Wave Optics Ques 37
37. 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 :
[2020]
(a) half
(b) four times
(c) one-fourth
(d) double
Show Answer
Answer:
Correct Answer: 37.(b)
Solution:
- (b) Fringe width $\beta=\frac{\lambda D}{d}$
Here, $\lambda=$ wavelength of light from coherent sources,
$D=$ distance of screen from the coherent sources,
$d=$ separation between coherent sources
When, $d^{\prime}=\frac{d}{2}$ and $D^{\prime}=2 D$
New Fringe width, $\beta^{\prime}=\frac{\lambda(2 D)}{d / 2}=\frac{4 \lambda D}{d}$
$\Rightarrow \beta^{\prime}=4 \beta$
Fringe width becomes 4 times.
BETA
