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

Answer: 1

Given amplitude $\propto$ slit width
Also intensity $\propto (\text{ Amplitude }{)}^2\propto (\text{ Slit width }{)}^2$
$\frac{I_1}{I_2}={\left(\frac{3}{1}\right)}^2=9$
$\Rightarrow I_1=9I_2$
$\frac{I_{\min }}{I_{\max }}={\left(\frac{\sqrt{I_1}-\sqrt{I_2}}{\sqrt{I_1}+\sqrt{I_2}}\right)}^2$
$={\left(\frac{3-1}{3+1}\right)}^2=\frac{1}{4}=\frac{x}{4}$
$\Rightarrow x=1.00$