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:

• A

  only $\frac{d}{4}$
• B

  only $\frac{d}{2}$
• C

  more than $\frac{d}{4}$ but less than $\frac{d}{2}$
• Correct answer
  D

  less than or equal to $\frac{d}{4}$

Solution

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) $$ f=\frac{d}{4} $$

Hence, the correct answer is option: 'D'.