PHYS 101 - ELEMENTS OF PHYSICS
- Lens and mirror equations
Question Bank - Set 4
Liberty University
Question 1
Question
A concave mirror has a focal length of 15 cm. An object is placed 25 cm from
the mirror. Determine the image distance and magnification produced by the
mirror.
Solution
Step 1: Given the object distance do=−25 cm (since it is in front of the mirror)
and the focal length f=−15 cm (since it is a concave mirror), we can use the
mirror equation to find the image distance di:
1
f=1
di
+1
do
Step 2: Substitute the given values into the mirror equation:
1
−15 =1
di
+1
−25
Step 3: Simplify the equation to solve for di:
−1
15 =1
di
−1
25
Step 4: Find a common denominator:
−25
375 =25
di
−15
375
Step 5: Combine the fractions:
−10
375 =25
di
Step 6: Solve for di:
di=375
−10 =−37.5 cm
Therefore, the image distance diis -37.5 cm.
Step 7: To find the magnification m, we use the formula:
m=−di
do
Step 8: Substitute the calculated values of diand do:
m=−
−37.5
−25 = 1.5
Therefore, the magnification mis 1.5.
Question 2
Question
A converging lens with a focal length of 15 cm is placed 30 cm to the left of
a diverging mirror. The combined optical system forms a final image that is
virtual and located 24 cm to the left of the lens. Determine the focal length of
the mirror.
Solution
Let flens be the focal length of the lens and fmirror be the focal length of the
mirror.
Step 1: Identify the given values and the unknown.
flens = 15 cm
Object distance for the lens, dobj,lens = 30 cm
Image distance from the lens, dimg,lens =−24 cm (negative for virtual
image on the same side as the object)
Image distance for the lens-mirror system, dimg,system =−30 cm (sum of
distances from the lens to the mirror and from the mirror to the final
image)
2
Step 2: Find the image distance for the mirror from the lens equation. The
lens equation relates the object distance, image distance, and focal length for a
lens: 1
flens
=1
dobj,lens
+1
dimg,lens
Substitute the given values:
1
15 =1
30 +1
−24
Solve for dimg,lens:
dimg,lens =−20 cm
Step 3: Calculate the image distance for the mirror. The image distance
for the lens-mirror system is the object distance for the mirror:
dobj,mirror =−dimg,lens = 20 cm
Step 4: Use the mirror equation to find the focal length of the mirror. The
mirror equation relates the object distance, image distance, and focal length for
a mirror: 1
fmirror
=1
dobj,mirror
+1
dimg,system
Substitute the known values:
1
fmirror
=1
20 +1
−30
Solve for fmirror:
1
fmirror
=1
20 −1
30 =3
60 −2
60 =1
60
fmirror = 60 cm
Question 3
Question
A concave mirror has a focal length of 15 cm. An object is placed 25 cm in
front of the mirror. Determine the image distance and the magnification of the
image.
3
Solution
Step 1: Given that f=−15 cm (focal length of the mirror) and do=−25
cm (object distance), we need to find the image distance diusing the mirror
equation:
1
f=1
di
+1
do
Plugging in the known values:
1
−15 =1
di
+1
−25
−1
15 =1
di
−1
25
−1
15 +1
25 =1
di
5
75 +3
75 =1
di
8
75 =1
di
di=75
8≈9.375 cm
Therefore, the image distance di≈9.375 cm.
Step 2: To find the magnification M, we use the formula:
M=−di
do
Plugging in the calculated values:
M=−9.375
−25 =9.375
25 = 0.375
Therefore, the magnification of the image is M= 0.375.
Question 4
Question
A concave mirror has a focal length of 10 cm. An object is placed 20 cm in
front of the mirror. Determine the position and size of the image formed by the
mirror.
4
Solution
Step 1: Identify the given values and mirror equation. Given: f=−10 cm,
do=−20 cm, and diand hiare to be found. The mirror equation is 1
f=1
do+1
di.
Step 2: Substitute the given values into the mirror equation. Substitute
f=−10 cm and do=−20 cm into the mirror equation: 1
−10 =1
−20 +1
di.
Step 3: Solve for di. Solving the equation, we get: 1
−10 =1
−20 +1
di
−1
10 =
−1
20 +1
di
−1
10 +1
20 =1
di
1
20 =1
didi= 20 cm
Step 4: Determine the size of the image using the magnification equation.
The magnification equation is M=−di
do=hi
ho. We are given di= 20 cm and
do=−20 cm. Substitute di= 20 cm and do=−20 cm into the magnification
equation: M=−20
−20 =hi
hoM=1= hi
ho
Step 5: Calculate the height of the image. Since M= 1 = hi
ho, the height of
the image is equal to the height of the object. Therefore, the size of the image
is the same as the size of the object.
Step 6: State the position and size of the image. The position of the image
is di= 20 cm in front of the mirror, and the size of the image is the same as
the size of the object.
Question 5
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, the magnification, and describe
the image formed by the mirror.
Solution
Step 1: Given that the focal length of the concave mirror is f=−15 cm, and
the object distance do=−30 cm. Step 2: We can use the mirror equation
1
f=1
do+1
dito find the image distance di. Step 3: Substituting the given values
into the mirror equation, we get 1
−15 =1
−30 +1
di. Step 4: Solving for di, we have
1
di=1
−15 −1
−30 . Step 5: Simplifying the equation, we find 1
di=−2
30 . Step 6:
Therefore, the image distance is di=−15 cm. Step 7: To find the magnification
M, we use the formula M=−di
do. Step 8: Substituting the values, we get
M=−
−15
−30 =1
2. Step 9: The positive magnification value indicates that the
image is upright compared to the object. Step 10: Since the magnification is
less than 1, the image is smaller than the object. Step 11: Therefore, the image
formed by the concave mirror is virtual, upright, and smaller than the object.
5
Question 6
Question
An object is placed 40 cm from a converging lens with a focal length of 20 cm.
Determine the image distance, magnification, and whether the image is real or
virtual.
Solution
Step 1: Identify the given values and the lens equation. Given: - Object distance
p=−40 cm - Focal length f= 20 cm The lens equation is given by:
1
f=1
p+1
q
Step 2: Substitute the given values into the lens equation to solve for the
image distance. Substitute p=−40 cm and f= 20 cm into the lens equation:
1
20 =1
−40 +1
q
Step 3: Solve for q.1
q=1
20 −1
−40 =3
40
q=40
3cm ≈13.33 cm
Step 4: Calculate the magnification. The magnification is given by:
m=−q
p
Substitute p=−40 cm and q=40
3cm into the magnification formula:
m=−
40
3
−40 =1
3
Step 5: Determine if the image is real or virtual. Since the magnification is
positive, the image is upright (virtual).
Therefore, the image distance is approximately 13.33 cm, the magnification
is 1/3, and the image is virtual.
Question 7
Question
An object is placed 10 cm in front of a concave lens of focal length 20 cm.
Determine the image distance and characterize the image in terms of size and
orientation.
6
Solution
Step 1: Identify the given values and the type of lens.
Object distance p=−10 cm (since the object is located in front of the
concave lens, the distance is negative).
Focal length f=−20 cm (since a concave lens has a negative focal length).
Since we are dealing with a concave lens, fis negative and will be used
as such in the lens formula 1
f=1
p+1
q.
Step 2: Calculate the image distance qusing the lens formula.
1
f=1
p+1
q
1
−20 =1
−10 +1
q
−1
20 =−1
10 +1
q
1
q=−1
20 +1
10
1
q=−1
20 +2
20
1
q=1
20
q= 20 cm
Step 3: Characterize the image in terms of size and orientation. Since the
image distance qis positive, the image is formed on the same side of the lens
as the object (i.e., the image is virtual). Since the magnification M=q
p, and
q > p, the image will be magnified. Therefore, the image formed by the concave
lens is virtual and magnified.
Question 8
Question
An object is placed 20 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 values and the mirror equation.
Object distance, do=−20 cm (negative because the object is in front of
the mirror)
7
Focal length, f=−15 cm (negative because the mirror is concave)
Mirror equation: 1
f=1
do+1
di
Step 2: Substitute the given values into the mirror equation and solve for
the image distance, di.
1
−15 =1
−20 +1
di
−1
15 =−1
20 +1
di
−1
15 +1
20 =1
di
4
60 −3
60 =1
di
1
60 =1
di
di= 60 cm
Therefore, the image distance is 60 cm.
Step 3: Calculate the magnification, M.
M=−di
do
=−60 cm
−20 cm
= 3
Therefore, the magnification produced by the mirror is 3.
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 the magnification produced by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance p=−20 cm (negative because the object is in front of
the mirror).
Focal length f=−15 cm (negative for a concave mirror).
Mirror equation: 1
f=1
p+1
q.
8
Step 2: Substitute the known values into the mirror equation and solve for
the image distance q.
1
−15 =1
−20 +1
q
−1
15 =−1
20 +1
q
1
q=−1
15 +1
20
1
q=−4+3
60
1
q=−1
60
q=−60 cm
Step 3: Calculate the magnification musing the formula m=−q
p.
m=−
−60
−20
m=−3
Final Answer: The image is formed 60 cm behind the mirror and the
magnification produced by the mirror is -3.
Question 10
Question
An object is placed 20 cm in front of a convex lens with a focal length of 10 cm.
Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values and the sign conventions.
Object distance, u=−20 cm (since the object is placed in front of the
lens, it is negative)
Focal length, f= +10 cm (for a convex lens, the focal length is positive)
Step 2: Apply the lens formula to find the image distance. The lens formula
is given by: 1
f=1
u+1
v
Plugging in the values: 1
10 =1
−20 +1
v
9
Solving for v:1
v=1
10 +1
20 =3
20
v=20
3cm = 6.6 cm
Step 3: Calculate the magnification using the formula:
m=−v
u
Plugging in the values:
m=−6.6
−20 =1
3
Therefore, the image distance is 6.6cmandthemagnificationis13.
Question 11
Question
A concave mirror with a focal length of 15 cm produces a real image 30 cm from
the mirror. Calculate the magnification of the image.
Solution
To calculate the magnification of the image, we can use the mirror equation:
1
f=1
do
+1
di
where: - fis the focal length of the mirror, - dois the object distance, and - di
is the image distance.
Step 1: Find the object distance Given that the image is real and 30
cm from the mirror, the image distance (di) is -30 cm (negative because it’s on
the same side as the object). Now we can rearrange the mirror equation to solve
for the object distance (do): 1
do
=1
f−1
di
1
do
=1
15 −1
−30
1
do
=1
15 +1
30
1
do
=2+1
30
1
do
=3
30 =1
10
10
do= 10 cm
Step 2: Calculate the magnification The magnification of the image is
given by:
m=−di
do
Substitute the values we found:
m=−
−30
10
m=−(−3)
m= 3
Therefore, the magnification of the image formed by the concave mirror with
a focal length of 15 cm is 3.
Question 12
Question
An object is placed 30 cm in front of a concave mirror of focal length 20 cm.
Determine the image distance and nature of the image formed.
Solution
Step 1: Given that the object distance p=−30 cm (negative since the object is
in front of the mirror) and the focal length f=−20 cm (negative for a concave
mirror). Step 2: Using the mirror equation 1
f=1
p+1
q, where qis the image
distance, we can solve for q. Step 3: Plug in the given values into the mirror
equation: 1
−20 =1
−30 +1
q. Step 4: Simplify the equation: −1
20 =−1
30 +1
q.
Step 5: Find a common denominator: −3
60 =−2
60 +1
q. Step 6: Combine the
fractions: −3
60 =1
60 +1
q. Step 7: Solve for q:−3
60 −1
60 =1
q. Step 8: Simplify
the equation: −4
60 =1
q. Step 9: To find q, take the reciprocal on both sides:
q=−60
4=−15 cm. Step 10: The negative sign indicates that the image is
formed behind the mirror, which confirms that the image is real. Step 11: Since
both the object and the image are on the same side of the mirror, the image is
upright. Step 12: Therefore, the image distance is −15 cm and the nature of
the image formed is real and upright.
Question 13
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 produced by the mirror.
11
Solution
Step 1: Identify the given values. The object distance, do, is -20 cm (since the
object is in front of the mirror, the distance is negative). The focal length, f,
is -15 cm (since the mirror is concave and the focal length is negative).
Step 2: Apply the mirror equation. The mirror equation is given by:
1
f=1
do
+1
di
Substitute the given values:
1
−15 =1
−20 +1
di
Step 3: Find the image distance, di. Solve for di:
−1
15 =−1
20 +1
di
1
di
=−1
15 +1
20
1
di
=−4
60 +3
60
1
di
=−1
60
di=−60 cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−di
do
Substitute the values:
M=−
−60
−20 = 3
Therefore, the image distance is -60 cm and the magnification is 3.
Question 14
Question
A concave mirror with a focal length of −20 cm forms an image that is 3/5
the size of the object. If the object distance is −30 cm, determine the image
distance.
12
Solution
Step 1: Recall the mirror equation for concave mirrors:
−1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Since the focal length is −20 cm and the object distance is −30 cm, we have:
−1
−20 =1
−30 +1
di
Step 2: Solve for the image distance diby first simplifying the equation:
1
20 =1
30 +1
di
Step 3: Find the LCD to combine the fractions on the right side:
3
60 =2
60 +1
di
Step 4: Combine the fractions:
3
60 =2 + 60
60di
Step 5: Simplify the equation:
3
60 =62
60di
Step 6: Cross multiply to solve for di:
3·60di= 62
180di= 62
Step 7: Solve for di:
di=62
180 =31
90 cm
Therefore, the image distance is 31
90 cm.
Question 15
Question
An object is placed 12 cm in front of a concave mirror, which has a focal length
of 8 cm. Determine the image distance, magnification, and whether the image
is real or virtual.
13
Solution
Step 1: Use the mirror equation 1
f=1
do
+1
di
to find the image distance.
Given: f=−8 cm, do=−12 cm
Plug in the values: 1
−8=1
−12 +1
di
Solve for di:di=−24 cm
Step 2: Use the magnification equation m=−di
do
to find the magnification.
Given: di=−24 cm, do=−12 cm
Plug in the values: m=−
−24
−12 = 2
Step 3: Determine if the image is real or virtual by examining the sign of di.
Since di=−24 cm, the image distance is negative.
Therefore, the image is real.
Question 16
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Calculate the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. We are given: -
Object distance, do=−20 cm (since it is in front of the mirror, we assign it a
negative sign). - Focal length of the mirror, f= 15 cm.
The mirror equation is: 1
f=1
do
+1
di
Step 2: Plug in the values and solve for the image distance. Substitute the
known values into the mirror equation:
1
15 =1
−20 +1
di
Solve for di:1
di
=1
15 −1
−20 =4
60 +3
60 =7
60
di=60
7
14
di≈8.57 cm
Therefore, the image distance is approximately 8.57 cm.
Step 3: Calculate the magnification. The magnification is given by the
formula:
m=−di
do
Substitute the values:
m=−8.57
−20 =8.57
20
m≈0.429
The magnification produced by the mirror is approximately 0.429.
Question 17
Question
An object is placed 12 cm in front of a convex lens with a focal length of 8 cm.
Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the required quantities. Given: Object
distance, do=−12 cm (since the object is placed in front of the lens)
Focal length of the lens, f= 8 cm
Required: Image distance, di
Magnification, m
Step 2: Apply the lens equation to find the image distance. The lens equation
is given by: 1
f=1
do
+1
di
Substitute the given values into the equation:
1
8=1
−12 +1
di
Solve for di:1
di
=1
8−1
−12
1
di
=1
8+1
12
1
di
=3
24 +2
24
1
di
=5
24
15
di=24
5= 4.8 cm
Step 3: Calculate the magnification. The magnification mis given by:
m=−di
do
Substitute the values of diand do:
m=−4.8
−12
m=4.8
12
m= 0.4
Step 4: State the final results. The image distance is 4.8 cm and the mag-
nification is 0.4.
Question 18
Question
An object is placed 15 cm in front of a concave mirror with a focal length of
10 cm. Determine the position and magnification of the image formed by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance p=−15 cm (since it is in front of the mirror)
Focal length f=−10 cm (since it is a concave mirror)
Mirror equation: 1
f=1
p+1
q
Step 2: Substitute the known values into the mirror equation and 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+2
30
1
q=−1
30
q=−30 cm
16
Step 3: Calculate the magnification Musing the magnification formula.
M=−q
p
M=−
−30
−15
M= 2
Step 4: Analyze the results.
The image is located at q=−30 cm, which means the image is real and
located on the same side as the object.
The magnification is M= 2, indicating that the image is enlarged.
Therefore, the image formed by the concave mirror is real, located 30 cm in
front of the mirror, and magnified by a factor of 2.
Question 19
Question
A converging lens with a focal length of 10 cm is placed 20 cm in front of a
diverging lens with a focal length of -15 cm. Determine the position of the final
image formed by the two lenses if an object is placed 40 cm in front of the
converging lens.
Solution
Step 1: Identify the given quantities and establish the sign conventions. Step 2:
Calculate the position of the image formed by the converging lens. Step 3: Use
the image formed by the converging lens as the object for the diverging lens.
Step 4: Calculate the final position of the image formed by the two lenses.
Step 1: Given quantities and sign conventions: - For converging lens: f1=
10 cm, u1=−20 cm, v1(unknown) - For diverging lens: f2=−15 cm, u2=v1,
v2(unknown)
Using the lens formula 1
f=1
v−1
uwhere fis the focal length, vis the image
distance, and uis the object distance.
Step 2: Calculating the position of the image formed by the converging
lens: For the converging lens:
1
f1
=1
v1
−1
u1
1
10 =1
v1
+1
20
v1=20
3= 6.67 cm (image distance for converging lens)
17
Step 3: Using v1as the object distance for the diverging lens:
u2=v1= 6.67 cm
Step 4: Calculating the final position of the image formed by the two lenses:
For the diverging lens: 1
f2
=1
v2
−1
u2
1
−15 =1
v2
+1
6.67
1
v2
=−1
15 −1
6.67
1
v2
=−0.067 −0.150 = −0.217
v2=−1
0.217 =−4.62 cm
Therefore, the final image is formed 4.62 cm in front of the diverging lens.
Question 20
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the lens equation. Given: - Object dis-
tance, do=−20 cm (negative because object is in front of lens) - Focal length,
f= 15 cm
The lens equation is: 1
f=1
do
+1
di
Step 2: Substitute the known values into the lens equation.
1
15 =1
−20 +1
di
Step 3: Solve for the image distance, di.
1
15 =−1
20 +1
di
1
di
=1
15 +1
20
18
1
di
=4
60 +3
60 =7
60
di=60
7cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−di
do
Substitute di=60
7cm and do=−20 cm into the equation:
M=−
60
7
−20 =3
7
Therefore, the image distance is 60
7cm and the magnification is 3
7.
Question 21
Question
An object is placed 10 cm in front of a converging lens with a focal length of 20
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the sign convention.
Given:
Object distance p=−10 cm (negative since it is in front of the lens)
Focal length f= 20 cm (positive for a converging lens)
Sign convention:
Distances to the left of the lens are taken as negative, to the right as
positive.
Focal length of a converging lens is positive.
Step 2: Apply the lens equation to find the image distance q. The lens
equation is given by: 1
f=1
p+1
q
Substitute the given values:
1
20 =1
−10 +1
q
19
Step 3: Solve for q.1
20 =−1
10 +1
q
1
q=1
20 +1
10
1
q=1
20 +2
20
1
q=3
20
q=20
3
q= 6.6 cm
Step 4: Calculate the magnification Musing the formula M=−q
p.
M=−6.6
−10
M= 0.666
Therefore, the image distance is approximately 6.67 cm and the magnifica-
tion is 0.666.
Question 22
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.
Solution
Step 1: Given that the object distance p=−30 cm (since it is in front of the
mirror, we take it as negative) and the focal length f=−20 cm (since the
mirror is concave), we can use the mirror equation to find the image distance q:
1
f=1
p+1
q
1
−20 =1
−30 +1
q
−1
20 +1
30 =1
q
3−2
60 =1
q
20
1
60 =1
q
q= 60 cm
Step 2: To find the magnification, we use the magnification formula:
m=−q
p
m=−60
−30
m= 2
Therefore, the image distance is 60 cm in front of the mirror (real image)
and the magnification is 2.
Question 23
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 magnification.
Solution
Step 1: Identify the given values.
f=−8 cm (focal length of the concave mirror)
do=−12 cm (object distance from the mirror)
Step 2: Calculate the image distance using the mirror equation.
1
f=1
di
+1
do
Substitute f=−8 cm and do=−12 cm into the equation:
1
−8=1
di
+1
−12
Solve for di:
−1
8=1
di
−1
12
1
di
=1
12 −1
8
1
di
=2−3
24
21
1
di
=−1
24
di=−24 cm
Step 3: Calculate the magnification using the magnification equation.
m=−di
do
Substitute di=−24 cm and do=−12 cm into the equation:
m=−
−24
−12
m=−2
Therefore, the image distance is −24 cm and the magnification is −2.
Question 24
Question
An object is placed 20 cm from a concave mirror of focal length 10 cm. Deter-
mine the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given quantities and the mirror formula.
Object distance p= -20 cm (because the object is placed in front of the
mirror)
Focal length f= -10 cm (negative for a concave mirror)
Mirror formula: 1
f=1
p+1
q
Step 2: Substitute the known values into the mirror formula to solve for q.
1
−10 =1
−20 +1
q
−1
10 =−1
20 +1
q
1
q=−1
10 +1
20
1
q=−2+1
20
1
q=−1
20
q=−20 cm
22
Step 3: Calculate the magnification using the formula m=−q
p.
m=−
−20
−20
m= 1
Therefore, the image distance produced by the concave mirror is 20 cm and
the magnification is 1.
Question 25
Question
An object is placed 30 cm in front of a concave mirror of focal length 20 cm.
Determine the image distance and magnification of the image formed.
Solution
Step 1: Given that the object distance do=−30 cm and the focal length
f=−20 cm, we can use the mirror equation to find the image distance di:
1
f=1
di
+1
do
Step 2: Substituting the values, we get:
1
−20 =1
di
+1
−30
Step 3: Solving for digives:
1
di
=1
−20 −1
−30
1
di
=−3
60 +2
60
1
di
=−1
60
di=−60 cm
Step 4: Therefore, the image distance diis -60 cm, indicating that the image
is formed on the same side as the object (virtual).
Step 5: To find the magnification M, we use the formula:
M=hi
ho
=−di
do
Step 6: Substituting the values, we get:
M=−(−60)
−30 = 2
Step 7: Thus, the magnification of the image formed by the concave mirror
is 2.
23
Question 26
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Calculate the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values. The object distance (do) is 20 cm, the focal
length (f) is −15 cm (negative for concave mirror).
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 into the mirror equation:
1
−15 =1
20 +1
di
Solve for the image distance (di).
Step 3: Calculate the magnification of the image. The magnification (M) is
given by:
M=−di
do
Substitute the calculated image distance and object distance to find the mag-
nification.
Therefore, the image distance and magnification of the image can be calcu-
lated using the mirror equation and the magnification formula.
Question 27
Question
A concave mirror with a focal length of 10 cm forms an image that is 4 times
the size of the object. If the object is placed 30 cm from the mirror, determine
the image distance and the magnification of the image.
Solution
Step 1: Recall the mirror equation for concave mirrors:
1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
24
Step 2: Given that the focal length of the concave mirror is 10 cm, the object
distance is 30 cm, and the magnification is 4, we can proceed to solve for the
image distance.
Step 3: Use the magnification formula:
M=−di
do
where Mis the magnification.
Step 4: Substitute the given values into the magnification formula to find
the image distance:
4 = −di
30
Step 5: Solve for the image distance:
di=−4×30 = −120 cm
Step 6: Now substitute the values of fand dointo the mirror equation to
solve for the image distance:
1
10 =1
30 +1
di
Step 7: Solve for di:
1
di
=1
10 −1
30 =2
30 =1
15
di= 15 cm
Step 8: The image distance is 15 cm and the magnification is -4.
Question 28
Question
An object is placed 40 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, do=−40 cm (since object is in front of mirror, the distance is negative)
- Focal length, f= 20 cm The mirror equation is given by:
1
f=1
di
+1
do
25
Step 2: Calculate the image distance using the mirror equation. Substitute
the known values into the mirror equation to solve for di:
1
20 =1
di
+1
−40
1
di
=1
20 −1
−40
1
di
=2−1
40
di= 40 cm
Step 3: Calculate the magnification. The magnification, M, can be calcu-
lated using the formula:
M=−di
do
Substitute di= 40 cm and do=−40 cm into the formula:
M=−40
−40 = 1
Step 4: Determine if the image is real or virtual. Since the magnification is
positive, the image is upright. Therefore, the image is virtual.
Thus, the image distance is 40 cm, the magnification is 1, and the image is
virtual.
Question 29
Question
An object is placed 50 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 equation. Given: - Object
distance (u) = -50 cm (negative sign indicates in front of the mirror) - Focal
length (f) = -30 cm (negative sign indicates concave mirror)
The mirror equation is given by:
1
f=1
u+1
v
Step 2: Calculate the image distance. Substitute the given values into the
mirror equation: 1
−30 =1
−50 +1
v
26
Step 3: Solve for the image distance.
1
v=1
−30 −1
−50
1
v=−50 + 30
−30 × −50
1
v=−20
1500
v=1500
−20
v=−75 cm
Therefore, the image distance is -75 cm, indicating that the image is formed
75 cm behind the mirror.
Step 4: Calculate the magnification of the image. The magnification (M) is
given by:
M=−v
u
Step 5: Substitute the values to find the magnification.
M=−(−75)
−50
M=75
50
M= 1.5
Therefore, the magnification of the image is 1.5.
Question 30
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and mirror equation.
Object distance (do) = -20 cm (the negative sign indicates that the object
is located in front of the mirror).
Focal length (f) = -10 cm (the negative sign indicates a concave mirror).
27
Mirror equation: 1
f=1
do+1
di
Step 2: Plug in the known values into the mirror equation and solve for the
image distance (di). 1
−10 =1
−20 +1
di
−1
10 =−1
20 +1
di
1
di
=−1
10 +1
20
1
di
=−2
20 +1
20 =−1
20
di=−20 cm
Step 3: Calculate the magnification using the magnification formula, m=
−di
do.
m=−
−20
−20 = 1
Therefore, the image is located 20 cm behind the mirror, and the magnifi-
cation is 1.
Question 31
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the position, nature, and magnification of the image formed by
the mirror.
Solution
Step 1: Identify the given values.
The object distance (do) is -20 cm (since it is in front of the mirror) and the
focal length (f) is -10 cm for a concave mirror.
Step 2: Apply the mirror formula to find the image distance.
The mirror formula is given by:
1
f=1
do
+1
di
Substitute f=−10 cm and do=−20 cm into the formula and solve for di:
1
−10 =1
−20 +1
di
−0.1 = −0.05 + 1
di
28
1
di
=−0.1+0.05 = −0.05
di=1
−0.05 =−20 cm
So, the image distance (di) is -20 cm.
Step 3: Determine the position of the image.
Since the image distance is negative, the image is formed on the same side as
the object, which means the image is virtual and erect.
Step 4: Calculate the magnification.
The magnification is given by:
M=−di
do
Substitute di=−20 cm and do=−20 cm into the formula:
M=−
−20
−20 = 1
The magnification is 1, indicating that the image is the same size as the object.
Step 5: Summarize the results.
The image is formed 20 cm in front of the concave mirror, on the same side as
the object. It is virtual, erect, and the same size as the object.
Question 32
Question
An object is placed 10 cm in front of a converging lens with a focal length of 20
cm. Determine the position and magnification of the image formed by the lens.
Solution
Step 1: Identify the given values and the focal length equation for the lens. The
given values are: Object distance (u) = -10 cm (negative since the object is
placed in front of the lens) Focal length (f) = 20 cm
The lens equation relates the object distance, image distance, and focal
length of the lens: 1
f=1
u+1
v
Step 2: Substitute the given values into the lens equation and solve for the
image distance. Substitute the given values into the lens equation:
1
20 =1
−10 +1
v
Solving for 1
v:
1
v=1
20 −1
−10
29
1
v=1
20 +1
10
1
v=3
20
Therefore, image distance, v=20
3cm.
Step 3: Calculate the magnification of the image. The magnification (m) of
the image is given by:
m=−v
u
Substitute the values of v and u into the magnification formula:
m=−
20
3
−10
m=2
3
The image is formed at a distance of 20
3cm from the lens and the magnifi-
cation of the image is 2
3.
Question 33
Question
An object is placed 20 cm from a concave mirror with a focal length of 10 cm.
Determine the position and nature of the image formed.
Solution
Step 1: Given the object distance p=−20 cm and the focal length f=−10
cm for a concave mirror, we can use the mirror equation:
1
f=1
p+1
q
where qis the image distance.
Step 2: Substituting f=−10 cm and p=−20 cm into the mirror equation,
we have: 1
−10 =1
−20 +1
q
Step 3: Solving for qgives:
−1
10 =−1
20 +1
q
1
q=−1
10 +1
20
30
1
q=−2+1
20
1
q=−1
20
q=−20 cm
Step 4: The negative sign for qindicates that the image is formed on the
same side as the object, which means it is a virtual image.
Step 5: Since q=−20 cm is negative, the image distance is measured to the
left of the mirror, indicating that the image is virtual and upright.
Therefore, the image formed by the concave mirror is virtual, upright, and
located 20 cm to the left of the mirror.
Question 34
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
Object distance, do=−20 cm (since the object is in front of the mirror)
Focal length, f= 10 cm
The mirror equation is: 1
f=1
do
+1
di
Step 2: Substitute the given values into the mirror equation.
1
10 =1
−20 +1
di
Step 3: Solve for the image distance (di).
1
di
=1
10 −1
−20
1
di
=1
10 +1
20
1
di
=3
20
di=20
3cm = 6.6 cm
31
Step 4: Calculate the magnification (m).
The magnification is given by:
m=−di
do
m=−6.6
−20
m=2
3
Answer: The image distance is 6.6 cm and the magnification is 2
3.
Question 35
Question
A concave mirror with a focal length of 15 cm forms an image of a tree 60 cm
tall. If the image is 30 cm tall, what is the distance of the tree from the mirror?
Solution
Step 1: Use the mirror equation 1
f=1
do+1
diwhere fis the focal length, dois
the object distance, and diis the image distance.
Step 2: Given that f=−15 cm (since it’s a concave mirror with a focal
length of 15 cm), and ho= 60 cm (height of the tree) and hi= 30 cm (height
of the image), we need to find do.
Step 3: Use the magnification formula M=−di
do=hi
hoand solve for do.
Step 4: Substituting the given values into the magnification formula, we
have −di
do=hi
ho.
Step 5: Substituting the values di=−15 cm, hi= 30 cm, and ho= 60 cm,
we get −
−15
do=30
60 .
Step 6: Simplifying, 15
do=1
2.
Step 7: Cross multiplying gives 30 = do.
Therefore, the distance of the tree from the mirror is 30 cm.
32
Step 5: Combine the fractions:
−10
375 =25
di
Step 6: Solve for di:
di=375
−10 =−37.5 cm
Therefore, the image distance diis -37.5 cm.
Step 7: To find the magnification m, we use the formula:
m=−di
do
Step 8: Substitute the calculated values of diand do:
m=−
−37.5
−25 = 1.5
Therefore, the magnification mis 1.5.
Question 2
Question
A converging lens with a focal length of 15 cm is placed 30 cm to the left of
a diverging mirror. The combined optical system forms a final image that is
virtual and located 24 cm to the left of the lens. Determine the focal length of
the mirror.
Solution
Let flens be the focal length of the lens and fmirror be the focal length of the
mirror.
Step 1: Identify the given values and the unknown.
flens = 15 cm
Object distance for the lens, dobj,lens = 30 cm
Image distance from the lens, dimg,lens =−24 cm (negative for virtual
image on the same side as the object)
Image distance for the lens-mirror system, dimg,system =−30 cm (sum of
distances from the lens to the mirror and from the mirror to the final
image)
2
Step 2: Find the image distance for the mirror from the lens equation. The
lens equation relates the object distance, image distance, and focal length for a
lens: 1
flens
=1
dobj,lens
+1
dimg,lens
Substitute the given values:
1
15 =1
30 +1
−24
Solve for dimg,lens:
dimg,lens =−20 cm
Step 3: Calculate the image distance for the mirror. The image distance
for the lens-mirror system is the object distance for the mirror:
dobj,mirror =−dimg,lens = 20 cm
Step 4: Use the mirror equation to find the focal length of the mirror. The
mirror equation relates the object distance, image distance, and focal length for
a mirror: 1
fmirror
=1
dobj,mirror
+1
dimg,system
Substitute the known values:
1
fmirror
=1
20 +1
−30
Solve for fmirror:
1
fmirror
=1
20 −1
30 =3
60 −2
60 =1
60
fmirror = 60 cm
Question 3
Question
A concave mirror has a focal length of 15 cm. An object is placed 25 cm in
front of the mirror. Determine the image distance and the magnification of the
image.
3
Solution
Step 1: Given that f=−15 cm (focal length of the mirror) and do=−25
cm (object distance), we need to find the image distance diusing the mirror
equation:
1
f=1
di
+1
do
Plugging in the known values:
1
−15 =1
di
+1
−25
−1
15 =1
di
−1
25
−1
15 +1
25 =1
di
5
75 +3
75 =1
di
8
75 =1
di
di=75
8≈9.375 cm
Therefore, the image distance di≈9.375 cm.
Step 2: To find the magnification M, we use the formula:
M=−di
do
Plugging in the calculated values:
M=−9.375
−25 =9.375
25 = 0.375
Therefore, the magnification of the image is M= 0.375.
Question 4
Question
A concave mirror has a focal length of 10 cm. An object is placed 20 cm in
front of the mirror. Determine the position and size of the image formed by the
mirror.
4
Solution
Step 1: Identify the given values and mirror equation. Given: f=−10 cm,
do=−20 cm, and diand hiare to be found. The mirror equation is 1
f=1
do+1
di.
Step 2: Substitute the given values into the mirror equation. Substitute
f=−10 cm and do=−20 cm into the mirror equation: 1
−10 =1
−20 +1
di.
Step 3: Solve for di. Solving the equation, we get: 1
−10 =1
−20 +1
di
−1
10 =
−1
20 +1
di
−1
10 +1
20 =1
di
1
20 =1
didi= 20 cm
Step 4: Determine the size of the image using the magnification equation.
The magnification equation is M=−di
do=hi
ho. We are given di= 20 cm and
do=−20 cm. Substitute di= 20 cm and do=−20 cm into the magnification
equation: M=−20
−20 =hi
hoM=1= hi
ho
Step 5: Calculate the height of the image. Since M= 1 = hi
ho, the height of
the image is equal to the height of the object. Therefore, the size of the image
is the same as the size of the object.
Step 6: State the position and size of the image. The position of the image
is di= 20 cm in front of the mirror, and the size of the image is the same as
the size of the object.
Question 5
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, the magnification, and describe
the image formed by the mirror.
Solution
Step 1: Given that the focal length of the concave mirror is f=−15 cm, and
the object distance do=−30 cm. Step 2: We can use the mirror equation
1
f=1
do+1
dito find the image distance di. Step 3: Substituting the given values
into the mirror equation, we get 1
−15 =1
−30 +1
di. Step 4: Solving for di, we have
1
di=1
−15 −1
−30 . Step 5: Simplifying the equation, we find 1
di=−2
30 . Step 6:
Therefore, the image distance is di=−15 cm. Step 7: To find the magnification
M, we use the formula M=−di
do. Step 8: Substituting the values, we get
M=−
−15
−30 =1
2. Step 9: The positive magnification value indicates that the
image is upright compared to the object. Step 10: Since the magnification is
less than 1, the image is smaller than the object. Step 11: Therefore, the image
formed by the concave mirror is virtual, upright, and smaller than the object.
5
Question 6
Question
An object is placed 40 cm from a converging lens with a focal length of 20 cm.
Determine the image distance, magnification, and whether the image is real or
virtual.
Solution
Step 1: Identify the given values and the lens equation. Given: - Object distance
p=−40 cm - Focal length f= 20 cm The lens equation is given by:
1
f=1
p+1
q
Step 2: Substitute the given values into the lens equation to solve for the
image distance. Substitute p=−40 cm and f= 20 cm into the lens equation:
1
20 =1
−40 +1
q
Step 3: Solve for q.1
q=1
20 −1
−40 =3
40
q=40
3cm ≈13.33 cm
Step 4: Calculate the magnification. The magnification is given by:
m=−q
p
Substitute p=−40 cm and q=40
3cm into the magnification formula:
m=−
40
3
−40 =1
3
Step 5: Determine if the image is real or virtual. Since the magnification is
positive, the image is upright (virtual).
Therefore, the image distance is approximately 13.33 cm, the magnification
is 1/3, and the image is virtual.
Question 7
Question
An object is placed 10 cm in front of a concave lens of focal length 20 cm.
Determine the image distance and characterize the image in terms of size and
orientation.
6
Solution
Step 1: Identify the given values and the type of lens.
Object distance p=−10 cm (since the object is located in front of the
concave lens, the distance is negative).
Focal length f=−20 cm (since a concave lens has a negative focal length).
Since we are dealing with a concave lens, fis negative and will be used
as such in the lens formula 1
f=1
p+1
q.
Step 2: Calculate the image distance qusing the lens formula.
1
f=1
p+1
q
1
−20 =1
−10 +1
q
−1
20 =−1
10 +1
q
1
q=−1
20 +1
10
1
q=−1
20 +2
20
1
q=1
20
q= 20 cm
Step 3: Characterize the image in terms of size and orientation. Since the
image distance qis positive, the image is formed on the same side of the lens
as the object (i.e., the image is virtual). Since the magnification M=q
p, and
q > p, the image will be magnified. Therefore, the image formed by the concave
lens is virtual and magnified.
Question 8
Question
An object is placed 20 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 values and the mirror equation.
Object distance, do=−20 cm (negative because the object is in front of
the mirror)
7
Focal length, f=−15 cm (negative because the mirror is concave)
Mirror equation: 1
f=1
do+1
di
Step 2: Substitute the given values into the mirror equation and solve for
the image distance, di.
1
−15 =1
−20 +1
di
−1
15 =−1
20 +1
di
−1
15 +1
20 =1
di
4
60 −3
60 =1
di
1
60 =1
di
di= 60 cm
Therefore, the image distance is 60 cm.
Step 3: Calculate the magnification, M.
M=−di
do
=−60 cm
−20 cm
= 3
Therefore, the magnification produced by the mirror is 3.
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 the magnification produced by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance p=−20 cm (negative because the object is in front of
the mirror).
Focal length f=−15 cm (negative for a concave mirror).
Mirror equation: 1
f=1
p+1
q.
8
Step 2: Substitute the known values into the mirror equation and solve for
the image distance q.
1
−15 =1
−20 +1
q
−1
15 =−1
20 +1
q
1
q=−1
15 +1
20
1
q=−4+3
60
1
q=−1
60
q=−60 cm
Step 3: Calculate the magnification musing the formula m=−q
p.
m=−
−60
−20
m=−3
Final Answer: The image is formed 60 cm behind the mirror and the
magnification produced by the mirror is -3.
Question 10
Question
An object is placed 20 cm in front of a convex lens with a focal length of 10 cm.
Determine the image distance and magnification produced by the lens.
Solution
Step 1: Identify the given values and the sign conventions.
Object distance, u=−20 cm (since the object is placed in front of the
lens, it is negative)
Focal length, f= +10 cm (for a convex lens, the focal length is positive)
Step 2: Apply the lens formula to find the image distance. The lens formula
is given by: 1
f=1
u+1
v
Plugging in the values: 1
10 =1
−20 +1
v
9
Solving for v:1
v=1
10 +1
20 =3
20
v=20
3cm = 6.6 cm
Step 3: Calculate the magnification using the formula:
m=−v
u
Plugging in the values:
m=−6.6
−20 =1
3
Therefore, the image distance is 6.6cmandthemagnificationis13.
Question 11
Question
A concave mirror with a focal length of 15 cm produces a real image 30 cm from
the mirror. Calculate the magnification of the image.
Solution
To calculate the magnification of the image, we can use the mirror equation:
1
f=1
do
+1
di
where: - fis the focal length of the mirror, - dois the object distance, and - di
is the image distance.
Step 1: Find the object distance Given that the image is real and 30
cm from the mirror, the image distance (di) is -30 cm (negative because it’s on
the same side as the object). Now we can rearrange the mirror equation to solve
for the object distance (do): 1
do
=1
f−1
di
1
do
=1
15 −1
−30
1
do
=1
15 +1
30
1
do
=2+1
30
1
do
=3
30 =1
10
10
do= 10 cm
Step 2: Calculate the magnification The magnification of the image is
given by:
m=−di
do
Substitute the values we found:
m=−
−30
10
m=−(−3)
m= 3
Therefore, the magnification of the image formed by the concave mirror with
a focal length of 15 cm is 3.
Question 12
Question
An object is placed 30 cm in front of a concave mirror of focal length 20 cm.
Determine the image distance and nature of the image formed.
Solution
Step 1: Given that the object distance p=−30 cm (negative since the object is
in front of the mirror) and the focal length f=−20 cm (negative for a concave
mirror). Step 2: Using the mirror equation 1
f=1
p+1
q, where qis the image
distance, we can solve for q. Step 3: Plug in the given values into the mirror
equation: 1
−20 =1
−30 +1
q. Step 4: Simplify the equation: −1
20 =−1
30 +1
q.
Step 5: Find a common denominator: −3
60 =−2
60 +1
q. Step 6: Combine the
fractions: −3
60 =1
60 +1
q. Step 7: Solve for q:−3
60 −1
60 =1
q. Step 8: Simplify
the equation: −4
60 =1
q. Step 9: To find q, take the reciprocal on both sides:
q=−60
4=−15 cm. Step 10: The negative sign indicates that the image is
formed behind the mirror, which confirms that the image is real. Step 11: Since
both the object and the image are on the same side of the mirror, the image is
upright. Step 12: Therefore, the image distance is −15 cm and the nature of
the image formed is real and upright.
Question 13
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 produced by the mirror.
11
Solution
Step 1: Identify the given values. The object distance, do, is -20 cm (since the
object is in front of the mirror, the distance is negative). The focal length, f,
is -15 cm (since the mirror is concave and the focal length is negative).
Step 2: Apply the mirror equation. The mirror equation is given by:
1
f=1
do
+1
di
Substitute the given values:
1
−15 =1
−20 +1
di
Step 3: Find the image distance, di. Solve for di:
−1
15 =−1
20 +1
di
1
di
=−1
15 +1
20
1
di
=−4
60 +3
60
1
di
=−1
60
di=−60 cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−di
do
Substitute the values:
M=−
−60
−20 = 3
Therefore, the image distance is -60 cm and the magnification is 3.
Question 14
Question
A concave mirror with a focal length of −20 cm forms an image that is 3/5
the size of the object. If the object distance is −30 cm, determine the image
distance.
12
Solution
Step 1: Recall the mirror equation for concave mirrors:
−1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
Since the focal length is −20 cm and the object distance is −30 cm, we have:
−1
−20 =1
−30 +1
di
Step 2: Solve for the image distance diby first simplifying the equation:
1
20 =1
30 +1
di
Step 3: Find the LCD to combine the fractions on the right side:
3
60 =2
60 +1
di
Step 4: Combine the fractions:
3
60 =2 + 60
60di
Step 5: Simplify the equation:
3
60 =62
60di
Step 6: Cross multiply to solve for di:
3·60di= 62
180di= 62
Step 7: Solve for di:
di=62
180 =31
90 cm
Therefore, the image distance is 31
90 cm.
Question 15
Question
An object is placed 12 cm in front of a concave mirror, which has a focal length
of 8 cm. Determine the image distance, magnification, and whether the image
is real or virtual.
13
Solution
Step 1: Use the mirror equation 1
f=1
do
+1
di
to find the image distance.
Given: f=−8 cm, do=−12 cm
Plug in the values: 1
−8=1
−12 +1
di
Solve for di:di=−24 cm
Step 2: Use the magnification equation m=−di
do
to find the magnification.
Given: di=−24 cm, do=−12 cm
Plug in the values: m=−
−24
−12 = 2
Step 3: Determine if the image is real or virtual by examining the sign of di.
Since di=−24 cm, the image distance is negative.
Therefore, the image is real.
Question 16
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Calculate the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given values and the mirror equation. We are given: -
Object distance, do=−20 cm (since it is in front of the mirror, we assign it a
negative sign). - Focal length of the mirror, f= 15 cm.
The mirror equation is: 1
f=1
do
+1
di
Step 2: Plug in the values and solve for the image distance. Substitute the
known values into the mirror equation:
1
15 =1
−20 +1
di
Solve for di:1
di
=1
15 −1
−20 =4
60 +3
60 =7
60
di=60
7
14
di≈8.57 cm
Therefore, the image distance is approximately 8.57 cm.
Step 3: Calculate the magnification. The magnification is given by the
formula:
m=−di
do
Substitute the values:
m=−8.57
−20 =8.57
20
m≈0.429
The magnification produced by the mirror is approximately 0.429.
Question 17
Question
An object is placed 12 cm in front of a convex lens with a focal length of 8 cm.
Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the required quantities. Given: Object
distance, do=−12 cm (since the object is placed in front of the lens)
Focal length of the lens, f= 8 cm
Required: Image distance, di
Magnification, m
Step 2: Apply the lens equation to find the image distance. The lens equation
is given by: 1
f=1
do
+1
di
Substitute the given values into the equation:
1
8=1
−12 +1
di
Solve for di:1
di
=1
8−1
−12
1
di
=1
8+1
12
1
di
=3
24 +2
24
1
di
=5
24
15
di=24
5= 4.8 cm
Step 3: Calculate the magnification. The magnification mis given by:
m=−di
do
Substitute the values of diand do:
m=−4.8
−12
m=4.8
12
m= 0.4
Step 4: State the final results. The image distance is 4.8 cm and the mag-
nification is 0.4.
Question 18
Question
An object is placed 15 cm in front of a concave mirror with a focal length of
10 cm. Determine the position and magnification of the image formed by the
mirror.
Solution
Step 1: Identify the given values and the mirror equation.
Object distance p=−15 cm (since it is in front of the mirror)
Focal length f=−10 cm (since it is a concave mirror)
Mirror equation: 1
f=1
p+1
q
Step 2: Substitute the known values into the mirror equation and 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+2
30
1
q=−1
30
q=−30 cm
16
Step 3: Calculate the magnification Musing the magnification formula.
M=−q
p
M=−
−30
−15
M= 2
Step 4: Analyze the results.
The image is located at q=−30 cm, which means the image is real and
located on the same side as the object.
The magnification is M= 2, indicating that the image is enlarged.
Therefore, the image formed by the concave mirror is real, located 30 cm in
front of the mirror, and magnified by a factor of 2.
Question 19
Question
A converging lens with a focal length of 10 cm is placed 20 cm in front of a
diverging lens with a focal length of -15 cm. Determine the position of the final
image formed by the two lenses if an object is placed 40 cm in front of the
converging lens.
Solution
Step 1: Identify the given quantities and establish the sign conventions. Step 2:
Calculate the position of the image formed by the converging lens. Step 3: Use
the image formed by the converging lens as the object for the diverging lens.
Step 4: Calculate the final position of the image formed by the two lenses.
Step 1: Given quantities and sign conventions: - For converging lens: f1=
10 cm, u1=−20 cm, v1(unknown) - For diverging lens: f2=−15 cm, u2=v1,
v2(unknown)
Using the lens formula 1
f=1
v−1
uwhere fis the focal length, vis the image
distance, and uis the object distance.
Step 2: Calculating the position of the image formed by the converging
lens: For the converging lens:
1
f1
=1
v1
−1
u1
1
10 =1
v1
+1
20
v1=20
3= 6.67 cm (image distance for converging lens)
17
Step 3: Using v1as the object distance for the diverging lens:
u2=v1= 6.67 cm
Step 4: Calculating the final position of the image formed by the two lenses:
For the diverging lens: 1
f2
=1
v2
−1
u2
1
−15 =1
v2
+1
6.67
1
v2
=−1
15 −1
6.67
1
v2
=−0.067 −0.150 = −0.217
v2=−1
0.217 =−4.62 cm
Therefore, the final image is formed 4.62 cm in front of the diverging lens.
Question 20
Question
An object is placed 20 cm in front of a converging lens with a focal length of 15
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the lens equation. Given: - Object dis-
tance, do=−20 cm (negative because object is in front of lens) - Focal length,
f= 15 cm
The lens equation is: 1
f=1
do
+1
di
Step 2: Substitute the known values into the lens equation.
1
15 =1
−20 +1
di
Step 3: Solve for the image distance, di.
1
15 =−1
20 +1
di
1
di
=1
15 +1
20
18
1
di
=4
60 +3
60 =7
60
di=60
7cm
Step 4: Calculate the magnification, M. The magnification is given by:
M=−di
do
Substitute di=60
7cm and do=−20 cm into the equation:
M=−
60
7
−20 =3
7
Therefore, the image distance is 60
7cm and the magnification is 3
7.
Question 21
Question
An object is placed 10 cm in front of a converging lens with a focal length of 20
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and the sign convention.
Given:
Object distance p=−10 cm (negative since it is in front of the lens)
Focal length f= 20 cm (positive for a converging lens)
Sign convention:
Distances to the left of the lens are taken as negative, to the right as
positive.
Focal length of a converging lens is positive.
Step 2: Apply the lens equation to find the image distance q. The lens
equation is given by: 1
f=1
p+1
q
Substitute the given values:
1
20 =1
−10 +1
q
19
Step 3: Solve for q.1
20 =−1
10 +1
q
1
q=1
20 +1
10
1
q=1
20 +2
20
1
q=3
20
q=20
3
q= 6.6 cm
Step 4: Calculate the magnification Musing the formula M=−q
p.
M=−6.6
−10
M= 0.666
Therefore, the image distance is approximately 6.67 cm and the magnifica-
tion is 0.666.
Question 22
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.
Solution
Step 1: Given that the object distance p=−30 cm (since it is in front of the
mirror, we take it as negative) and the focal length f=−20 cm (since the
mirror is concave), we can use the mirror equation to find the image distance q:
1
f=1
p+1
q
1
−20 =1
−30 +1
q
−1
20 +1
30 =1
q
3−2
60 =1
q
20
1
60 =1
q
q= 60 cm
Step 2: To find the magnification, we use the magnification formula:
m=−q
p
m=−60
−30
m= 2
Therefore, the image distance is 60 cm in front of the mirror (real image)
and the magnification is 2.
Question 23
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 magnification.
Solution
Step 1: Identify the given values.
f=−8 cm (focal length of the concave mirror)
do=−12 cm (object distance from the mirror)
Step 2: Calculate the image distance using the mirror equation.
1
f=1
di
+1
do
Substitute f=−8 cm and do=−12 cm into the equation:
1
−8=1
di
+1
−12
Solve for di:
−1
8=1
di
−1
12
1
di
=1
12 −1
8
1
di
=2−3
24
21
1
di
=−1
24
di=−24 cm
Step 3: Calculate the magnification using the magnification equation.
m=−di
do
Substitute di=−24 cm and do=−12 cm into the equation:
m=−
−24
−12
m=−2
Therefore, the image distance is −24 cm and the magnification is −2.
Question 24
Question
An object is placed 20 cm from a concave mirror of focal length 10 cm. Deter-
mine the image distance and magnification produced by the mirror.
Solution
Step 1: Identify the given quantities and the mirror formula.
Object distance p= -20 cm (because the object is placed in front of the
mirror)
Focal length f= -10 cm (negative for a concave mirror)
Mirror formula: 1
f=1
p+1
q
Step 2: Substitute the known values into the mirror formula to solve for q.
1
−10 =1
−20 +1
q
−1
10 =−1
20 +1
q
1
q=−1
10 +1
20
1
q=−2+1
20
1
q=−1
20
q=−20 cm
22
Step 3: Calculate the magnification using the formula m=−q
p.
m=−
−20
−20
m= 1
Therefore, the image distance produced by the concave mirror is 20 cm and
the magnification is 1.
Question 25
Question
An object is placed 30 cm in front of a concave mirror of focal length 20 cm.
Determine the image distance and magnification of the image formed.
Solution
Step 1: Given that the object distance do=−30 cm and the focal length
f=−20 cm, we can use the mirror equation to find the image distance di:
1
f=1
di
+1
do
Step 2: Substituting the values, we get:
1
−20 =1
di
+1
−30
Step 3: Solving for digives:
1
di
=1
−20 −1
−30
1
di
=−3
60 +2
60
1
di
=−1
60
di=−60 cm
Step 4: Therefore, the image distance diis -60 cm, indicating that the image
is formed on the same side as the object (virtual).
Step 5: To find the magnification M, we use the formula:
M=hi
ho
=−di
do
Step 6: Substituting the values, we get:
M=−(−60)
−30 = 2
Step 7: Thus, the magnification of the image formed by the concave mirror
is 2.
23
Question 26
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 15
cm. Calculate the image distance and magnification of the image formed.
Solution
Step 1: Identify the given values. The object distance (do) is 20 cm, the focal
length (f) is −15 cm (negative for concave mirror).
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 into the mirror equation:
1
−15 =1
20 +1
di
Solve for the image distance (di).
Step 3: Calculate the magnification of the image. The magnification (M) is
given by:
M=−di
do
Substitute the calculated image distance and object distance to find the mag-
nification.
Therefore, the image distance and magnification of the image can be calcu-
lated using the mirror equation and the magnification formula.
Question 27
Question
A concave mirror with a focal length of 10 cm forms an image that is 4 times
the size of the object. If the object is placed 30 cm from the mirror, determine
the image distance and the magnification of the image.
Solution
Step 1: Recall the mirror equation for concave mirrors:
1
f=1
do
+1
di
where fis the focal length, dois the object distance, and diis the image distance.
24
Step 2: Given that the focal length of the concave mirror is 10 cm, the object
distance is 30 cm, and the magnification is 4, we can proceed to solve for the
image distance.
Step 3: Use the magnification formula:
M=−di
do
where Mis the magnification.
Step 4: Substitute the given values into the magnification formula to find
the image distance:
4 = −di
30
Step 5: Solve for the image distance:
di=−4×30 = −120 cm
Step 6: Now substitute the values of fand dointo the mirror equation to
solve for the image distance:
1
10 =1
30 +1
di
Step 7: Solve for di:
1
di
=1
10 −1
30 =2
30 =1
15
di= 15 cm
Step 8: The image distance is 15 cm and the magnification is -4.
Question 28
Question
An object is placed 40 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, do=−40 cm (since object is in front of mirror, the distance is negative)
- Focal length, f= 20 cm The mirror equation is given by:
1
f=1
di
+1
do
25
Step 2: Calculate the image distance using the mirror equation. Substitute
the known values into the mirror equation to solve for di:
1
20 =1
di
+1
−40
1
di
=1
20 −1
−40
1
di
=2−1
40
di= 40 cm
Step 3: Calculate the magnification. The magnification, M, can be calcu-
lated using the formula:
M=−di
do
Substitute di= 40 cm and do=−40 cm into the formula:
M=−40
−40 = 1
Step 4: Determine if the image is real or virtual. Since the magnification is
positive, the image is upright. Therefore, the image is virtual.
Thus, the image distance is 40 cm, the magnification is 1, and the image is
virtual.
Question 29
Question
An object is placed 50 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 equation. Given: - Object
distance (u) = -50 cm (negative sign indicates in front of the mirror) - Focal
length (f) = -30 cm (negative sign indicates concave mirror)
The mirror equation is given by:
1
f=1
u+1
v
Step 2: Calculate the image distance. Substitute the given values into the
mirror equation: 1
−30 =1
−50 +1
v
26
Step 3: Solve for the image distance.
1
v=1
−30 −1
−50
1
v=−50 + 30
−30 × −50
1
v=−20
1500
v=1500
−20
v=−75 cm
Therefore, the image distance is -75 cm, indicating that the image is formed
75 cm behind the mirror.
Step 4: Calculate the magnification of the image. The magnification (M) is
given by:
M=−v
u
Step 5: Substitute the values to find the magnification.
M=−(−75)
−50
M=75
50
M= 1.5
Therefore, the magnification of the image is 1.5.
Question 30
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the image distance and magnification.
Solution
Step 1: Identify the given values and mirror equation.
Object distance (do) = -20 cm (the negative sign indicates that the object
is located in front of the mirror).
Focal length (f) = -10 cm (the negative sign indicates a concave mirror).
27
Mirror equation: 1
f=1
do+1
di
Step 2: Plug in the known values into the mirror equation and solve for the
image distance (di). 1
−10 =1
−20 +1
di
−1
10 =−1
20 +1
di
1
di
=−1
10 +1
20
1
di
=−2
20 +1
20 =−1
20
di=−20 cm
Step 3: Calculate the magnification using the magnification formula, m=
−di
do.
m=−
−20
−20 = 1
Therefore, the image is located 20 cm behind the mirror, and the magnifi-
cation is 1.
Question 31
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Determine the position, nature, and magnification of the image formed by
the mirror.
Solution
Step 1: Identify the given values.
The object distance (do) is -20 cm (since it is in front of the mirror) and the
focal length (f) is -10 cm for a concave mirror.
Step 2: Apply the mirror formula to find the image distance.
The mirror formula is given by:
1
f=1
do
+1
di
Substitute f=−10 cm and do=−20 cm into the formula and solve for di:
1
−10 =1
−20 +1
di
−0.1 = −0.05 + 1
di
28
1
di
=−0.1+0.05 = −0.05
di=1
−0.05 =−20 cm
So, the image distance (di) is -20 cm.
Step 3: Determine the position of the image.
Since the image distance is negative, the image is formed on the same side as
the object, which means the image is virtual and erect.
Step 4: Calculate the magnification.
The magnification is given by:
M=−di
do
Substitute di=−20 cm and do=−20 cm into the formula:
M=−
−20
−20 = 1
The magnification is 1, indicating that the image is the same size as the object.
Step 5: Summarize the results.
The image is formed 20 cm in front of the concave mirror, on the same side as
the object. It is virtual, erect, and the same size as the object.
Question 32
Question
An object is placed 10 cm in front of a converging lens with a focal length of 20
cm. Determine the position and magnification of the image formed by the lens.
Solution
Step 1: Identify the given values and the focal length equation for the lens. The
given values are: Object distance (u) = -10 cm (negative since the object is
placed in front of the lens) Focal length (f) = 20 cm
The lens equation relates the object distance, image distance, and focal
length of the lens: 1
f=1
u+1
v
Step 2: Substitute the given values into the lens equation and solve for the
image distance. Substitute the given values into the lens equation:
1
20 =1
−10 +1
v
Solving for 1
v:
1
v=1
20 −1
−10
29
1
v=1
20 +1
10
1
v=3
20
Therefore, image distance, v=20
3cm.
Step 3: Calculate the magnification of the image. The magnification (m) of
the image is given by:
m=−v
u
Substitute the values of v and u into the magnification formula:
m=−
20
3
−10
m=2
3
The image is formed at a distance of 20
3cm from the lens and the magnifi-
cation of the image is 2
3.
Question 33
Question
An object is placed 20 cm from a concave mirror with a focal length of 10 cm.
Determine the position and nature of the image formed.
Solution
Step 1: Given the object distance p=−20 cm and the focal length f=−10
cm for a concave mirror, we can use the mirror equation:
1
f=1
p+1
q
where qis the image distance.
Step 2: Substituting f=−10 cm and p=−20 cm into the mirror equation,
we have: 1
−10 =1
−20 +1
q
Step 3: Solving for qgives:
−1
10 =−1
20 +1
q
1
q=−1
10 +1
20
30
1
q=−2+1
20
1
q=−1
20
q=−20 cm
Step 4: The negative sign for qindicates that the image is formed on the
same side as the object, which means it is a virtual image.
Step 5: Since q=−20 cm is negative, the image distance is measured to the
left of the mirror, indicating that the image is virtual and upright.
Therefore, the image formed by the concave mirror is virtual, upright, and
located 20 cm to the left of the mirror.
Question 34
Question
An object is placed 20 cm in front of a concave mirror with a focal length of 10
cm. Calculate the image distance and magnification.
Solution
Step 1: Identify the given values and the mirror equation.
Given:
Object distance, do=−20 cm (since the object is in front of the mirror)
Focal length, f= 10 cm
The mirror equation is: 1
f=1
do
+1
di
Step 2: Substitute the given values into the mirror equation.
1
10 =1
−20 +1
di
Step 3: Solve for the image distance (di).
1
di
=1
10 −1
−20
1
di
=1
10 +1
20
1
di
=3
20
di=20
3cm = 6.6 cm
31
Step 4: Calculate the magnification (m).
The magnification is given by:
m=−di
do
m=−6.6
−20
m=2
3
Answer: The image distance is 6.6 cm and the magnification is 2
3.
Question 35
Question
A concave mirror with a focal length of 15 cm forms an image of a tree 60 cm
tall. If the image is 30 cm tall, what is the distance of the tree from the mirror?
Solution
Step 1: Use the mirror equation 1
f=1
do+1
diwhere fis the focal length, dois
the object distance, and diis the image distance.
Step 2: Given that f=−15 cm (since it’s a concave mirror with a focal
length of 15 cm), and ho= 60 cm (height of the tree) and hi= 30 cm (height
of the image), we need to find do.
Step 3: Use the magnification formula M=−di
do=hi
hoand solve for do.
Step 4: Substituting the given values into the magnification formula, we
have −di
do=hi
ho.
Step 5: Substituting the values di=−15 cm, hi= 30 cm, and ho= 60 cm,
we get −
−15
do=30
60 .
Step 6: Simplifying, 15
do=1
2.
Step 7: Cross multiplying gives 30 = do.
Therefore, the distance of the tree from the mirror is 30 cm.
32
