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# An object of size 7.0 cm is placed at a distance of 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of the i...

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An object of size 7.0 cm is placed at a distance of 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of the image.

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Given: h1 = 7 cm, u = -27 cm, f = -18 cm, v = ?

Applying the formula $$\frac{1}{f}$$ = $$\frac{1}{v}$$ + $$\frac{1}{u}$$

we get,

$$\frac{1}{-18}$$ = $$\frac{1}{v}$$ + $$\frac{1}{-27}$$

⇒ $$\frac{1}{-18}$$ + $$\frac{1}{27}$$ = $$\frac{1}{v}$$

⇒ $$\frac{-3 + 2}{54}$$ = $$\frac{1}{v}$$

⇒ $$\frac{-1}{54}$$ = $$\frac{1}{v}$$

⇒ v = -54 cm

Hence, the screen should be placed at a distance of 54 cm from the mirror.

Now, applying the formula m = -$$\frac{v}{u}$$ = $$\frac{h_2}{h_1}$$,

we get

⇒ $$\frac{-(-54)}{-27}$$ = $$\frac{h_2}{7}$$

⇒ h2 = -2 x 7 = -14 cm

∴ Size of the image is 14 cm.

Nature of image: Negative sign shows that the image is inverted and enlarged.

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