The width of one of the two slits in a Young’s double slit experiment is three times the other slit.
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

Solution:

Given amplitude $\propto$ slit width

Also intensity $\propto(\text { Amplitude })^{2} \propto(\text { Slit width })^{2}$

$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{3}{1}\right)^{2}=9 \Rightarrow \mathrm{I}_{1}=9 \mathrm{I}_{2}$

$\frac{I_{\min }}{I_{\max }}=\left(\frac{\sqrt{I_{1}}-\sqrt{I_{2}}}{\sqrt{I_{1}}+\sqrt{I_{2}}}\right)^{2}=\left(\frac{3-1}{3+1}\right)^{2}=\frac{1}{4}=\frac{x}{4}$

$\Rightarrow x=1.00$

 

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