A bulb is located on a wall. Its image is to… - NEET UG 2002 Physics PYQ

NEET UG 2002 — Physics Optics

2002mcqhard

A bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distance between parallel walls is ' $d$ ' then the required focal length of the lens placed in between the walls is:

Official previous-year question

Held on 30 Apr 2002 · Verified 9 Jul 2026.

Options

A
only $\frac{d}{4}$
B
only $\frac{d}{2}$
C
more than $\frac{d}{4}$ but less than $\frac{d}{2}$
D
less than or equal to $\frac{d}{4}$

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Solution

NEET 2002 Physics, Ray Optics — solution figure Since $v+u=d \Rightarrow v=d-u$ From lens formula $\begin{aligned} \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \quad \frac{1}{f} & =\frac{1}{d-u}-\frac{1}{-u} \\ & =\frac{d}{(d-u) u} \\ \Rightarrow \quad f & =\frac{d u-u^2}{d} \ldots(\mathrm{i}) \end{aligned}$ Differentiating eq. (i) we have $\frac{d f}{d u}=\frac{1}{d}(d-2 u)$ for maximum value of $f_1=\frac{d f}{d u}=0$ $\begin{aligned} & \Rightarrow \frac{1}{d}(d-2 u)=0 \Rightarrow d=24 \\ & \Rightarrow u=\frac{d}{2} \end{aligned}$ putting these value in eq. (i) $f=\frac{d}{4}$

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