# In Young's double-slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent courses is doubled, the fringe width becomes Option: 1 double Option: 2 half Option: 3 four times   Option: 4 one-fourth

The fringe width, $\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
Where,  $\lambda=$ the wavelength of light

d=distance of slit

D=distance of the screen from the slit.
According to the question,

$\mathrm{d}^{\prime}=\frac{\mathrm{d}}{2}\\ D=2 D\\ So \ \beta^{\prime}=\frac{\lambda \mathrm{D}^{\prime}}{\mathrm{d}^{\prime}} =\frac{\lambda *2 \mathrm{D}}{\mathrm{d} / 2}=4 \beta$
So, the fringe width will be quadrupled.

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