JEE PHYSICS MCQ: WAVE OPTICS

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In Young's double slit experiment, slits are separated by

2 \mathrm{~mm}

and the screen is placed at a distance of

1.2 \mathrm{~m}

from the slits. Light consisting of two wavelengths

6500\,\,\mathop {\text{A}}\limits^ \circ

and

5200\,\,\mathop {\text{A}}\limits^ \circ

are used to obtain interference fringes. Then, the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

View answer and explanation

Correct option: A

0.312 \mathrm{~mm}

Option A (

0.312 \mathrm{~mm}

) is correct because it is the calculated separation between the fourth bright fringes of the two wavelengths.

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