Optical Instruments & the Eye question from NEET 1995 (AIPMT)

Answer: (c)

1. Light reaching film $\propto$ aperture area $\propto \dfrac{1}{(f\text{-number})^2}$, so exposure time $t \propto (f\text{-number})^2$.
2. $\dfrac{t_2}{t_1} = \left(\dfrac{5.6}{2.8}\right)^2 = 4$
3. $t_2 = 4 \times \dfrac{1}{200} = 0.02\,\text{s}$
4. So the answer is (c) 0.02 second.