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Question: A convex lens of focal length f is placed somewhere in between an object and a screen. The distance ...

A convex lens of focal length f is placed somewhere in between an object and a screen. The distance between the object and the screen is x. If the numerical value of the magnification produced by the lens is m, then the focal length of the lens is –

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(m−1)2mx\frac { ( m - 1 ) ^ { 2 } } { m } x

Answer

Explanation

Solution

v−u\frac { v } { - u } = – m and u + v = x Ž u = x1+m\frac { \mathrm { x } } { 1 + \mathrm { m } }

By lens formula v = mu = mxm+1\frac { \mathrm { mx } } { \mathrm { m } + 1 }

1f=1v−1u\frac { 1 } { \mathrm { f } } = \frac { 1 } { \mathrm { v } } - \frac { 1 } { \mathrm { u } }

1f=1(mxm+1)−1−x(m+1)\frac { 1 } { f } = \frac { 1 } { \left( \frac { m x } { m + 1 } \right) } - \frac { 1 } { \frac { - x } { ( m + 1 ) } } = m+1 m+m+1x\frac { \mathrm { m } + 1 } { \mathrm {~m} } + \frac { \mathrm { m } + 1 } { \mathrm { x } }

= (m+1)x⋅(1+m)m\frac { ( m + 1 ) } { x } \cdot \frac { ( 1 + m ) } { m }

f=mx(m+1)2\mathrm { f } = \frac { \mathrm { mx } } { ( \mathrm { m } + 1 ) ^ { 2 } }