$O A$ is an object with size $4 c m$ and $O^{'} A^{'}$ is the image of the object. The height of the image is and it is.

erect

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inverted

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3 c m

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6 c m

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2 c m

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Solution

The correct option is E $2 c m$
Given below,
Object size = $4 c m$
Object distance = $30 c m$
Image distance = $15 c m$

We know that,
Magnification = $\frac{v}{u}$ = $\frac{h^{i}}{h}$

$\Rightarrow$ $\frac{+ 15}{- 30}$ = $\frac{h^{i}}{(4)}$

$h^{i}$ = $\frac{- 4}{2}$ = $- 2 c m$

Image height is $2 c m$ and the negative sign shows it is inverted.

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