A lens forms real and virtual images of an object when the object is at u 1 and u 2 distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (Take the magnification of the real image as ' m ' )

Question

Aicharya · Accepted Answer

The formula of focal length for a convex lens is,

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ , and formula for magnification, $m = \frac{v}{u}$ .

For real image,

$u_{1} (- v e), v_{1} (+ v e) \Rightarrow \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{v_{1}} + \frac{1}{u_{1}} \Rightarrow \frac{u_{1}}{f} = \frac{u_{1}}{v_{1}} + 1 \Rightarrow \frac{u_{1}}{f} = \frac{1}{m} + 1 . . . E q (1)$

Similarly, for virtual image case, we can write

$u_{2} (- v e), v_{2} (- v e) \Rightarrow \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = - \frac{1}{v_{2}} + \frac{1}{u_{2}} \Rightarrow \frac{u_{2}}{f} = - \frac{u_{2}}{v_{2}} + 1 \Rightarrow \frac{u_{2}}{f} = \frac{1}{- 2 m} + 1 \Rightarrow \frac{- u_{2}}{f} = \frac{1}{2 m} - 1 . . . E q (2)$

From both equations, we can write

$\frac{u_{1} - u_{2}}{f} = \frac{1}{m} + 1 + \frac{1}{2 m} - 1 \Rightarrow \frac{u_{1} - u_{2}}{f} = \frac{3}{2 m} \Rightarrow f = \frac{2 (u_{1} - u_{2}) m}{3}$

A lens forms real and virtual images of an object when the object is at u 1 and u 2 distances respectively.…

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