Problems on Concave Mirror:
Example Problem:
- An object 4cm in size, is placed at 25cm in front of a concave mirror of a focal length of 15cm. At what distance from the mirror would a screen be placed in order to obtain a sharp image? Find the nature and size of the image.
Ans: Given Data (according to sign convention):
Object height (ho) = +4cm
Object distance (u) = - 25cm
Focal length (f) = - 15cm
Image distance (v) = ?
Image height (hi) = ?
Substitute the above values in the mirror formula,
Mirror Formula: 1/f = 1/u + 1/v
1/(-15) = 1/v + 1/(-25)
1/(-15) = 1/v - 1/25
1/v = -1/15 + 1/25
1/v = -5+3 / 75
1/v = -2/75
v = -75/2
v = -37.5cm
Hence screen should be placed at a distance of 37.5cm in front of the mirror. The image is real.
Magnification m = hi / ho = -v / u
by substituting the above values, we get hi / 4 = -(-37.5) / (-25)
hi = -(37.5 × 4) / 25
hi = -150 / 25
hi = -6cm
So the image is inverted and enlarged.
WORKOUT PROBLEMS:
1. Find the distance of the image when an object is placed on the principal axis at a distance of 10cm in front of a concave mirror whose radius of curvature is 8cm.
Sol: Given Data (according to sign convention):
Image distance (v) =?
Object distance (u) = -10cm
Radius of Curvature (R) = -8cm (since concave mirror)
we know that R = 2f => f = R/2 => f = -8/2 => f = -4cm
Applying Mirror Formula: 1/f = 1/v + 1/u
1/(-4) = 1/v + 1/(-10)
-1/4 = 1/v - 1/10
1/v = -1/4 +1/10
1/v = -5+2/20 => 1/v = -3/20
v = -20/3
v = -6.6cm
Hence the image is at a distance of 6.6cm in front of the mirror. Therefore the image is real.
2. An object is placed 20cm from a concave mirror with a radius of curvature of 60cm. Find the position of the image and its magnification.
Sol: Given Data (according to sign convention):
Object distance (u) = -20cm
Radius of Curvature (R) = -60cm (since concave mirror)
Image distance (v) =?
we know that R = 2f => f = R/2 => f = -60/2 => f = -30cm
Using mirror formula: 1/f = 1/v + 1/u
1/(-30) = 1/v + 1/(-20)
-1/30 = 1/v - 1/20
1/v = -1/30 + 1/20
1/v = -2+3/60 => 1/v = 1/60
v = 60cm
Position: Hence the image is placed at a distance of 15cm behind the mirror. Therefore the image is virtual.
Magnification (m) = -v/u
= -60/(-20) = 3 => m = 3
Characteristics of image: The image is virtual, erect, and enlarged.
3. The focal length of a concave mirror is 30cm. Find the position of the object in front of the mirror so that the image is 3 times the size of the object.
Sol: Given Data (according to sign convention):
Focal length (f) = -30cm (since concave mirror)
Magnification (m) = 3
Case: 1 - If the image is real and inverted:
m = -v/u = -3
v = 3u
Using mirror formula: 1/f = 1/v + 1/u
1/(-30) = 1/3u + 1/u
-1/30 = 4/3u
-3u = 120
u = -40cm
Therefore v = 3(-40) = -120cm
Hence the object is placed at a distance of 40cm in front of the concave mirror.
Case: 2 - If the image is virtual and erect:
m = -v/u = 3
v = -3u
Using mirror formula: 1/f = 1/v + 1/u
1/(-30) = 1/(-3u) + 1/u
-1/30 = -1/3u +1/u
-1/30 = -1+3/3u
-1/30 = 2/3u => -3u = 60
u = -20cm
Therefore v = -3(-20) = +60cm
4. An object, 4cm in size, is placed 25cm in front of a concave mirror of focal length 15cm. At what distance from the mirror, should a screen be placed in order to obtain a sharp image? Find the nature and size of the image.
Sol: Given Data (according to sign convention):
Height of the object (ho) = 4cm
Object distance (u) = -25cm
Focal length (f) = -15cm (since concave mirror)
Image distance (v) =?
Using mirror formula: 1/f = 1/v + 1/u
1/(-15) = 1/v + 1/(-25)
-1/15 = 1/v - 1/25
1/v = -1/15 + 1/25
1/v = -5+3/75 => 1/v = -2/75 => v = -75/2
v = -37.5cm
Position: Hence the screen should be placed at 37.5cm from the mirror. The image is real.
Magnification (m) = hi / ho = -v/u
hi = -v × ho / u
hi = -(-37.5) × 4 / (-25)
hi = +150 / (-25)
Size: hi = - 6cm
Characteristics of image: The image is real, inverted, and enlarged.
5. Focal length of a concave mirror is 20cm. Where should an object be placed in front of the concave mirror so as to obtain an image which is real, inverted as of the same size as that of the object?
Sol: Given Data (according to sign convention):
Focal length (f) = -20cm
Object distance (u) =?
To obtain an image that is real, inverted, and of the same size as that of the object, the object should be placed at the centre of curvature (C).
Also, we observed that in a concave mirror, the centre of curvature is at a distance twice the focal length,
Therefore 2f = 2 × (-20)
2f = -40cm (since concave mirror)
Hence object should be placed at a distance of 40cm from the concave mirror.
6. An object of height 5cm is placed at a 30cm distance on the principal axis in front of a concave mirror of focal length 20cm. Find the image distance and size of the image.
Ans: 
Given Data (according to sign convention):
Object height (ho) = 5cm
Object distance (u) = -30cm
Focal length (f) = -20cm (since concave mirror)
Image distance (v) =?
Using mirror formula: 1/f = 1/v + 1/u
1/(-20) = 1/v + 1/(-30)
-1/20 = 1/v -1/30
1/v = -1/20 + 1/30
1/v = -3+2 / 60 => 1/v = -1/60
v = -60cm
Hence the image is placed at a distance of 60cm from the mirror. Therefore the image is real.
Magnification (m) = hi/ho = -v/u
hi/5 = -(-60)/(-30)
hi/5 = -2
hi = -10cm
Characteristics of image: The image is real, inverted and enlarged.
7. An object of 6cm is placed at a distance of 30cm in front of a concave mirror of focal length of 10cm. At what distance from the mirror, will the image be formed? What are the characteristics of the image?
Sol: Given Data (according to sign convention):
Object height (ho) = 6cm
Object distance (u) = -30cm
Focal length (f) = -10cm (since concave mirror)
Image distance (v) =?
Using mirror formula: 1/f = 1/v + 1/u
1/(-10) = 1/v + 1/(-30)
-1/10 = 1/v -1/30
1/v = -1/10 + 1/30
1/v = -3+1 / 30 => 1/v = -2/30 => 1/v = -1/15
v = -15cm
Hence the image is formed at a distance of 15cm from the concave mirror. Therefore the image is real.
Magnification (m) = hi/ho = -v/u
hi/6 = -(-15) / (-30)
hi/6 = -1/2
Size: hi = -3cm
Characteristics of image: The image is real, inverted and diminished.
8. A spherical mirror produces an image of magnification -1 on a screen placed at a distance of 50cm from the mirror.
(a) Write the type of mirror.
(b) FInd the distance of the image from the object.
(c) What is the focal length of the mirror.
(d) Draw the ray diagram to show the image formation
Sol: (a) Type of mirror: It is a concave mirror as its magnification is negative and the image formed is real.
(b) Distance of the object from the mirror:
Magnification, m = -v/u =-1
Object distance, u = -50cm
Substitute 'u' value in magnification formula, we get
-v/(-50) = -1
v = -50cm
Hence the distance of the object from the mirror is 50cm
(c) Focal Length of the mirror:
By using mirror formula, 1/f = 1/v + 1/u
substituting 'u' and 'v' values in the above formula
1/f = 1/(-50) + 1/(-50)
1/f = -1/50 - 1/50
1/f = -2/50 => 1/f = -1/25
f = -25cm
(d) Ray Diagram:
9. Mahira wants to project the image of a candle flame on a screen 60cm in front of a mirror by keeping the flame at a distance of 15cm from its pole.
(a) Write the type of mirror she should use.
(b) Find the linear magnification of the image produced.
(c) What is the distance between the object and its image.
(d) Find its focal length and radius of curvature.
Ans: (a) Type of mirror: It is a concave mirror as it forms a real image that presents on the same side of the mirror.
(b) Magnification of the image produced:
Given object distance, u = -15cm and Image distance, v = -60cm
Magnification, m = -v/u
m = -(-60) / (-15)
m = -4cm (-ve sign indicates that image formed is real and inverted)
(c) Distance between object and its image:
v-u = 60-15 (as we taking the difference between distances, so no need to use sign convention)
v-u = 45cm
(d) Finding Focal Length and Radius of Curvature:
Given object distance, u = -15cm and Image distance, v = -60cm
By using the mirror formula, 1/f = 1/v + 1/u
1/f = 1/(-60) + 1/(-15)
1/f = -1/60 - 1/15
1/f = -1-4 / 60
1/f = -5/60
1/f = -1/12
f = -12cm
we know that the Radius of curvature, R = 2f
R = 2(-12)
R = -24cm
10. Find the distance at which an object to be placed for getting a real, inverted and enlarged image at 45cm using a concave mirror of focal length 20cm.
Sol: Given Data (according to sign convention):
Image distance, v = -45cm
focal length, f = -20cm (since concave mirror)
Using mirror formula: 1/f = 1/v + 1/u
1/(-20) = 1/(-45) + 1/u
-1/20 = -1/45 + 1/u
1/u = -9+4 / 180
1/u = -5/180 => u = -180/5
u = -36cm
Problems on Convex Mirror:
WORKOUT PROBLEMS:
1. A 4.5cm candle is placed 12cm away from a convex mirror of focal length 15cm. Give the location of the image and magnification. Describe what happens to the image as the needle is moved farther from the mirror.
Sol: Given Data (according to sign convention):
Object height (ho) = 4.5cm
Object distance (u) = -12cm
Focal length (f) = +15cm (since convex mirror)
Image distance (v) =?
Magnification (m) =?
Using mirror formula: 1/f = 1/v + 1/u
1/(15) = 1/v + 1/(-12)
1/15 = 1/v - 1/12
1/v = 1/15 + 1/12
1/v = 4+5 / 60
1/v = 9/60
v = 6.6cm
Position: The image is placed at a distance of 6.6cm behind the mirror.
Magnification (m) = hi/ho = -v/u
m = -v/u = -6.6/(-12) = 0.5
m = 0.6
If we move the object farther from the mirror, the image will be diminished
Also m = hi/ho
0.6 = hi/4.5
hi = 0.6 × 4.5
hi = 2.7cm
Characteristics of the image: The image is virtual, erect and diminished.
2. A convex mirror with a radius of curvature of 3m is used as a rearview mirror for a vehicle. If a bus is located at 5m from this mirror, find the position, nature and size of the image.
Sol: Given Data (according to sign convention):
Radius of curvature (R) = +3m (since convex mirror)
Object distance (u) = -5m
Image distance (v) =?
Magnification (m) =?
we know that, R = 2f
f = R/2 = 3/2
f = +1.5m (since convex mirror)
Using mirror formula: 1/f = 1/v + 1/u
1/1.5 = 1/v + 1/(-5)
1/1.5 = 1/v - 1/5
1/v = 1/1.5 +1/5
1/v = 10+3 / 15
1/v = 13/15
v = 15/13
v = +1.15m
Position: The image is formed at a distance of 1.15m inside the mirror. As v is positive, it indicates that the image is virtual.
Size: Magnification (m) = hi/ho = -v/u
m = -v/u
m = -1.15/(-5)
m = 0.23
As the magnification of the image is positive and less than zero, the image is erect and smaller than the object (diminished).
Characteristics of the image: The image is virtual, erect and diminished.
3. A 2.5cm candle is placed 12cm away from a convex mirror of focal length 30cm. Find the position of the image and its magnification.
Sol: Given Data (according to sign convention):
Object height (ho) = 2.5cm
Object distance (u) = -12cm
Focal length (f) = +30cm (since convex mirror)
Image distance (v) =?
Magnification (m) =?
Using mirror formula: 1/f = 1/v + 1/u
1/(30) = 1/v + 1/(-12)
1/30 = 1/v - 1/12
1/v = 1/30 + 1/12
1/v = 2+5 / 60
1/v = 7/60
v = 60/7
v = +8.6cm
Position: The image is formed at a distance of 8.6cm inside the mirror. As v is positive, it indicates that the image is virtual.
Magnification (m) = hi/ho = -v/u
m = -v/u
m = -8.6/(-12)
m = 0.7
As the magnification of the image is positive and less than zero, the image is erect and smaller than the object (diminished).
Characteristics of the image: The image is virtual, erect and diminished.
