Monochromatic light from a narrow slit illuminates two parallel narrow slits producing an interference pattern on a screen. The separation between the two slits is now doubled and the distance between the screen and the slits is reduced to half. The fringe width

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becomes one-fourth

is doubled

remains the same

becomes four times

answer is C.

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Detailed Solution

Fringe width, $β = \frac{λ D}{d} \Rightarrow β \propto \frac{D}{d}$

$\frac{β_{2}}{β_{1}} = \frac{D_{2}}{d_{2}} \times \frac{d_{1}}{D_{1}} = \frac{\frac{1}{2} D_{1}}{2 d_{1}} \times \frac{d_{1}}{D_{1}} = \frac{1}{4} \Rightarrow β_{2} = \frac{β_{1}}{4}$

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