An object and a screen are fixed at a distance D apart. When a lens of focal length f is moved between the object and the screen, sharp images of the object are formed on the screen for two positions of the lens. The magnifications produced at these two positions are $M_{1}$ and $M_{2}$ and separation between these two positions is d. Then

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$| M_{1} | - | M_{2} | = \frac{d}{f}$

$D > 2 f$

$D > 4 f$

$M_{1} M_{2} = 1$

answer is B, C, D.

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Detailed Solution

$\frac{1}{+ (D - u)} - \frac{1}{(- u)} = \frac{1}{+ f} \Rightarrow u^{2} - D u + D f = 0 \Rightarrow u = \frac{1}{2} (D \pm \sqrt{D^{2} - 4 D f}) S o v = D - u = \frac{1}{2} (D \mp \sqrt{D^{2} - 4 D f}) F r o m a b o v e r e s u l t s, w e c a n w r i t e u_{1} = v_{2} a n d u_{2} = v_{1} N o w M_{1} = \frac{v_{1}}{u_{1}}, M_{2} = \frac{v_{2}}{u_{2}} = \frac{u_{1}}{v_{1}} a n d a l s o f o r a c o n v e x l e n s \frac{1}{f} = \frac{1}{+ v_{1}} - \frac{1}{(- u_{1})} = \frac{u_{1} + v_{1}}{u_{1} v_{1}} T h e r e f o r e |M_{1}| - |M_{2}| = \frac{v_{1}}{u_{1}} - \frac{u_{1}}{v_{1}} = \frac{(v_{1} - u_{1}) (V_{1} + u_{1})}{v_{1} u_{1}} = \frac{(u_{2} - u_{1})}{f} = \frac{d}{f}$

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