Question

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is $x: 4$ where $x$ is ___

Answer 1

Given amplitude $\propto$ slit width
Also intensity $\propto(\text { Amplitude })^{2} \propto(\text { Slit width })^{2}$
$\frac{ I _{1}}{ I _{2}}=\left(\frac{3}{1}\right)^{2}=9 $
$\Rightarrow I _{1}=9 I _{2}$
$\frac{ I _{\min }}{ I _{\max }}=\left(\frac{\sqrt{ I _{1}}-\sqrt{ I _{2}}}{\sqrt{ I _{1}}+\sqrt{ I _{2}}}\right)^{2}$
$=\left(\frac{3-1}{3+1}\right)^{2}=\frac{1}{4}=\frac{ x }{4}$
$\Rightarrow x =1.00$