Question 114: Wave Optics

Question

In the Young's double slit experiment, the distance between the slits varies in time as
d(t) = d
0
+ a
0
sin$$\omega$$t; where d
0
, $$\omega$$ and a
0
are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

Accepted Answer

Fringe Width, $$\beta  = {{\lambda D} \over d}$$
$${\beta _{\max }} \Rightarrow {d_{\min }}$$ and $${\beta _{\min }} \Rightarrow {d_{\max }}$$
$$d = {d_0} + {a_0}\sin \omega t$$
$${d_{\max }} = {d_0} + {a_0}$$ and $${d_{\min }} = {d_0} - {a_0}$$
$$	herefore$$ $${\beta _{\min }} = {{\lambda D} \over {{d_0} + {a_0}}}$$ and
$$	herefore$$ $${\beta _{\max }} = {{\lambda D} \over {{d_0} - {a_0}}}$$
$${\beta _{\max }} - {\beta _{\min }} = {{\lambda D} \over {{d_0} - {a_0}}} - {{\lambda D} \over {{d_0} + {a_0}}} = {{2\lambda D{a_0}} \over {d_0^2 - a_0^2}}$$

Answer

$${{2\lambda D({d_0})} \over {(d_0^2 - a_0^2)}}$$

Answer

$${{2\lambda D{a_0}} \over {(d_0^2 - a_0^2)}}$$

Answer

$${{\lambda D} \over {d_0^2}}{a_0}$$

Answer

$${{\lambda D} \over {{d_0} + {a_0}}}$$

Wave Optics

In the Young's double slit experiment, the distance between the slits varies in time as d(t) = d 0 + a 0 sin$\omegat; where d 0 , \omega$ and a 0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

Choose the correct answer: