In a double slit experiment, the distance between slits is increased 10 times whereas their distance from screen is halved, then what is the fringe width?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → become \( \displaystyle\frac{1}{20} \)
In Young’s double slit experiment the fringe width is \( \beta = \frac{\lambda D}{d} \), where D is the distance of the screen and d the slit separation. For the same light, \( \beta \propto D/d \).
\( \displaystyle \beta' = \frac{\lambda(D/2)}{10d} = \frac{1}{20}\cdot\frac{\lambda D}{d} = \frac{\beta}{20} \)
So the fringe width becomes \( \frac{1}{20} \) of its original value.