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Question

An object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of focal length 15.0 cm.
(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?
(ii) Find the size of the image.
(iii) Draw a ray diagram to show the formation of image in this case.

Solution
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(i)
Given:
The size of object = $$4.0cm$$
Distance of the object = $$25cm$$ (-ve as it is in front of mirrors)
focal length of concave mirror = $$15cm$$ (-ve)
Distance of the image is given by-

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

$$\dfrac{1}{v}+\dfrac{1}{-25}=\dfrac{1}{-15}$$

$$\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{25}$$

$$=\dfrac{-10+6}{150}$$

$$\dfrac{-4}{150}\Rightarrow v=-37.5cm$$
The screen should be at $$37.5cm$$ in front of the lens.

(ii)
The size of the image is given by-
$$\dfrac{h_i}{h_0}=\dfrac{-v}{u}$$
$$\Rightarrow h_i=\dfrac{-v\times h_0}{u}=\dfrac{-(-37.5)\times 4}{-25}$$
$$h_i=-\dfrac{150}{25}=-6cm$$
The image will be $$6cm$$ high and it will be inverted.

(iii)
Refer image 1
The image will be formed at a distance of $$37.5cm$$ from the mirror.
It will be an inverted image.

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