PHYS 101 - ELEMENTS OF PHYSICS
- Lens and mirror equations
Question Bank - Set 1
Liberty University
Question 1
Question
A concave mirror with a focal length of 10 cm forms a real image that is 2/3
the size of the object. If the object is placed 30 cm from the mirror, determine
the image distance and the magnification.
Solution
Step 1: Given that the focal length of the concave mirror is f=−10 cm, the
object distance is do=−30 cm, and the magnification is m=−2/3, we can use
the mirror equation and the magnification equation to find the image distance
di.
Step 2: The mirror equation is given by: 1
f=1
di
+1
do
.
Substitute the known values into the mirror equation: 1
−10 =1
di
+1
−30.
Step 3: Simplify the equation: 1
−10 =1
di
−1
30.
Step 4: Solve for di:3
30 =1
di
di= 10 cm.
Step 5: The image distance is di= 10 cm. Now, let’s calculate the magnifi-
cation using the magnification equation m=−di
do
.
Step 6: Substitute the known values into the magnification equation: m=
−10
−30 =1
3.
Step 7: Therefore, the image distance is 10 cm and the magnification is 1/3.
Question 2
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 produced by the lens.
Solution
Step 1: Identify the given values.
Object distance, do=−20 cm (negative because it is in front of the lens).
Focal length of the lens, f= 15 cm.
Step 2: Use the lens equation to calculate the image distance, di. The lens
equation is given by: 1
f=1
do
+1
di
Plugging in the given values:
1
15 =1
−20 +1
di
1
di
=1
15 −1
−20
1
di
=4
60 −3
60 =1
60
di= 60 cm
Therefore, the image distance is 60 cm.
Step 3: Calculate the magnification, m, using the formula:
m=−di
do
Plugging in the values:
m=−60
−20 = 3
The magnification produced by the lens is 3.
Question 3
Question
An object is placed 20 cm from a convex lens with a focal length of 10 cm.
Calculate the image distance and magnification produced by the lens.
2
Solution
Step 1: Identify the given values and the focal length.
Given: Object distance, do=−20 cm Focal length, f= 10 cm
Step 2: Use the lens equation to find the image distance.
The lens equation is given by:
1
f=1
do
+1
di
Substitute the given values:
1
10 =1
−20 +1
di
Solving for di:1
di
=1
10 −1
−20
1
di
=2
20 +1
20
1
di
=3
20
di=20
3cm ≈6.67 cm
Therefore, the image distance is approximately 6.67 cm.
Step 3: Calculate the magnification using the magnification formula.
The magnification of a lens is given by:
m=−di
do
Substitute the values of diand do:
m=−6.67
−20
m≈0.333
Therefore, the magnification produced by the lens is approximately 0.333.
Question 4
Question
An object is placed 20 cm in front of a concave mirror of focal length 15 cm.
Determine the position, nature, and size of the image formed.
3
Solution
Step 1: Given that: The object distance, u=−20 cm (negative since the object
is in front of the mirror). The focal length, f=−15 cm (negative since the
mirror is concave).
Step 2: Applying the mirror equation:
1
f=1
v+1
u
1
−15 =1
v+1
−20
−1
15 =1
v−1
20
−1
15 +1
20 =1
v
4
60 +3
60 =1
v
7
60 =1
v
v=60
7cm
Step 3: Calculating the magnification, M:
M=−v
u
M=−60/7
−20
M=60
140 =3
7
Since M < 0, the image is inverted.
Step 4: Determining the nature and position of the image based on the
magnification: If M= 3/7, the image is diminished. Additionally, since the
magnification is positive, the image is virtual. Therefore, the image is virtual,
erect, diminished, and formed 60
7cm behind the mirror.
Question 5
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Determine the image distance, image height, magnification, and whether
the image is real or virtual.
4
Solution
Step 1: Identify the given values. The object distance dois 30 cm, and the
focal length fis -20 cm for a concave mirror (since the focal length of a concave
mirror is negative).
Step 2: Apply the mirror equation to find the image distance. The mirror
equation is given by: 1
f=1
do
+1
di
Substitute the given values:
1
−20 =1
30 +1
di
−1
20 −1
30 =1
di
−1
12 =1
di
Therefore, di=−12 cm.
Step 3: Calculate the magnification. The magnification Mis given by:
M=−di
do
Substitute the values:
M=−
−12
30
M=2
5
Step 4: Determine if the image is real or virtual. Since diis negative, the
image is formed on the same side as the object, which makes it a virtual image.
Step 5: Calculate the image height. The image height hican be determined
using the magnification formula:
M=hi
ho
Since the object height hois not given, we cannot calculate the image height
without additional information.
Therefore, the image distance is -12 cm, the magnification is 2/5, and the im-
age formed is virtual. The image height cannot be determined without knowing
the object height.
Question 6
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Determine the image distance, image height, and magnification.
5
Solution
Step 1: Use the lens equation to find the image distance.
1
f=1
do
+1
di
Given f= 15 cm and do=−20 cm (since it is in front of the lens), we can plug
these values in and solve for di.
1
15 =1
−20 +1
di
1
di
=1
15 +1
20
1
di
=4
60 +3
60
1
di
=7
60
di=60
7≈8.57 cm
Step 2: Calculate the magnification using the formula m=−di
do.
m=−8.57
−20
m≈0.429
Step 3: Find the image height using the magnification formula m=hi
ho.
Given that the object height ho= 2 cm, we can find the image height hi.
0.429 = hi
2
hi= 0.429 ×2
hi≈0.857 cm
Therefore, the image distance is approximately 8.57 cm, the magnification
is 0.429, and the image height is approximately 0.857 cm.
Question 7
Question
A converging lens with a focal length of 15 cm is placed 30 cm to the left of a
diverging lens with a focal length of -20 cm. If an object is placed 40 cm to the
left of the converging lens, determine the location and magnification of the final
image formed by the system.
6
Solution
Step 1: Find the image created by the converging lens. The lens equation for
the converging lens is 1
f=1
do+1
di, where fis the focal length, dois the object
distance, and diis the image distance. Given f= 15 cm and do=−40 cm (since
the object distance is measured from the left of the lens and the object is placed
to the left), we can solve for di:
1
15 =1
−40 +1
di
1
di
=1
15 −1
−40
1
di
=8
120
di= 15 cm
Therefore, the image formed by the converging lens is located 15 cm to the
right of the lens.
Step 2: Find the object distance for the diverging lens based on the image
created by the converging lens. Since the diverging lens is 30 cm to the right of
the converging lens, the object distance for the diverging lens is 15 cm−30 cm =
−15 cm.
Step 3: Find the image created by the diverging lens. Using the lens equation
for the diverging lens, which is 1
f=1
do+1
di, where f=−20 cm and do=−15 cm,
we can solve for di:1
−20 =1
−15 +1
di
1
di
=1
−20 −1
−15
1
di
=−3
60
di=−20 cm
Thus, the final image formed by the system is located 20 cm to the left of
the diverging lens.
Step 4: Find the magnification of the final image. The magnification Mis
given by di
do, where di=−20 cm and do=−15 cm.
M=−20
−15 =4
3
Therefore, the final image is formed 20 cm to the left of the diverging lens
with a magnification of 4
3.
7
Question 8
Question
A concave mirror has a focal length of 10 cm. An object is placed 15 cm from
the mirror. Determine the position and magnification of the image formed by
the mirror.
Solution
Step 1: Identify the given quantities and the mirror equation. The mirror
equation relating the object distance do, image distance di, and focal length f
for a mirror is given by: 1
f=1
do
+1
di
Given: f=−10 cm (concave mirror), do=−15 cm (object distance).
Step 2: Solve for the image distance di. Substitute the given values into the
mirror equation: 1
−10 =1
−15 +1
di
Solving for di:1
di
=1
−10 −1
−15 =3
30 −2
30 =1
30
di= 30 cm
Step 3: Calculate the magnification. The magnification Mis given by:
M=−di
do
Substitute the values of diand dointo the equation:
M=−30
−15 = 2
Step 4: Interpret the results. The image is formed 30 cm from the mirror
on the same side as the object. The magnification of the image is 2, indicating
that the image is upright and enlarged.
Question 9
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.
8
Solution
Step 1: Identify known values and the mirror equation.
Object distance, do=−20 cm (negative since it is in front of the mirror)
Focal length, f=−15 cm (negative for concave mirror)
Mirror equation: 1
f=1
do+1
di
Step 2: Solve for the image distance, di.
1
−15 =1
−20 +1
di
−2
30 =−3
60 +1
di
−2
30 +3
60 =1
di
1
di
=4
60
di=60
4
di= 15 cm
So, the image distance is 15 cm.
Step 3: Calculate the magnification, M.
M=−di
do
M=−15
−20
M= 0.75
Therefore, the image distance is 15 cm and the magnification is 0.75.
Question 10
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: Identify the given values: The object distance, do, is 20 cm (positive
because it is in front of the mirror) and the focal length, f, is -10 cm (negative
for a concave mirror).
9
Step 2: Apply the mirror equation: The mirror equation is given by:
1
f=1
do
+1
di
Substitute the values into the equation:
1
−10 =1
20 +1
di
Step 3: Solve for the image distance, di:
−1
10 =1
20 +1
di
−1
di
=−1
10 −1
20
−1
di
=−3
20
di=−20
3cm
Step 4: Analyze the image distance: The negative value of the image distance
indicates that the image is formed behind the mirror, which means it is a real
image.
Step 5: Determine the position of the image: Since the image distance is
negative, the image is located 6.67 cm behind the mirror.
Step 6: Determine the nature of the image: Since the image is real and
formed on the same side as the object, it is a real and inverted image.
Question 11
Question
A converging lens with a focal length of 15 cm is placed 25 cm in front of
a diverging mirror with a focal length of -10 cm. Determine the final image
distance produced by the combination of the lens and the mirror. Is the final
image real or virtual?
Solution
Step 1: Calculate the image distance produced by the converging lens using the
lens equation: 1
flens
=1
do
+1
di
where flens = 15 cm and do=−25 cm (since the object is placed 25 cm in front
of the lens).
10
Substitute the given values into the equation and solve for di:
1
15 =1
−25 +1
di
1
di
=1
15 +1
25
1
di
=5+3
75
1
di
=8
75
di=75
8cm
So, the image distance produced by the lens is di= 9.375 cm.
Step 2: Use the diverging mirror equation to calculate the final image dis-
tance: 1
fmirror
=1
di
+1
d′
o
where fmirror =−10 cm, di= 9.375 cm, and we want to find d′
o.
Substitute the given values into the equation and solve for d′
o:
1
−10 =1
9.375 +1
d′
o
−1
10 =8
75 +1
d′
o
−3
30 −8
75 =1
d′
o
−9
75 =1
d′
o
d′
o=−75
9cm
Therefore, the final image distance produced by the combination of the lens
and the mirror is d′
o=−8.333 cm. Since the image distance is negative, the
final image is virtual.
Question 12
Question
An object is placed 20 cm in front of a concave mirror of focal length 15 cm.
Determine the position and nature of the image formed.
11
Solution
Step 1: Identify the given values The given values are: Object distance, u=
−20 cm (negative sign indicates that the object is located in front of the mirror)
Focal length, f=−15 cm (negative sign for a concave mirror)
Step 2: Apply the mirror formula The mirror formula relates the object
distance (u), image distance (v), and focal length (f) as:
1
f=1
v+1
u
Substitute f=−15 cm and u=−20 cm into the mirror formula:
1
−15 =1
v+1
−20
Step 3: Solve for v
1
−15 =1
v−1
20 =⇒1
v=1
−15 +1
20
=⇒1
v=−4+3
60 =⇒1
v=−1
60
=⇒v=−60 cm
Therefore, the image is located 60 cm behind the mirror.
Step 4: Determine the nature of the image Since the image distance is nega-
tive, the image is formed on the same side as the object (in front of the mirror).
This indicates that the image is virtual and upright.
Thus, the image formed by the concave mirror is virtual and located 60 cm
in front of the mirror.
Question 13
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
10 cm. Calculate the position and magnification of the image formed by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given values: -
Object distance, do=−20 cm - Focal length, f=−10 cm (note the negative
sign for concave mirror)
The mirror equation is: 1
f=1
do
+1
di
12
Step 2: Substitute the known values into the mirror equation and solve for
image distance. Substitute f=−10 cm and do=−20 cm into the mirror
equation: 1
−10 =1
−20 +1
di
Solving for di, we get:
1
di
=1
−10 −1
−20 =1
−10 +1
20 =2−1
20 =1
20
So, di= 20 cm.
Step 3: Calculate the magnification. The magnification, m, in terms of
object and image distances is given by:
m=−di
do
Substitute di= 20 cm and do=−20 cm into the magnification formula:
m=−20
−20 = 1
Step 4: Determine the position of the image. The negative sign indicates
that the image is in front of the mirror, which means the image is formed on
the same side as the object. Therefore, the image is located 20 cm in front of
the mirror.
Step 5: Conclusion The image formed by the concave mirror is located 20
cm in front of the mirror and the magnification is 1.
Question 14
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 of the image formed by
the lens.
Solution
Step 1: Identify the given values and the lens equation. Step 2: Apply the lens
equation to find the image distance. Step 3: Determine the magnification of the
image.
Step 1: Given values: Object distance, do=−30 cm (since the object is
placed in front of the lens) Focal length, f= 20 cm
The lens equation is: 1
f=1
di
+1
do
13
Step 2: Apply the lens equation to find the image distance:
1
20 =1
di
+1
−30
1
di
=1
20 +1
30
1
di
=3
60 +2
60
1
di
=5
60
di=60
5
di= 12 cm
Therefore, the image distance, di, is 12 cm.
Step 3: Determine the magnification of the image: The magnification, m,
is given by:
m=−di
do
Substitute di= 12 cm and do=−30 cm into the formula:
m=−12
−30 =2
5
Therefore, the magnification of the image is 2
5.
Question 15
Question
A concave mirror has a focal length of 15 cm. An object is placed 25 cm in
front of the mirror. Calculate the image distance and magnification produced
by the mirror.
Solution
Step 1: Recall the mirror equation for concave mirrors:
1
f=1
do
+1
di
where: - fis the focal length of the mirror, - dois the object distance, and - di
is the image distance.
14
Step 2: Given that f=−15 cm and do=−25 cm (since the object is in
front of the mirror), we can substitute these values into the mirror equation to
find di:1
−15 =1
−25 +1
di
Step 3: Simplify the equation:
−2
30 =−1
25 +1
di
Step 4: Find a common denominator:
−2
30 =−1·30
25 ·30 +1
di
Step 5: Simplify further:
−2
30 =−30
750 +1
di
Step 6: Add the fractions:
−2
30 =−30
750 +1
di
−2
30 =−30
750 +1
di
−2
30 =−1
25 +1
di
Step 7: Find the common denominator:
−50
750 =−30
750 +1
di
Step 8: Simplify the equation:
−50
750 =−30
750 +1
di
−20
750 =1
di
Step 9: Solve for di:
−20di= 750
di=750
−20
di=−37.5 cm
Step 10: The image distance, di, is -37.5 cm.
15
Step 11: The magnification, M, is given by:
M=−di
do
Step 12: Substitute the values of diand dointo the magnification formula:
M=−37.5
−25
M= 1.5
Step 13: The magnification produced by the mirror is 1.5.
Question 16
Question
An object is placed 20 cm in front of a concave mirror of focal length 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 sign indicates it is in front of
the mirror) - Focal length f=−15 cm (negative sign indicates it is a concave
mirror)
The mirror equation relates the object distance, image distance, and focal
length:
−1
di
+1
do
=1
f
Step 2: Substitute the values into the mirror equation and solve for the im-
age distance.
−1
di
+1
−20 =1
−15
−1
di
−1
20 =−1
15
−1
di
=−1
15 +1
20
−1
di
=−4
60 +3
60
−1
di
=−1
60
di=−60 cm
16
Therefore, the image distance is di=−60 cm. The negative sign indicates
that the image is formed behind the mirror.
Step 3: Calculate the magnification using the magnification formula.
The magnification Mis given by:
M=−di
do
Substitute di=−60 cm and do=−20 cm into the formula:
M=−
−60
−20 =60
20 = 3
Therefore, the magnification is 3, indicating that the image is magnified and
upright.
Question 17
Question
A concave mirror has a focal length of -15 cm. An object is placed 20 cm from
the mirror. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and mirror equation.
Given: f=−15 cm, do= 20 cm
Mirror equation: 1
f=1
di+1
do
Step 2: Solve for diusing the mirror equation.
1
−15 =1
di
+1
20
−1
15 =1
di
+1
20
−1
15 −1
20 =1
di
−4
60 −3
60 =1
di
−7
60 =1
di
di=60
−7
di=−8.57 cm
17
Step 3: Calculate the magnification, m, using the formula m=−di
do.
m=−
−8.57
20
m= 0.43
Therefore, the image distance is −8.57 cm and the magnification is 0.43.
Question 18
Question
An object is placed 10 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 known quantities and the mirror equation.
Object distance, do=−10 cm (negative because it is in front of the mirror)
Focal length, f=−15 cm (negative for a concave mirror)
Mirror equation: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation to find the
image distance.
1
−15 =1
−10 +1
di
−1
15 =−1
10 +1
di
1
di
=−1
15 +1
10
1
di
=−2
30 +3
30
1
di
=1
30
di= 30 cm
Therefore, the image distance is 30 cm.
Step 3: Calculate the magnification using the magnification equation.
M=−di
do
M=−30
−10
M= 3
18
The magnification is 3.
Step 4: Determine if the image is real or virtual. Since the magnification is
positive, the image is upright. Therefore, the image formed is virtual.
In conclusion, the image distance is 30 cm, the magnification is 3, and the
image is virtual.
Question 19
Question
A thin converging lens has a focal length of 15 cm. An object is placed 30 cm
from the lens. Determine the position and size of the final image formed by the
lens.
Solution
Step 1: Use the lens equation to find the image distance.
1
f=1
do
+1
di
Given that the focal length f= 15 cm, and the object distance do= 30 cm.
Substitute these values into the lens equation to solve for di.
Step 2: Solve for di.1
15 =1
30 +1
di
1
di
=1
15 −1
30
1
di
=2
30 −1
30
1
di
=1
30
di= 30 cm
Step 3: Calculate the magnification (m) using the formula m=−di
do.
m=−30
30
m=−1
Step 4: Determine the size of the image. Since the magnification is -1, the
image is the same size as the object but inverted. Thus, the size of the image
is equal to the size of the object.
Step 5: Write the final answer. The final image is located 30 cm from the
lens on the same side as the object, and it is the same size as the object but
inverted.
19
Question 20
Question
A thin convex lens has a focal length of 10 cm. An object placed 20 cm from
the lens forms a real image. Determine the image distance and magnification
produced by the lens.
Solution
Step 1: Identify the given values and the lens formula.
Object distance, u=−20 cm (since the object is placed to the left of the
lens)
Focal length, f= 10 cm (for a convex lens, fis positive)
Using the lens formula: 1
f=1
v+1
u
Step 2: Substitute the values into the lens formula and solve for the image
distance, v.
1
10 =1
v+1
−20
1
10 =1
v−1
20
1
v=1
10 +1
20 =3
20
v=20
3≈6.67 cm
Step 3: Calculate the magnification, m, using the formula m=−v
u.
m=−6.67
−20 =1
3
Therefore, the image distance produced by the lens is approximately 6.67
cm (real image), and the magnification is 1
3.
Question 21
Question
An object is placed 25 cm in front of a converging lens with a focal length of 15
cm. Calculate the position and size of the image formed.
20
Solution
Step 1: Identify the given values and the lens formula. Given:
u=−25 cm
f= 15 cm
Using the lens formula: 1
f=1
v+1
u
Step 2: Calculate the position of the image. Substitute the given values into
the lens formula: 1
15 =1
v+1
−25
1
v=1
15 +1
25
1
v=5+3
75
1
v=8
75
v=75
8
v= 9.375 cm
Step 3: Calculate the size of the image. Use the magnification formula:
M=−v
u
M=−9.375
−25
M= 0.375
Step 4: Interpret the result. The image is formed at a distance of 9.375 cm
from the lens on the same side as the object, and it is smaller than the object
with a magnification of 0.375.
Question 22
Question
An object is placed 20 cm in front of a convex lens of focal length 10 cm.
Calculate the image distance and magnification produced by the lens.
21
Solution
Step 1: Identify the given values.
Object distance (u) = -20 cm (negative sign indicates object is on the
same side as the incident light)
Focal length (f) = 10 cm
Step 2: Use the lens equation to find the image distance.
1
f=1
v+1
u
Plugging in the values: 1
10 =1
v+1
−20
Solving for v:1
v=1
10 +1
20
1
v=3
20
v=20
3
v= 6.67 cm
Therefore, the image distance produced by the lens is 6.67 cm.
Step 3: Calculate the magnification using the magnification formula.
Magnification (M) = −v
u
Plugging in the values:
Magnification (M) = −6.67
−20
Magnification (M) = 0.3333
Therefore, the magnification produced by the lens is 0.3333.
Question 23
Question
An object is placed 30 cm from a converging lens of focal length 20 cm. Calculate
the position of the image formed by the lens.
22
Solution
Given: Object distance, u=−30 cm (because the object is on the side opposite
to the incident light), Focal length of the lens, f= 20 cm.
We will use the lens formula to determine the position of the image:
1
f=1
v+1
u
Step 1: Substitute the given values into the lens formula:
1
20 =1
v+1
−30
Step 2: Simplify the equation:
1
v=1
20 +1
30
Step 3: Find the sum on the right side of the equation:
1
v=3+2
60 =5
60 =1
12
Step 4: Take the reciprocal of both sides to solve for v:
v= 12 cm
Step 5: The positive value of vindicates that the image is formed on the
same side as the incident light, which means the image is virtual and upright,
12 cm away from the lens.
Question 24
Question
A converging lens with a focal length of 15 cm is placed 30 cm away from an
object. At what distance from the lens should the image be formed? Then, a
concave mirror with a focal length of 10 cm is placed 20 cm away from the same
object. At what distance from the mirror should the image be formed?
Solution
Step 1: For the converging lens Given the object distance u=−30 cm (since it
is to the left of the lens), and the focal length f= 15 cm. We can use the lens
formula: 1
f=1
v+1
u
where vis the image distance. Substitute the values into the lens formula:
1
15 =1
v+1
−30
23
1
15 =1
v−1
30
v=30
2= 15 cm
The image is formed 15 cm to the right of the lens.
Step 2: For the concave mirror Given the object distance u=−20 cm and
the focal length f=−10 cm (negative for a concave mirror). We can use the
mirror formula: 1
f=1
v+1
u
where vis the image distance. Substitute the values into the mirror formula:
1
−10 =1
v+1
−20
1
−10 =1
v−1
20
v=−20 cm
The image is formed 20 cm to the left of the mirror.
Question 25
Question
An object is placed 20 cm in front of a concave mirror whose focal length is
30 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
Solution
Step 1: Identify the given values. The object distance dois 20 cm and the focal
length fis -30 cm (since it is a concave mirror).
Step 2: Use the mirror equation to find the image distance. The mirror
equation is given by:
−1
do
+1
di
=1
f
Substitute do= 20 cm and f=−30 cm into the equation:
−1
20 +1
di
=1
−30
Solving for di, we get: 1
di
=1
−30 +1
20
1
di
=−1
60
24
di=−60 cm
Step 3: Calculate the magnification. The magnification is given by the
formula:
m=−di
do
Substitute di=−60 cm and do= 20 cm into the formula:
m=−
−60
20
m= 3
Step 4: Determine if the image is real or virtual. Since the image distance
diis negative, the image is formed on the same side as the object. Therefore,
the image is real.
Therefore, the image distance is -60 cm, the magnification is 3, and the
image is real.
Question 26
Question
A concave mirror with a focal length of 15 cm forms a real image that is 10
times larger than the object. Determine the object distance from the mirror.
Solution
Step 1: Identify the given information.
The focal length of the concave mirror, f=−15 cm (negative because it is a
concave mirror). The magnification of the mirror, m=−10 (negative because
the image is real and inverted).
Step 2: Recall the mirror equation.
The mirror equation relates the object distance (do), image distance (di), and
focal length (f) of a mirror: 1
f=1
do
+1
di
Step 3: Substitute the given values into the mirror equation.
Since m=−di
do, we can rewrite the mirror equation as:
1
f=1
do
+1
−m·do
Substitute f=−15 cm and m=−10 into the equation:
1
−15 =1
do
+1
10 ·do
25
Step 4: Solve for the object distance (do).
Combine the fractions on the right side of the equation:
−1
15 =1+1
10do
−1
15 =2
10do
−1
15 =1
5do
Solve for do:
do=5
−1/15 =−75 cm
Step 5: Answer
The object distance from the mirror is −75 cm.
Question 27
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 produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance, do=−15 cm (negative since the object is in front of the
mirror)
Focal length, f=−10 cm (negative for a concave mirror)
Mirror equation: 1
f=1
di
+1
do
Step 2: Substitute the values into the mirror equation to find the image
distance.
1
−10 =1
di
+1
−15
−1
10 =1
di
−1
15
1
di
=1
15 −1
10
1
di
=2−3
30
1
di
=−1
30
di=−30 cm
26
Step 3: Calculate the magnification using the magnification equation, m=
−di
do
.
m=−
−30
−15
m= 2
Thus, the image distance is -30 cm and the magnification is 2.
Question 28
Question
A concave mirror has a focal length of 20 cm. An object is placed 30 cm in
front of the mirror. Calculate the position of the image formed by the mirror.
Solution
Step 1: Identify the given information: The given focal length of the concave
mirror is f=−20 cm (note the negative sign for concave mirrors). The object
distance dois 30 cm.
Step 2: Calculate the image distance using the mirror equation: The mirror
equation relates the object distance (do), image distance (di), and focal length
(f) of the mirror: 1
f=1
di
+1
do
Substitute the given values into the equation:
1
−20 =1
di
+1
30
Simplify:
−1
20 =1
di
+1
30
Step 3: Solve for di:
−1
20 =1
di
+1
30
Find a common denominator:
−3
60 =2
60di
+2
60
Combine the fractions:
−3
60 =2+2di
60
Solve for di:
2+2di=−3
27
2di=−5
di=−5
2=−2.5 cm
Step 4: Interpret the result: The negative sign indicates that the image is
formed on the same side as the object (in this case, between the mirror and the
object). The image is located 2.5 cm in front of the mirror.
Question 29
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 the given values. The object distance (u) is 20 cm and the focal
length (f) is -15 cm (since the mirror is concave).
Step 2: Apply the mirror formula. The mirror formula relates the object
distance (u), image distance (v), and focal length (f) of a mirror:
1
f=1
v+1
u
Substitute the given values:
1
−15 =1
v+1
20
Step 3: Solve for the image distance.
1
−15 =1
v+1
20
−1
15 =1
v+1
20
−1
v=−1
15 −1
20
−1
v=−4
60 −3
60
−1
v=−7
60
v=−60
7cm
Step 4: Analyze the result. The negative sign indicates that the image is
formed on the same side as the object, which means the image is virtual. The
image distance is −60
7cm, which is approximately -8.57 cm. Thus, the image is
formed at 8.57 cm from the mirror on the same side as the object.
28
Question 30
Question
An object is placed 12 cm in front of a concave mirror with a focal length of 8
cm. Determine the image distance and describe the nature of the image.
Solution
Step 1: Identify the given values. The object distance, do, is 12 cm (positive
since it is in front of the mirror) and the focal length, f, is 8 cm.
Step 2: Apply the mirror equation to find the image distance, di. The mirror
equation relates the object distance, image distance, and focal length:
1
f=1
di
+1
do
Substitute f= 8 cm and do= 12 cm into the equation:
1
8=1
di
+1
12
Step 3: Solve for the image distance.
1
di
=1
8−1
12 =3−2
24 =1
24
di= 24 cm
Step 4: Determine the nature of the image. Since the image distance is
positive, the image is real and formed on the same side as the object (in front
of the mirror). Given that di>2f(24 cm ¿ 16 cm), the image is beyond the
focal point and is therefore real, inverted, and diminished.
Therefore, the image distance is 24 cm and the nature of the image is real,
inverted, and diminished.
Question 31
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 magnification.
Solution
Step 1: Identify the given values. The object distance, o, is 20 cm. The focal
length of the concave mirror, f, is 30 cm.
29
Step 2: Apply the mirror equation to find the image distance. The mirror
equation is given by: 1
f=1
o+1
i
Substitute the values of fand ointo the mirror equation:
1
30 =1
20 +1
i
Solve for i:1
i=1
30 −1
20
1
i=2
60 −3
60
1
i=−1
60
i=−60 cm
Step 3: Calculate the magnification using the magnification formula:
m=−i
o
Substitute the values of iand ointo the magnification formula:
m=−
−60
20
m= 3
Therefore, the image distance is −60 cm and the magnification is 3.
Question 32
Question
A converging lens with a focal length of 20 cm is placed 40 cm away from an
object. Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values.
Focal length of the lens, f= 20 cm (given as 20 cm)
Object distance, do=−40 cm (negative to indicate it is in front of the
lens)
30
Step 2: Apply the lens equation to find the image distance di. The lens
equation is given by: 1
f=1
di
+1
do
Substitute the given values:
1
20 =1
di
+1
−40
Solve for di:
di=−40
3cm
Step 3: Calculate the magnification musing the formula:
m=−di
do
Substitute the values of diand do:
m=−
−40/3
−40 =1
3
Therefore, the image distance is −40/3 cm and the magnification is 1/3.
Question 33
Question
A concave mirror with a focal length of 20 cm forms an image which is six times
the size of the object. Calculate the object distance from the mirror.
Solution
Step 1: Recall the mirror equation:
1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Step 2: Given that the magnification (M) is -6, we have:
M=−di
do
Substitute −6 for Mand solve for di:
−6 = −di
do
=⇒di= 6do
31
Step 3: Recall that for a concave mirror, the focal length is negative. Thus,
f=−20 cm.
Step 4: Substitute the given values into the mirror equation to solve for do:
1
−20 =1
do
+1
6do
Step 5: Simplify the equation:
−1
20 =6
6do
−1
do
Step 6: Combine the fractions on the right-hand side of the equation:
−1
20 =5
6do
Step 7: Solve for do:
do=5×20
6=100
6= 16.6 cm
Thus, the object distance from the mirror is approximately 16.67 cm.
Question 34
Question
A concave mirror with a focal length of 15 cm forms a real image 30 cm from
the mirror. An object is placed 45 cm in front of the mirror. Determine the
magnification of the image.
Solution
Step 1: Identify the given values and the desired quantity to find. Step 2: Recall
the mirror equation for concave mirrors: 1
f=1
do+1
di, and the magnification
equation: m=−di
do. Step 3: Substitute the given values into the mirror equation
to find di. Step 4: Substitute the calculated diand dointo the magnification
equation to find the magnification of the image. Step 5: Analyze the sign of the
magnification to determine the nature of the image (whether it is inverted or
upright).
Step 1: Given: Focal length, f=−15 cm (negative sign for concave mirror)
Image distance, di=−30 cm Object distance, do=−45 cm
We need to find the magnification of the image, m.
Step 2: The mirror equation for concave mirrors is:
−1
f=1
do
+1
di
32
The magnification equation is:
m=−di
do
Step 3: Substitute the given values into the mirror equation:
−1
−15 =1
−45 +1
−30
0.067 = 0.022 + 0.033
0.067 = 0.055
Step 4: Now, substitute the calculated diand dointo the magnification
equation:
m=−
−30
−45
m=−2
3
m= 0.67
Step 5: Since the magnification is positive, the image is upright.
Therefore, the magnification of the image is 0.67, and the image is upright.
Question 35
Question
An object is placed 20 cm in front of a convex lens with a focal length of 15 cm.
Calculate the position and magnification of the image formed by the lens.
Solution
Step 1: Identify the given values and known quantities. Given:
Object distance u=−20 cm (negative sign indicates the object is placed
in front of the lens)
Focal length f= 15 cm
Step 2: Use the lens equation to find the image distance. The lens equation
is given by: 1
f=1
v+1
u
Substitute the given values:
1
15 =1
v+1
−20
33
Question 2
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 produced by the lens.
Solution
Step 1: Identify the given values.
Object distance, do=−20 cm (negative because it is in front of the lens).
Focal length of the lens, f= 15 cm.
Step 2: Use the lens equation to calculate the image distance, di. The lens
equation is given by: 1
f=1
do
+1
di
Plugging in the given values:
1
15 =1
−20 +1
di
1
di
=1
15 −1
−20
1
di
=4
60 −3
60 =1
60
di= 60 cm
Therefore, the image distance is 60 cm.
Step 3: Calculate the magnification, m, using the formula:
m=−di
do
Plugging in the values:
m=−60
−20 = 3
The magnification produced by the lens is 3.
Question 3
Question
An object is placed 20 cm from a convex lens with a focal length of 10 cm.
Calculate the image distance and magnification produced by the lens.
2
Solution
Step 1: Identify the given values and the focal length.
Given: Object distance, do=−20 cm Focal length, f= 10 cm
Step 2: Use the lens equation to find the image distance.
The lens equation is given by:
1
f=1
do
+1
di
Substitute the given values:
1
10 =1
−20 +1
di
Solving for di:1
di
=1
10 −1
−20
1
di
=2
20 +1
20
1
di
=3
20
di=20
3cm ≈6.67 cm
Therefore, the image distance is approximately 6.67 cm.
Step 3: Calculate the magnification using the magnification formula.
The magnification of a lens is given by:
m=−di
do
Substitute the values of diand do:
m=−6.67
−20
m≈0.333
Therefore, the magnification produced by the lens is approximately 0.333.
Question 4
Question
An object is placed 20 cm in front of a concave mirror of focal length 15 cm.
Determine the position, nature, and size of the image formed.
3
Solution
Step 1: Given that: The object distance, u=−20 cm (negative since the object
is in front of the mirror). The focal length, f=−15 cm (negative since the
mirror is concave).
Step 2: Applying the mirror equation:
1
f=1
v+1
u
1
−15 =1
v+1
−20
−1
15 =1
v−1
20
−1
15 +1
20 =1
v
4
60 +3
60 =1
v
7
60 =1
v
v=60
7cm
Step 3: Calculating the magnification, M:
M=−v
u
M=−60/7
−20
M=60
140 =3
7
Since M < 0, the image is inverted.
Step 4: Determining the nature and position of the image based on the
magnification: If M= 3/7, the image is diminished. Additionally, since the
magnification is positive, the image is virtual. Therefore, the image is virtual,
erect, diminished, and formed 60
7cm behind the mirror.
Question 5
Question
An object is placed 30 cm in front of a concave mirror with a focal length of 20
cm. Determine the image distance, image height, magnification, and whether
the image is real or virtual.
4
Solution
Step 1: Identify the given values. The object distance dois 30 cm, and the
focal length fis -20 cm for a concave mirror (since the focal length of a concave
mirror is negative).
Step 2: Apply the mirror equation to find the image distance. The mirror
equation is given by: 1
f=1
do
+1
di
Substitute the given values:
1
−20 =1
30 +1
di
−1
20 −1
30 =1
di
−1
12 =1
di
Therefore, di=−12 cm.
Step 3: Calculate the magnification. The magnification Mis given by:
M=−di
do
Substitute the values:
M=−
−12
30
M=2
5
Step 4: Determine if the image is real or virtual. Since diis negative, the
image is formed on the same side as the object, which makes it a virtual image.
Step 5: Calculate the image height. The image height hican be determined
using the magnification formula:
M=hi
ho
Since the object height hois not given, we cannot calculate the image height
without additional information.
Therefore, the image distance is -12 cm, the magnification is 2/5, and the im-
age formed is virtual. The image height cannot be determined without knowing
the object height.
Question 6
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Determine the image distance, image height, and magnification.
5
Solution
Step 1: Use the lens equation to find the image distance.
1
f=1
do
+1
di
Given f= 15 cm and do=−20 cm (since it is in front of the lens), we can plug
these values in and solve for di.
1
15 =1
−20 +1
di
1
di
=1
15 +1
20
1
di
=4
60 +3
60
1
di
=7
60
di=60
7≈8.57 cm
Step 2: Calculate the magnification using the formula m=−di
do.
m=−8.57
−20
m≈0.429
Step 3: Find the image height using the magnification formula m=hi
ho.
Given that the object height ho= 2 cm, we can find the image height hi.
0.429 = hi
2
hi= 0.429 ×2
hi≈0.857 cm
Therefore, the image distance is approximately 8.57 cm, the magnification
is 0.429, and the image height is approximately 0.857 cm.
Question 7
Question
A converging lens with a focal length of 15 cm is placed 30 cm to the left of a
diverging lens with a focal length of -20 cm. If an object is placed 40 cm to the
left of the converging lens, determine the location and magnification of the final
image formed by the system.
6
Solution
Step 1: Find the image created by the converging lens. The lens equation for
the converging lens is 1
f=1
do+1
di, where fis the focal length, dois the object
distance, and diis the image distance. Given f= 15 cm and do=−40 cm (since
the object distance is measured from the left of the lens and the object is placed
to the left), we can solve for di:
1
15 =1
−40 +1
di
1
di
=1
15 −1
−40
1
di
=8
120
di= 15 cm
Therefore, the image formed by the converging lens is located 15 cm to the
right of the lens.
Step 2: Find the object distance for the diverging lens based on the image
created by the converging lens. Since the diverging lens is 30 cm to the right of
the converging lens, the object distance for the diverging lens is 15 cm−30 cm =
−15 cm.
Step 3: Find the image created by the diverging lens. Using the lens equation
for the diverging lens, which is 1
f=1
do+1
di, where f=−20 cm and do=−15 cm,
we can solve for di:1
−20 =1
−15 +1
di
1
di
=1
−20 −1
−15
1
di
=−3
60
di=−20 cm
Thus, the final image formed by the system is located 20 cm to the left of
the diverging lens.
Step 4: Find the magnification of the final image. The magnification Mis
given by di
do, where di=−20 cm and do=−15 cm.
M=−20
−15 =4
3
Therefore, the final image is formed 20 cm to the left of the diverging lens
with a magnification of 4
3.
7
Question 8
Question
A concave mirror has a focal length of 10 cm. An object is placed 15 cm from
the mirror. Determine the position and magnification of the image formed by
the mirror.
Solution
Step 1: Identify the given quantities and the mirror equation. The mirror
equation relating the object distance do, image distance di, and focal length f
for a mirror is given by: 1
f=1
do
+1
di
Given: f=−10 cm (concave mirror), do=−15 cm (object distance).
Step 2: Solve for the image distance di. Substitute the given values into the
mirror equation: 1
−10 =1
−15 +1
di
Solving for di:1
di
=1
−10 −1
−15 =3
30 −2
30 =1
30
di= 30 cm
Step 3: Calculate the magnification. The magnification Mis given by:
M=−di
do
Substitute the values of diand dointo the equation:
M=−30
−15 = 2
Step 4: Interpret the results. The image is formed 30 cm from the mirror
on the same side as the object. The magnification of the image is 2, indicating
that the image is upright and enlarged.
Question 9
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.
8
Solution
Step 1: Identify known values and the mirror equation.
Object distance, do=−20 cm (negative since it is in front of the mirror)
Focal length, f=−15 cm (negative for concave mirror)
Mirror equation: 1
f=1
do+1
di
Step 2: Solve for the image distance, di.
1
−15 =1
−20 +1
di
−2
30 =−3
60 +1
di
−2
30 +3
60 =1
di
1
di
=4
60
di=60
4
di= 15 cm
So, the image distance is 15 cm.
Step 3: Calculate the magnification, M.
M=−di
do
M=−15
−20
M= 0.75
Therefore, the image distance is 15 cm and the magnification is 0.75.
Question 10
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: Identify the given values: The object distance, do, is 20 cm (positive
because it is in front of the mirror) and the focal length, f, is -10 cm (negative
for a concave mirror).
9
Step 2: Apply the mirror equation: The mirror equation is given by:
1
f=1
do
+1
di
Substitute the values into the equation:
1
−10 =1
20 +1
di
Step 3: Solve for the image distance, di:
−1
10 =1
20 +1
di
−1
di
=−1
10 −1
20
−1
di
=−3
20
di=−20
3cm
Step 4: Analyze the image distance: The negative value of the image distance
indicates that the image is formed behind the mirror, which means it is a real
image.
Step 5: Determine the position of the image: Since the image distance is
negative, the image is located 6.67 cm behind the mirror.
Step 6: Determine the nature of the image: Since the image is real and
formed on the same side as the object, it is a real and inverted image.
Question 11
Question
A converging lens with a focal length of 15 cm is placed 25 cm in front of
a diverging mirror with a focal length of -10 cm. Determine the final image
distance produced by the combination of the lens and the mirror. Is the final
image real or virtual?
Solution
Step 1: Calculate the image distance produced by the converging lens using the
lens equation: 1
flens
=1
do
+1
di
where flens = 15 cm and do=−25 cm (since the object is placed 25 cm in front
of the lens).
10
Substitute the given values into the equation and solve for di:
1
15 =1
−25 +1
di
1
di
=1
15 +1
25
1
di
=5+3
75
1
di
=8
75
di=75
8cm
So, the image distance produced by the lens is di= 9.375 cm.
Step 2: Use the diverging mirror equation to calculate the final image dis-
tance: 1
fmirror
=1
di
+1
d′
o
where fmirror =−10 cm, di= 9.375 cm, and we want to find d′
o.
Substitute the given values into the equation and solve for d′
o:
1
−10 =1
9.375 +1
d′
o
−1
10 =8
75 +1
d′
o
−3
30 −8
75 =1
d′
o
−9
75 =1
d′
o
d′
o=−75
9cm
Therefore, the final image distance produced by the combination of the lens
and the mirror is d′
o=−8.333 cm. Since the image distance is negative, the
final image is virtual.
Question 12
Question
An object is placed 20 cm in front of a concave mirror of focal length 15 cm.
Determine the position and nature of the image formed.
11
Solution
Step 1: Identify the given values The given values are: Object distance, u=
−20 cm (negative sign indicates that the object is located in front of the mirror)
Focal length, f=−15 cm (negative sign for a concave mirror)
Step 2: Apply the mirror formula The mirror formula relates the object
distance (u), image distance (v), and focal length (f) as:
1
f=1
v+1
u
Substitute f=−15 cm and u=−20 cm into the mirror formula:
1
−15 =1
v+1
−20
Step 3: Solve for v
1
−15 =1
v−1
20 =⇒1
v=1
−15 +1
20
=⇒1
v=−4+3
60 =⇒1
v=−1
60
=⇒v=−60 cm
Therefore, the image is located 60 cm behind the mirror.
Step 4: Determine the nature of the image Since the image distance is nega-
tive, the image is formed on the same side as the object (in front of the mirror).
This indicates that the image is virtual and upright.
Thus, the image formed by the concave mirror is virtual and located 60 cm
in front of the mirror.
Question 13
Question
An object is placed 20 cm in front of a concave mirror with a focal length of
10 cm. Calculate the position and magnification of the image formed by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation. Given values: -
Object distance, do=−20 cm - Focal length, f=−10 cm (note the negative
sign for concave mirror)
The mirror equation is: 1
f=1
do
+1
di
12
Step 2: Substitute the known values into the mirror equation and solve for
image distance. Substitute f=−10 cm and do=−20 cm into the mirror
equation: 1
−10 =1
−20 +1
di
Solving for di, we get:
1
di
=1
−10 −1
−20 =1
−10 +1
20 =2−1
20 =1
20
So, di= 20 cm.
Step 3: Calculate the magnification. The magnification, m, in terms of
object and image distances is given by:
m=−di
do
Substitute di= 20 cm and do=−20 cm into the magnification formula:
m=−20
−20 = 1
Step 4: Determine the position of the image. The negative sign indicates
that the image is in front of the mirror, which means the image is formed on
the same side as the object. Therefore, the image is located 20 cm in front of
the mirror.
Step 5: Conclusion The image formed by the concave mirror is located 20
cm in front of the mirror and the magnification is 1.
Question 14
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 of the image formed by
the lens.
Solution
Step 1: Identify the given values and the lens equation. Step 2: Apply the lens
equation to find the image distance. Step 3: Determine the magnification of the
image.
Step 1: Given values: Object distance, do=−30 cm (since the object is
placed in front of the lens) Focal length, f= 20 cm
The lens equation is: 1
f=1
di
+1
do
13
Step 2: Apply the lens equation to find the image distance:
1
20 =1
di
+1
−30
1
di
=1
20 +1
30
1
di
=3
60 +2
60
1
di
=5
60
di=60
5
di= 12 cm
Therefore, the image distance, di, is 12 cm.
Step 3: Determine the magnification of the image: The magnification, m,
is given by:
m=−di
do
Substitute di= 12 cm and do=−30 cm into the formula:
m=−12
−30 =2
5
Therefore, the magnification of the image is 2
5.
Question 15
Question
A concave mirror has a focal length of 15 cm. An object is placed 25 cm in
front of the mirror. Calculate the image distance and magnification produced
by the mirror.
Solution
Step 1: Recall the mirror equation for concave mirrors:
1
f=1
do
+1
di
where: - fis the focal length of the mirror, - dois the object distance, and - di
is the image distance.
14
Step 2: Given that f=−15 cm and do=−25 cm (since the object is in
front of the mirror), we can substitute these values into the mirror equation to
find di:1
−15 =1
−25 +1
di
Step 3: Simplify the equation:
−2
30 =−1
25 +1
di
Step 4: Find a common denominator:
−2
30 =−1·30
25 ·30 +1
di
Step 5: Simplify further:
−2
30 =−30
750 +1
di
Step 6: Add the fractions:
−2
30 =−30
750 +1
di
−2
30 =−30
750 +1
di
−2
30 =−1
25 +1
di
Step 7: Find the common denominator:
−50
750 =−30
750 +1
di
Step 8: Simplify the equation:
−50
750 =−30
750 +1
di
−20
750 =1
di
Step 9: Solve for di:
−20di= 750
di=750
−20
di=−37.5 cm
Step 10: The image distance, di, is -37.5 cm.
15
Step 11: The magnification, M, is given by:
M=−di
do
Step 12: Substitute the values of diand dointo the magnification formula:
M=−37.5
−25
M= 1.5
Step 13: The magnification produced by the mirror is 1.5.
Question 16
Question
An object is placed 20 cm in front of a concave mirror of focal length 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 sign indicates it is in front of
the mirror) - Focal length f=−15 cm (negative sign indicates it is a concave
mirror)
The mirror equation relates the object distance, image distance, and focal
length:
−1
di
+1
do
=1
f
Step 2: Substitute the values into the mirror equation and solve for the im-
age distance.
−1
di
+1
−20 =1
−15
−1
di
−1
20 =−1
15
−1
di
=−1
15 +1
20
−1
di
=−4
60 +3
60
−1
di
=−1
60
di=−60 cm
16
Therefore, the image distance is di=−60 cm. The negative sign indicates
that the image is formed behind the mirror.
Step 3: Calculate the magnification using the magnification formula.
The magnification Mis given by:
M=−di
do
Substitute di=−60 cm and do=−20 cm into the formula:
M=−
−60
−20 =60
20 = 3
Therefore, the magnification is 3, indicating that the image is magnified and
upright.
Question 17
Question
A concave mirror has a focal length of -15 cm. An object is placed 20 cm from
the mirror. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and mirror equation.
Given: f=−15 cm, do= 20 cm
Mirror equation: 1
f=1
di+1
do
Step 2: Solve for diusing the mirror equation.
1
−15 =1
di
+1
20
−1
15 =1
di
+1
20
−1
15 −1
20 =1
di
−4
60 −3
60 =1
di
−7
60 =1
di
di=60
−7
di=−8.57 cm
17
Step 3: Calculate the magnification, m, using the formula m=−di
do.
m=−
−8.57
20
m= 0.43
Therefore, the image distance is −8.57 cm and the magnification is 0.43.
Question 18
Question
An object is placed 10 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 known quantities and the mirror equation.
Object distance, do=−10 cm (negative because it is in front of the mirror)
Focal length, f=−15 cm (negative for a concave mirror)
Mirror equation: 1
f=1
do
+1
di
Step 2: Substitute the known values into the mirror equation to find the
image distance.
1
−15 =1
−10 +1
di
−1
15 =−1
10 +1
di
1
di
=−1
15 +1
10
1
di
=−2
30 +3
30
1
di
=1
30
di= 30 cm
Therefore, the image distance is 30 cm.
Step 3: Calculate the magnification using the magnification equation.
M=−di
do
M=−30
−10
M= 3
18
The magnification is 3.
Step 4: Determine if the image is real or virtual. Since the magnification is
positive, the image is upright. Therefore, the image formed is virtual.
In conclusion, the image distance is 30 cm, the magnification is 3, and the
image is virtual.
Question 19
Question
A thin converging lens has a focal length of 15 cm. An object is placed 30 cm
from the lens. Determine the position and size of the final image formed by the
lens.
Solution
Step 1: Use the lens equation to find the image distance.
1
f=1
do
+1
di
Given that the focal length f= 15 cm, and the object distance do= 30 cm.
Substitute these values into the lens equation to solve for di.
Step 2: Solve for di.1
15 =1
30 +1
di
1
di
=1
15 −1
30
1
di
=2
30 −1
30
1
di
=1
30
di= 30 cm
Step 3: Calculate the magnification (m) using the formula m=−di
do.
m=−30
30
m=−1
Step 4: Determine the size of the image. Since the magnification is -1, the
image is the same size as the object but inverted. Thus, the size of the image
is equal to the size of the object.
Step 5: Write the final answer. The final image is located 30 cm from the
lens on the same side as the object, and it is the same size as the object but
inverted.
19
Question 20
Question
A thin convex lens has a focal length of 10 cm. An object placed 20 cm from
the lens forms a real image. Determine the image distance and magnification
produced by the lens.
Solution
Step 1: Identify the given values and the lens formula.
Object distance, u=−20 cm (since the object is placed to the left of the
lens)
Focal length, f= 10 cm (for a convex lens, fis positive)
Using the lens formula: 1
f=1
v+1
u
Step 2: Substitute the values into the lens formula and solve for the image
distance, v.
1
10 =1
v+1
−20
1
10 =1
v−1
20
1
v=1
10 +1
20 =3
20
v=20
3≈6.67 cm
Step 3: Calculate the magnification, m, using the formula m=−v
u.
m=−6.67
−20 =1
3
Therefore, the image distance produced by the lens is approximately 6.67
cm (real image), and the magnification is 1
3.
Question 21
Question
An object is placed 25 cm in front of a converging lens with a focal length of 15
cm. Calculate the position and size of the image formed.
20
Solution
Step 1: Identify the given values and the lens formula. Given:
u=−25 cm
f= 15 cm
Using the lens formula: 1
f=1
v+1
u
Step 2: Calculate the position of the image. Substitute the given values into
the lens formula: 1
15 =1
v+1
−25
1
v=1
15 +1
25
1
v=5+3
75
1
v=8
75
v=75
8
v= 9.375 cm
Step 3: Calculate the size of the image. Use the magnification formula:
M=−v
u
M=−9.375
−25
M= 0.375
Step 4: Interpret the result. The image is formed at a distance of 9.375 cm
from the lens on the same side as the object, and it is smaller than the object
with a magnification of 0.375.
Question 22
Question
An object is placed 20 cm in front of a convex lens of focal length 10 cm.
Calculate the image distance and magnification produced by the lens.
21
Solution
Step 1: Identify the given values.
Object distance (u) = -20 cm (negative sign indicates object is on the
same side as the incident light)
Focal length (f) = 10 cm
Step 2: Use the lens equation to find the image distance.
1
f=1
v+1
u
Plugging in the values: 1
10 =1
v+1
−20
Solving for v:1
v=1
10 +1
20
1
v=3
20
v=20
3
v= 6.67 cm
Therefore, the image distance produced by the lens is 6.67 cm.
Step 3: Calculate the magnification using the magnification formula.
Magnification (M) = −v
u
Plugging in the values:
Magnification (M) = −6.67
−20
Magnification (M) = 0.3333
Therefore, the magnification produced by the lens is 0.3333.
Question 23
Question
An object is placed 30 cm from a converging lens of focal length 20 cm. Calculate
the position of the image formed by the lens.
22
Solution
Given: Object distance, u=−30 cm (because the object is on the side opposite
to the incident light), Focal length of the lens, f= 20 cm.
We will use the lens formula to determine the position of the image:
1
f=1
v+1
u
Step 1: Substitute the given values into the lens formula:
1
20 =1
v+1
−30
Step 2: Simplify the equation:
1
v=1
20 +1
30
Step 3: Find the sum on the right side of the equation:
1
v=3+2
60 =5
60 =1
12
Step 4: Take the reciprocal of both sides to solve for v:
v= 12 cm
Step 5: The positive value of vindicates that the image is formed on the
same side as the incident light, which means the image is virtual and upright,
12 cm away from the lens.
Question 24
Question
A converging lens with a focal length of 15 cm is placed 30 cm away from an
object. At what distance from the lens should the image be formed? Then, a
concave mirror with a focal length of 10 cm is placed 20 cm away from the same
object. At what distance from the mirror should the image be formed?
Solution
Step 1: For the converging lens Given the object distance u=−30 cm (since it
is to the left of the lens), and the focal length f= 15 cm. We can use the lens
formula: 1
f=1
v+1
u
where vis the image distance. Substitute the values into the lens formula:
1
15 =1
v+1
−30
23
1
15 =1
v−1
30
v=30
2= 15 cm
The image is formed 15 cm to the right of the lens.
Step 2: For the concave mirror Given the object distance u=−20 cm and
the focal length f=−10 cm (negative for a concave mirror). We can use the
mirror formula: 1
f=1
v+1
u
where vis the image distance. Substitute the values into the mirror formula:
1
−10 =1
v+1
−20
1
−10 =1
v−1
20
v=−20 cm
The image is formed 20 cm to the left of the mirror.
Question 25
Question
An object is placed 20 cm in front of a concave mirror whose focal length is
30 cm. Determine the image distance, magnification, and whether the image is
real or virtual.
Solution
Step 1: Identify the given values. The object distance dois 20 cm and the focal
length fis -30 cm (since it is a concave mirror).
Step 2: Use the mirror equation to find the image distance. The mirror
equation is given by:
−1
do
+1
di
=1
f
Substitute do= 20 cm and f=−30 cm into the equation:
−1
20 +1
di
=1
−30
Solving for di, we get: 1
di
=1
−30 +1
20
1
di
=−1
60
24
di=−60 cm
Step 3: Calculate the magnification. The magnification is given by the
formula:
m=−di
do
Substitute di=−60 cm and do= 20 cm into the formula:
m=−
−60
20
m= 3
Step 4: Determine if the image is real or virtual. Since the image distance
diis negative, the image is formed on the same side as the object. Therefore,
the image is real.
Therefore, the image distance is -60 cm, the magnification is 3, and the
image is real.
Question 26
Question
A concave mirror with a focal length of 15 cm forms a real image that is 10
times larger than the object. Determine the object distance from the mirror.
Solution
Step 1: Identify the given information.
The focal length of the concave mirror, f=−15 cm (negative because it is a
concave mirror). The magnification of the mirror, m=−10 (negative because
the image is real and inverted).
Step 2: Recall the mirror equation.
The mirror equation relates the object distance (do), image distance (di), and
focal length (f) of a mirror: 1
f=1
do
+1
di
Step 3: Substitute the given values into the mirror equation.
Since m=−di
do, we can rewrite the mirror equation as:
1
f=1
do
+1
−m·do
Substitute f=−15 cm and m=−10 into the equation:
1
−15 =1
do
+1
10 ·do
25
Step 4: Solve for the object distance (do).
Combine the fractions on the right side of the equation:
−1
15 =1+1
10do
−1
15 =2
10do
−1
15 =1
5do
Solve for do:
do=5
−1/15 =−75 cm
Step 5: Answer
The object distance from the mirror is −75 cm.
Question 27
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 produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance, do=−15 cm (negative since the object is in front of the
mirror)
Focal length, f=−10 cm (negative for a concave mirror)
Mirror equation: 1
f=1
di
+1
do
Step 2: Substitute the values into the mirror equation to find the image
distance.
1
−10 =1
di
+1
−15
−1
10 =1
di
−1
15
1
di
=1
15 −1
10
1
di
=2−3
30
1
di
=−1
30
di=−30 cm
26
Step 3: Calculate the magnification using the magnification equation, m=
−di
do
.
m=−
−30
−15
m= 2
Thus, the image distance is -30 cm and the magnification is 2.
Question 28
Question
A concave mirror has a focal length of 20 cm. An object is placed 30 cm in
front of the mirror. Calculate the position of the image formed by the mirror.
Solution
Step 1: Identify the given information: The given focal length of the concave
mirror is f=−20 cm (note the negative sign for concave mirrors). The object
distance dois 30 cm.
Step 2: Calculate the image distance using the mirror equation: The mirror
equation relates the object distance (do), image distance (di), and focal length
(f) of the mirror: 1
f=1
di
+1
do
Substitute the given values into the equation:
1
−20 =1
di
+1
30
Simplify:
−1
20 =1
di
+1
30
Step 3: Solve for di:
−1
20 =1
di
+1
30
Find a common denominator:
−3
60 =2
60di
+2
60
Combine the fractions:
−3
60 =2+2di
60
Solve for di:
2+2di=−3
27
2di=−5
di=−5
2=−2.5 cm
Step 4: Interpret the result: The negative sign indicates that the image is
formed on the same side as the object (in this case, between the mirror and the
object). The image is located 2.5 cm in front of the mirror.
Question 29
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 the given values. The object distance (u) is 20 cm and the focal
length (f) is -15 cm (since the mirror is concave).
Step 2: Apply the mirror formula. The mirror formula relates the object
distance (u), image distance (v), and focal length (f) of a mirror:
1
f=1
v+1
u
Substitute the given values:
1
−15 =1
v+1
20
Step 3: Solve for the image distance.
1
−15 =1
v+1
20
−1
15 =1
v+1
20
−1
v=−1
15 −1
20
−1
v=−4
60 −3
60
−1
v=−7
60
v=−60
7cm
Step 4: Analyze the result. The negative sign indicates that the image is
formed on the same side as the object, which means the image is virtual. The
image distance is −60
7cm, which is approximately -8.57 cm. Thus, the image is
formed at 8.57 cm from the mirror on the same side as the object.
28
Question 30
Question
An object is placed 12 cm in front of a concave mirror with a focal length of 8
cm. Determine the image distance and describe the nature of the image.
Solution
Step 1: Identify the given values. The object distance, do, is 12 cm (positive
since it is in front of the mirror) and the focal length, f, is 8 cm.
Step 2: Apply the mirror equation to find the image distance, di. The mirror
equation relates the object distance, image distance, and focal length:
1
f=1
di
+1
do
Substitute f= 8 cm and do= 12 cm into the equation:
1
8=1
di
+1
12
Step 3: Solve for the image distance.
1
di
=1
8−1
12 =3−2
24 =1
24
di= 24 cm
Step 4: Determine the nature of the image. Since the image distance is
positive, the image is real and formed on the same side as the object (in front
of the mirror). Given that di>2f(24 cm ¿ 16 cm), the image is beyond the
focal point and is therefore real, inverted, and diminished.
Therefore, the image distance is 24 cm and the nature of the image is real,
inverted, and diminished.
Question 31
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 magnification.
Solution
Step 1: Identify the given values. The object distance, o, is 20 cm. The focal
length of the concave mirror, f, is 30 cm.
29
Step 2: Apply the mirror equation to find the image distance. The mirror
equation is given by: 1
f=1
o+1
i
Substitute the values of fand ointo the mirror equation:
1
30 =1
20 +1
i
Solve for i:1
i=1
30 −1
20
1
i=2
60 −3
60
1
i=−1
60
i=−60 cm
Step 3: Calculate the magnification using the magnification formula:
m=−i
o
Substitute the values of iand ointo the magnification formula:
m=−
−60
20
m= 3
Therefore, the image distance is −60 cm and the magnification is 3.
Question 32
Question
A converging lens with a focal length of 20 cm is placed 40 cm away from an
object. Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values.
Focal length of the lens, f= 20 cm (given as 20 cm)
Object distance, do=−40 cm (negative to indicate it is in front of the
lens)
30
Step 2: Apply the lens equation to find the image distance di. The lens
equation is given by: 1
f=1
di
+1
do
Substitute the given values:
1
20 =1
di
+1
−40
Solve for di:
di=−40
3cm
Step 3: Calculate the magnification musing the formula:
m=−di
do
Substitute the values of diand do:
m=−
−40/3
−40 =1
3
Therefore, the image distance is −40/3 cm and the magnification is 1/3.
Question 33
Question
A concave mirror with a focal length of 20 cm forms an image which is six times
the size of the object. Calculate the object distance from the mirror.
Solution
Step 1: Recall the mirror equation:
1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Step 2: Given that the magnification (M) is -6, we have:
M=−di
do
Substitute −6 for Mand solve for di:
−6 = −di
do
=⇒di= 6do
31
Step 3: Recall that for a concave mirror, the focal length is negative. Thus,
f=−20 cm.
Step 4: Substitute the given values into the mirror equation to solve for do:
1
−20 =1
do
+1
6do
Step 5: Simplify the equation:
−1
20 =6
6do
−1
do
Step 6: Combine the fractions on the right-hand side of the equation:
−1
20 =5
6do
Step 7: Solve for do:
do=5×20
6=100
6= 16.6 cm
Thus, the object distance from the mirror is approximately 16.67 cm.
Question 34
Question
A concave mirror with a focal length of 15 cm forms a real image 30 cm from
the mirror. An object is placed 45 cm in front of the mirror. Determine the
magnification of the image.
Solution
Step 1: Identify the given values and the desired quantity to find. Step 2: Recall
the mirror equation for concave mirrors: 1
f=1
do+1
di, and the magnification
equation: m=−di
do. Step 3: Substitute the given values into the mirror equation
to find di. Step 4: Substitute the calculated diand dointo the magnification
equation to find the magnification of the image. Step 5: Analyze the sign of the
magnification to determine the nature of the image (whether it is inverted or
upright).
Step 1: Given: Focal length, f=−15 cm (negative sign for concave mirror)
Image distance, di=−30 cm Object distance, do=−45 cm
We need to find the magnification of the image, m.
Step 2: The mirror equation for concave mirrors is:
−1
f=1
do
+1
di
32
The magnification equation is:
m=−di
do
Step 3: Substitute the given values into the mirror equation:
−1
−15 =1
−45 +1
−30
0.067 = 0.022 + 0.033
0.067 = 0.055
Step 4: Now, substitute the calculated diand dointo the magnification
equation:
m=−
−30
−45
m=−2
3
m= 0.67
Step 5: Since the magnification is positive, the image is upright.
Therefore, the magnification of the image is 0.67, and the image is upright.
Question 35
Question
An object is placed 20 cm in front of a convex lens with a focal length of 15 cm.
Calculate the position and magnification of the image formed by the lens.
Solution
Step 1: Identify the given values and known quantities. Given:
Object distance u=−20 cm (negative sign indicates the object is placed
in front of the lens)
Focal length f= 15 cm
Step 2: Use the lens equation to find the image distance. The lens equation
is given by: 1
f=1
v+1
u
Substitute the given values:
1
15 =1
v+1
−20
33
Solve for v:1
v=1
15 −1
20
1
v=4
60 −3
60 =1
60
v= 60 cm
Step 3: Calculate the magnification. The magnification of the lens is given
by:
m=−v
u
Substitute the values of uand vto find the magnification m:
m=−60
−20 = 3
Step 4: Interpretation of the results. The image is formed at a distance
of 60 cm from the lens on the same side as the object (virtual image). The
magnification of the image is 3, indicating that the image is enlarged three
times compared to the object.
34
