In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to
Solution
$\beta$ = ${{\lambda D} \over d}$
<br><br>and $\theta$ = ${\beta \over D}$
<br><br>$\Rightarrow$ $\theta$ = ${\lambda \over d}$
<br><br>= ${{500 \times {{10}^{ - 9}}} \over {0.05 \times {{10}^{ - 3}}}}$
<br><br>= 0.01 rad
<br><br>= 0.57<sup>o</sup>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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