Wave Optics - Result Question 7
7. 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
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Answer:
Correct Answer: 7. (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.
