MCQ
In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
- Aone$-$fourth
- Bdouble
- Chalf
- ✓four times
✓
Answer
Correct option: D.
four times
d
$\beta=\frac{\lambda D}{d}$
$\beta=\frac{\lambda D}{d}$
$\beta^{\prime}=\frac{\lambda D^{\prime}}{d^{\prime}}$
$D ^{\prime}=2 D , d ^{\prime}=\frac{ d }{2}$
$\beta^{\prime}=\frac{\lambda \times 2 D}{d / 2}=\frac{4 \lambda D}{d}$
$\beta^{\prime}=4 \beta$
Fringe with becomes $4$ times
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