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

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According to the question;

Object distance (u) = -27cm; 

Focal length (f) = -18cm; 

Image distance = v; 

By mirror formula; 

\(\frac1v\,+\,\frac1u\,=\frac1f\)

\(\frac1v\,+\,\frac{1}{-27}\,=\frac{1}{-18}\)

⇒ \(\frac1v\,=\,\frac1{27}\,-\,\frac1{18}\)

\(\frac1v\,=\,\frac{-3+2}{54}\)

\(\frac1v\,=\,-\frac{1}{54}\)

⇒ v = -54cm.

Thus, screen should be placed 54cm in front of the mirror to obtain the sharp focused image. 

Height of object h1= 7cm; 

Magnification =\(\frac{h_2}{h_1}\)\(-\frac{v}{u}\)

Putting values of v and u

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

\(\frac{h_2}{7}\) = -2

⇒ h2 = 7 ×-2 = -14. 

Height of image is 14 cm. 

Negative sign means image is real and inverted. 

Thus real, inverted image of 14cm size is formed.

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