Question

An eraser having a height of 4 cm when placed 4 cm away from a mirror, produces the image at a distance of 8 cm. How tall (in cm) will it appear on the other side of the mirror?

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Answer 1

Answer:

Correct option is

concave mirror of radius of curvature 12 cm

Given,

h

0

=1cm;h

i

=3cm

u=−4cm

m=

h

0

h

1

=

1

3

=3

m=

v

v

=3⇒v=3u

f

1

=

v

1

+

u

1

=

12

1

−

4

1

=

12

1−3

=

12

−2

=

6

1

f=−6cm

The mirror must be concave mirror of focal length 6 cm i.e.,

R=2f=12cm

Answer 2

The height of image is $4 cm$.

Explanation:

• The mirror discussed here is a plane mirror.
• According to laws of reflection for plane mirror, the distance of image from mirror is equal to the distance of object from the mirror.
• Distance of the image from the object is $8 cm$ which is double the distance of object from mirror.
• The image is formed behind the mirror and is formed at the same distance from the mirror as the object distance from mirror.
• Object height is $4 cm$ which is kept in front of a plane mirror.
• Image formed in plane mirror is laterally inverted, virtual i.e. behind the mirror and of the same size as object and at the same distance as the object from mirror.
• So, the height of the image is $4 cm$.