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What will be the size of the image, formed when an object of $3 cm$ in height is placed at a distance of $10 cm$ in front of a converging mirror of focal length of $20 cm$ ?

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large

small

same size

infinitely large

answer is A.

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Detailed Solution

The size of the image, formed when an object of

3 cm

in height is placed at a distance of

10 cm

in front of a converging mirror of focal length of

20 cm

, will be large.
Given,
Distance of the object from the mirror,

u = - 10 cm

(for a concave mirror)
The focal length (

f

) of the mirror,

f = - 20 cm

Height of the object

{, h}_{0} = 3 cm

By using the mirror formula, we know that,

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

\frac{1}{v} + \frac{1}{(- 10)} = \frac{1}{(- 20)}

\frac{1}{v} = \frac{1}{(- 20)} - \frac{1}{(- 10)}

\frac{1}{v} = \frac{1}{20}

v = + 20 cm

Now by using the magnification formula.

m = \frac{h_{i}}{h_{o}}

\frac{h_{i}}{h_{o}} = - \frac{v}{u}

\frac{h_{i}}{3} = + \frac{20}{(- 10)}

h_{i} = - \frac{2 \times 3}{(- 1)}

h_{i} = + 6 cm

So, the height of the image will be 6 cm, and the positive sign implies that the image formed is above the principal axis and its size is large.

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