Four students Ram, Shyam, Ghanshyam and Radheshyam perform an optical bench experiment to final focal length of a convex lens. Ram forms image of a distant luminous object on a small screen mounted on the optical bench. Other 3 students perform usual optical bench experiment. Their position of object needle, lens and image needle are given below. Which of the student is most accurate in his result ?
Name Object Lens Image
Ram ----- 15.0 cm 30.0 cm
Shyam 15.0 cm 75.0 cm 95.0 cm
Ghanshyam 15.0 cm 45.0 cm 75.0 cm
Radheshyam 15.0 cm 60.0 cm 82.5 cm

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Ghanshyam

Ram

Radheshyam

All are equally accurate

answer is B.

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Detailed Solution

$\frac{d f}{f^{2}} = \frac{d u}{u^{2}} + \frac{d v}{v^{2}}, d u = d v$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{u^{2}} + \frac{1}{v^{2}} - \frac{2}{u v} = \frac{1}{f^{2}}$

$\frac{d f}{f^{2}} = d u (\frac{1}{f^{2}} + \frac{2}{u v}) = d u (\frac{1}{f^{2}} + \frac{2}{u} (\frac{1}{f} + \frac{1}{u}))$

$d f = d u (1 + \frac{2 f}{u} + \frac{2 f^{2}}{u^{2}}) \Rightarrow d f$ is minimum when $0 = - \frac{2 f}{u^{2}} - \frac{4 f^{2}}{u^{3}}$

$u = - 2 f$ (for $u = \infty,$ 2nd derivative is zero so it is not a minimum)

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