Q.21. In Young’s double slit experiment the separation between the slits is doubled without changing other setting of the experiment to obtain same fringe width, the distance ‘D’ of the screen from slit should be made
A. \frac{\mathrm{D}}{2}
B. \frac{\mathrm{D}}{\sqrt{2}}
C. 2\mathrm{D}
D. 4\mathrm{D}
Answer: C. 2\mathrm{D}
Explanation :-
Fringe width, \mathrm{W} = \frac{\lambda \mathrm{D}}{\mathrm{~d}}
Given: \mathrm{d} = 2\mathrm{~d}
\Rightarrow \mathrm{W}^{\prime} = \frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}}Since given,
\mathrm{W} = \mathrm{W}^{\prime}
\therefore \frac{\lambda \mathrm{D}}{\mathrm{~d}} = \frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}}