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JEE PHYSICS MCQ: WAVE OPTICS

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Application +4 | -1

In a Young’s double slit experiment, the slit separation is

0.2\;cm

, the distance between the screen and slit is

1\;m

. Wavelength of the light used is

5000\mathop \mathrm{A}\limits^ \circ

. The distance between two consecutive dark fringes (in

mm

) is

View answer and explanation

Correct option: A

0.25

Option A, 0.25, is correct because it is the calculated fringe width using the formula

\beta = \frac{{\lambda D}}{d}

with the given values.

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