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

Answers (1)

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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