Question

An object 5cm tall is placed in front of a spherical mirror at 20 cm distance from the mirror. If virtual image of 10 cm tall was formed behind the mirror, find the focal length of the mirror and the position of the images. Name the type of mirror used

Answer 1

HEYA MATE, HERE IS UR ANSWER

$<b><i>=5 cm$

H'=10 cm (image is virtual)

The image is enlarged, so the mirror used is Concave mirror. Convex mirror never has an enlarged image.

U=-20 cm (placed before the mirror )

M=$\frac {H'}{H}$

M=$\frac{10}{5}$

M=+2

M=$\frac {-V}{U}$

2=$\frac {-V}{-20}$

By cross multiplication

2×-20=-v

-40=-v

v=40 cm

Position of the image =40 cm behind the mirror

Using mirror formula

$\frac {1}{f}$=$\frac {1}{u}$+$\frac {1}{v}$

$\frac {1}{f}$=$\frac {1}{-20}$+$\frac {1}{30}$

$\frac {1}{f}$=$\frac {-3+2}{60}$

$\frac {1}{f}$=$\frac {-1}{60}$

Focal length =-60 cm

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@laughterqueen

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Answer 2

Explanation:

It is given that,

Height of the object, h = 5 cm

Object distance, u = -20 cm

Height of the object, h' = 10 cm

Magnification of the mirror, $m=\dfrac{-v}{u}=\dfrac{h'}{h}$

$\dfrac{-v}{-20}=\dfrac{10}{5}$

v = 40 cm

So, the image is formed behind the mirror at a distance of 40 cm. The mirror is concave mirror.

Using mirror's equation :

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{f}=\dfrac{1}{-20}+\dfrac{1}{40}$

f = -40 cm

Hence, this is the required solution.