Question

In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of light from one slit is double of that from the other slit. If $I_m$ is the maximum intensity, what is the resultant intensity when they interfere at phase difference $Q$?

• A. $\frac{I_m}{9}(1-8{\cos }^2\frac{Q}{2})$
• B. $\frac{I_m}{9}(1+8{\cos }^2\frac{Q}{2})$
• C. $\frac{I_m}{9}(1-8{\cos }^{2}Q)$
• D. $\frac{I_m}{9}(1-{\sin }^2\frac{Q}{2})$

Answer 1

The correct option is B $\frac{I_m}{9}(1+8{\cos }^2\frac{Q}{2})$

Intensity of light is directly proportional to the square of amplitude i.e $I\propto A^2$
Given : $A^{\prime }=2A$
$⟹I^{\prime }=4I$
Maximum intensity $I_m=(\sqrt{I_1}+\sqrt{I_2}{)}^2=(\sqrt{I}+\sqrt{4I}{)}^2=9I$
Resultant intensity of interference of lights having phase difference $Q$,
$I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos Q$
$I_R=I+4I+2\sqrt{\left(I\right)\left(4I\right)}\cos Q$
$I_R=I+4I+4I\cos Q$
$I_R=I+4I(1+\cos Q)$
$I_R=I+4I\left(2{\cos }^2\frac{Q}{2}\right)$
$I_R=I(1+8{\cos }^2\frac{Q}{2})$
$I_R=\frac{I_m}{9}(1+8{\cos }^2\frac{Q}{2})$