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Question: Answered & Verified by Expert
A small candle, $2.5 \mathrm{~cm}$ in size is placed at $27 \mathrm{~cm}$ in front of a concave mirror of radius of curvature $36 \mathrm{~cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
PhysicsRay Optics
Solution:
1793 Upvotes Verified Answer
Here, $u=-27 \mathrm{~cm}, \mathrm{R}=-36 \mathrm{~cm}$, $\mathrm{f}=\frac{-\mathrm{R}}{2}=-\frac{36}{2}=-18 \mathrm{~cm}$
Height of object, $\mathrm{h}=2.5 \mathrm{~cm}, \mathrm{v}=$ ?
As, $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}$
$$
=\frac{1}{-18}-\frac{1}{(-27)}=-\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}=-\frac{1}{54}
$$
$\therefore \mathrm{v}=-54 \mathrm{~cm}$.
The screen should be placed at $54 \mathrm{~cm}$ from the mirror on the same side as the object.
Magnification, $\mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{\mathrm{v}}{\mathrm{u}}$
or, $h^{\prime}=-\frac{\mathrm{v}}{\mathrm{u}} . \mathrm{h}=-\frac{(-54)}{(-27)} \times 2.5=-2 \times 2.5=-5 \mathrm{~cm}$.
The image is real and inverted. If the candle is moved closer to the mirror, the screen has to be moved away from the mirror.

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