10 cm.
Question 25
Question
A concave mirror with a focal length of 15 cm forms an image of an object located
30 cm from the mirror. If the magnification of the image is -2, determine the
image distance and characterize the image in terms of size and orientation.
Solution
Step 1: Identify the given quantities and sign conventions.
f=−15 cm (focal length of the concave mirror)
do=−30 cm (object distance, negative since the object is located in front
of the mirror)
m=−2 (magnification, negative indicating an inverted image)
di(image distance, to be determined)
Step 2: Use the mirror equation 1
f=1
di+1
doto find di.
1
f=1
di
+1
do
1
−15 =1
di
+1
−30
−1
15 =1
di
−1
30
−2
30 =1
di
di=−15 cm
Step 3: Analyze the characteristics of the image based on the sign and
magnitude of the image distance and magnification.
Since di=−15 cm (<0), the image is formed in front of the mirror.
23
The negative magnification value m=−2 indicates an inverted image.
The magnitude of the magnification (|m|= 2) greater than 1, suggests
that the image is enlarged.
Step 4: Summarize the characteristics of the image.
The image is located 15 cm in front of the mirror.
The image is inverted.
The image is enlarged.
Question 26
Question
A concave mirror with a focal length of 20 cm forms an image that is 1/3 the
size of the object and located 30 cm in front of the mirror. Determine the object
distance (do) and the image distance (di).
Solution
Step 1: Given that the focal length (f) of the concave mirror is 20 cm, the
magnification (M) is 1/3, and the object distance (do) is 30 cm.
Step 2: Recall that the magnification for a mirror is given by the equation
M=−di
do, where diis the image distance.
Step 3: Substituting the given values into the magnification formula, we
have 1
3=−di
30 .
Step 4: Solving for di, we find di=−10 cm.
Step 5: Now, we can use the mirror equation 1
f=1
do+1
dito find do.
Step 6: Substituting the known values into the mirror equation, we get
1
20 =1
do+1
−10 .
Step 7: Solving for do, we find do=−15 cm.
Step 8: Finally, we note that the negative sign for doand diindicates that
they are on the same side as the object, as per the sign convention for mirror
equations. Thus, the object distance is 15 cm and the image distance is 10 cm.
Question 27
Question
A converging lens with a focal length of 10 cm is placed 20 cm from an object.
Calculate the image distance and magnification produced by the lens.
24
Solution
Step 1: Given that the object distance do=−20 cm and the focal length f= 10
cm, we can use the lens equation 1
f=1
do+1
dito find the image distance di.
Step 2: Substitute the known values into the lens equation:
1
10 =1
−20 +1
di
Step 3: Simplify the equation:
1
10 =−1
20 +1
di
Step 4: Solve for 1
di:
1
di
=1
10 +1
20 =3
20
Step 5: Find the value of diby taking the reciprocal:
di=20
3cm
Step 6: Now we can calculate the magnification Musing the magnification
formula M=−di
do.
Step 7: Substitute the known values into the magnification formula:
M=−
20
3
−20 =1
3
Step 8: Therefore, the image distance is 20
3cm and the magnification is 1
3.
Question 28
Question
An object is placed 25 cm in front of a concave mirror of focal length 15 cm.
Determine the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given quantities and the mirror equation. The given quan-
tities are: - Object distance, do=−25 cm (since the object is placed in front
of the mirror) - Focal length, f=−15 cm (for a concave mirror, focal length is
negative) The mirror equation is given by:
−1
f=1
di
+1
do
where diis the image distance.
25
Step 2: Substitute the given values into the mirror equation.
−1
−15 =1
di
+1
−25
Step 3: Solve for di. Simplifying the equation, we get:
1
15 =1
di
−1
25
1
di
=1
15 +1
25
1
di
=5+3
75
1
di
=8
75
di=75
8
di= 9.375 cm
Therefore, the image distance is 9.375 cm.
Step 4: Calculate the magnification. The magnification, M, is given by:
M=−di
do
Substitute the values of diand do:
M=−9.375
−25
M= 0.375
Therefore, the magnification produced by the mirror is 0.375.
Question 29
Question
An object is placed 30 cm in front of a thin lens with a focal length of 20 cm.
Determine the position and magnification of the image formed by the lens.
26
Solution
Step 1: Given the object distance (do=−30 cm) and the focal length of the
lens (f= 20 cm), we can use the lens equation to find the image distance:
1
f=1
do
+1
di
Step 2: Plug in the values:
1
20 =1
−30 +1
di
Step 3: Solve for the image distance, di:
1
di
=1
20 −1
30 =3−2
60 =1
60
di= 60 cm
Step 4: Calculate the magnification (M) using the magnification equation:
M=−di
do
Step 5: Plug in the values:
M=−60
−30 = 2
Step 6: Therefore, the image is formed 60 cm behind the lens and the image
is magnified by a factor of 2.
Question 30
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 produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given: - Object
distance (do) = -30 cm (negative since the object is in front of the mirror) -
Focal length (f) = -20 cm (negative since the concave mirror has a focus in
front of it) The mirror equation is:
1
f=1
do
+1
di
27
Step 2: Substitute the known values into the mirror equation and solve for
the image distance (di). 1
−20 =1
−30 +1
di
−3
60 =−2
60 +1
di
−1
60 =1
di
di=−60cm
Step 3: Calculate the magnification using the magnification equation:
m=−di
do
m=−−60
−30 = 2
Therefore, the image distance is 60 cm in front of the mirror and the mag-
nification produced by the mirror is 2.
Question 31
Question
A converging lens of focal length 15 cm is placed 25 cm to the left of a diverging
lens of focal length -20 cm. An object is placed 40 cm to the left of the converging
lens. Calculate the position of the final image formed.
Solution
Step 1: Calculate the position of the image formed by the converging lens. Step
2: Use this position as the object distance for the diverging lens to find the final
image position.
Step 1: Given: f1= 15 cm (focal length of converging lens) u1=−40 cm
(object distance for converging lens)
Using the lens formula: 1
f1
=1
v1
+1
u1
1
15 =1
v1
+1
−40
1
v1
=1
15 +1
40
1
v1
=40 + 15
600
28
v1=600
55 ≈10.91 cm
So, the image formed by the converging lens is approximately 10.91 cm to
the right of the lens.
Step 2: Now, we use v1as the object distance for the diverging lens.
Given: f2=−20 cm (focal length of diverging lens) u2=v1= 10.91 cm
Using the lens formula for the diverging lens:
1
f2
=1
v2
+1
u2
1
−20 =1
v2
+1
10.91
1
v2
=−1
20 −1
10.91
1
v2
=−10.91 + 20
218
v2=218
9.09 ≈23.98 cm
Therefore, the final image is formed approximately 23.98 cm to the right of
the diverging lens.
Question 32
Question
An object located 40 cm in front of a concave mirror produces a real image
three times the size of the object. If the image is located at a distance of 30 cm
from the mirror, determine the focal length of the mirror.
Solution
Step 1: Given that the object distance is p=−40 cm and the image distance is
q=−30 cm. The magnification mis given by
m=−3
since the image is three times the size of the object.
Step 2: The magnification formula for mirrors is given by
m=−q
p.
Step 3: Substituting the given values into the magnification formula, we
have
−3 = −−30
−40.
29
Step 4: Solving the equation above, we get the object distance as p=−40 cm
and the image distance as q=−30 cm.
Step 5: The mirror equation is given by
1
f=1
p+1
q.
Step 6: Substituting the values of pand q, we have
1
f=1
−40 +1
−30.
Step 7: Simplifying the equation above gives
1
f=−3
120 −4
120 =−7
120.
Step 8: Solving for f, we find
f=−120
7cm ≈ −17.14 cm.
Step 9: Therefore, the focal length of the concave mirror is approximately
−17.14 cm.
Question 33
Question
An object is placed 25 cm from a concave mirror with a focal length of 15 cm.
Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the mirror formula.
Given values:
f=−15 cm
do= 25 cm
Mirror formula: 1
f=1
do
+1
di
Step 2: Calculate the image distance using the mirror formula.
Substitute the given values into the mirror formula:
1
−15 =1
25 +1
di
Solve for di:
−1
15 =1
25 +1
di
30
1
di
=−1
15 −1
25
1
di
=−8
75
di=−75
8
di=−9.375 cm
Step 3: Calculate the magnification.
The magnification is given by:
M=−di
do
Substitute the calculated values:
M=−−9.375
25
M= 0.375
Therefore, the image distance is -9.375 cm and the magnification is 0.375.
Question 34
Question
A concave mirror with a focal length of 20 cm forms an inverted image that is
three times the size of the object. If the object is placed 30 cm from the mirror,
what is the object distance, image distance, and the magnification of the image?
Solution
Step 1: Given data: The focal length of the mirror, f=−20 cm (concave mirror
has negative focal length) The magnification, m=−3 (negative sign indicates
an inverted image) The object distance, u=−30 cm
Step 2: Lens and mirror equations: The mirror equation relates the object
distance, image distance, and focal length:
1
f=1
u+1
v
The magnification equation is:
m=−v
u
Step 3: Calculate the image distance: Substitute the given values into the
mirror equation to solve for v:
1
−20 =1
−30 +1
v
31
1
v=1
−20 −1
−30
1
v=3
60
v=−60 cm
Step 4: Calculate the magnification of the image: Substitute the calculated
image distance into the magnification equation:
m=−(−60)
−30
m= 2
Step 5: Calculate the object distance: Use the mirror equation with the
calculated image distance to find the object distance:
1
−20 =1
−30 +1
−60
1
u=1
−20 −1
−30
1
u=3
60
u=−20 cm
Therefore, the object distance is -20 cm, the image distance is -60 cm, and
the magnification of the image is 2.
Question 35
Question
An object is placed 30 cm in front of a concave mirror of focal length 20 cm.
Determine the image distance, magnification, and whether the image is real or
virtual.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
Object distance, p=−30 cm (since the object is in front of the mirror)
Focal length, f=−20 cm (negative for a concave mirror)
32
The mirror equation is: 1
f=1
p+1
q
Step 2: Substitute the given values into the mirror equation.
1
−20 =1
−30 +1
q
Step 3: Solve for the image distance, q.
1
q=1
−20 −1
−30
1
q=−3
60
q=−20 cm
Step 4: Calculate the magnification, M.
The magnification formula is:
M=−q
p
Substitute the values:
M=−−20
−30 =2
3
Step 5: Determine if the image is real or virtual.
For a mirror, if qis negative, the image is real; if qis positive, the image is
virtual. Since qis negative in this case, the image is real.
Therefore, the image distance is −20 cm, the magnification is 2
3, and the
image is real.
33
