In the Young's double slit experiment, the distance between the slits varies in time as
d(t) = d0 + a0 sin$\omega$t; where d0, $\omega$ and a0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :
Solution
Fringe Width, $\beta = {{\lambda D} \over d}$<br><br>${\beta _{\max }} \Rightarrow {d_{\min }}$ and ${\beta _{\min }} \Rightarrow {d_{\max }}$<br><br>$d = {d_0} + {a_0}\sin \omega t$<br><br>${d_{\max }} = {d_0} + {a_0}$ and ${d_{\min }} = {d_0} - {a_0}$<br><br>$\therefore$ ${\beta _{\min }} = {{\lambda D} \over {{d_0} + {a_0}}}$ and <br><br>$\therefore$ ${\beta _{\max }} = {{\lambda D} \over {{d_0} - {a_0}}}$<br><br>$${\beta _{\max }} - {\beta _{\min }} = {{\lambda D} \over {{d_0} - {a_0}}} - {{\lambda D} \over {{d_0} + {a_0}}} = {{2\lambda D{a_0}} \over {d_0^2 - a_0^2}}$$
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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