In Young's double slit experiment, the fringe width is $12 \mathrm{~mm}$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$, then the fringe width becomes (in mm):
Solution
<p>$B = 12 \times {10^{ - 3}}$</p>
<p>$\beta ' = {\beta \over \mu } = {{12 \times {{10}^{ - 3}}} \over {{4 \over 3}}}$</p>
<p>$= 9 \times {10^{ - 3}}$ m = 9 mm</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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