Optical Instruments: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Optical Instruments. Understanding the underlying formula is key to arriving at the correct answer below:

For the objective, $u_o = -200 cm$, $f_o = +40 cm$. Using $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$, we get $\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{4}{200}$, so $v_o = 50 cm$. In normal adjustment, the intermediate image forms at the focus of the eyepiece, so $u_e = f_e = 4 cm$. The required separation is $L = v_o + f_e = 50 + 4 = 54 cm$.