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An object of height $6 cm$ is placed perpendicular to the principal axis of a concave lens of focal length $5 cm$ . and the distance of the object from the lens is $10 cm$ . The size of the image will be__

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2 cm

3 cm

4 cm

6 cm

answer is A.

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Detailed Solution

The size of the image will be

2 cm

.
Given,
The height of the object,

h_{o} = + 6 cm

Focal length,

f = - 5 cm

[Negative: as the lens is concave]
The distance of the object,

u = - 10 cm

Let the distance of the image,

= v

By using the lens formula,

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

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

\Rightarrow \frac{1}{v} = \frac{1}{- 5} + \frac{1}{- 10}

\Rightarrow \frac{1}{v} = \frac{1}{- 5} - \frac{1}{10}

\Rightarrow \frac{1}{v} = \frac{- 2 - 1}{10}

\Rightarrow v = - \frac{10}{3}

Let

h_{i}

be the size of the image, then by the formula of magnification(

m

) for the lens,

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

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

\Rightarrow \frac{h_{i}}{6} = \frac{- 10 / 3}{- 10}

\Rightarrow h_{i} = 2 cm

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