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:

  1. (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.



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