JEE PHYSICS MCQ: WAVE OPTICS

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Application +4 | -1

In a Young's double slit experiment, the slits are

2 \mathrm{~mm}

apart and are illuminated with a mixture of two wavelength

\lambda_0=750 \mathrm{~nm}

and

\lambda=900 \mathrm{~nm}

. The minimum distance from the common central bright fringe on a screen

2 m

from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

View answer and explanation

Correct option: C

4.5 \mathrm{~mm}

Option C (

4.5 \mathrm{~mm}

) is correct because the minimum distance is found by equating the path differences for the two wavelengths and solving for the order of the fringe.

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