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  • In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double that form the other slit. If <img alt="\mathrm{I_m}" src=

In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double that form the other slit. If \mathrm{I_m} be the maximum intensity, the resultant intensity when they interfere at phase difference \mathrm{\phi} is given by:

Option: 1

\mathrm{\frac{I_m}{3}\left(1+2 \cos ^2 \frac{\phi}{2}\right)}


Option: 2

\mathrm{\frac{I_m}{5}\left(1+4 \cos ^2 \frac{\phi}{2}\right)}


Option: 3

\mathrm{\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)}


Option: 4

\mathrm{\frac{I_m}{9}\left(8+\cos ^2 \frac{\phi}{2}\right)}


Answers (1)

best_answer

Here, \mathrm{A_2=2 A_1}

  \mathrm{\because \quad \text { Intensity } \propto(\text { Amplitude })^2}

\mathrm{ \therefore \frac{\mathrm{I}_2}{\mathrm{I}_1}=\left(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\right)^2=\left(\frac{2 \mathrm{~A}_1}{\mathrm{~A}_1}\right)^2=4 }

\mathrm{ \mathrm{I}_2=4 \mathrm{I}_1 }

Maximum intensity,
\mathrm{ \begin{aligned} & I_m=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=\left(\sqrt{I_1}+\sqrt{4 I_1}\right)^2=\left(3 \sqrt{I_1}\right)^2=9 I_1 \\ & \text { or } \quad I_1=\frac{I_m}{9} \end{aligned} }
Resultant intensity, \mathrm{I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi}

\mathrm{ \begin{aligned} & =\mathrm{I}_1+4 \mathrm{I}_1+2 \sqrt{\mathrm{I}_1\left(4 \mathrm{I}_1\right)} \cos \phi \\\\ & =5 \mathrm{I}_1+4 \mathrm{I}_1 \cos \phi=\mathrm{I}_1+4 \mathrm{I}_1+4 \mathrm{I}_1 \cos \phi \\\\ & =\mathrm{I}_1+4 \mathrm{I}_1(1+\cos \phi) \\\\ & =\mathrm{I}_1+8 \mathrm{I}_1 \cos ^2 \frac{\phi}{2} \quad\left(\because 1+\cos \phi=2 \cos ^2 \frac{\phi}{2}\right) \\\\ & =\mathrm{I}_1\left(1+8 \cos ^2 \frac{\phi}{2}\right) \\ & \end{aligned} }
Putting the value of \mathrm{\mathrm{I}_1} from eqn. (i), we get

\mathrm{ I=\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right) }

Posted by

Ritika Jonwal

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