Question

In YDSE, the width of one slit is different from other, so that the amplitude of light from one slit is double that from the other. If $I_m$ is the maximum intensity, the intensity when they interfere with a phase difference of $\Phi$ is

• A. $\frac{I_m}{9}(4+5cos\Phi )$
• B. $\frac{I_m}{3}(1+2co{s}^2\frac{\Phi }{2})$
• C. $\frac{I_m}{5}(1+4co{s}^2\frac{\Phi }{2})$
• D. $\frac{I_m}{9}(1+8co{s}^2\frac{\Phi }{2})$

Answer 1

The correct option is D $\frac{I_m}{9}(1+8co{s}^2\frac{\Phi }{2})$

$Let{I}_1=I;I_2=4I$
$I_m=(\sqrt{I_1}+\sqrt{I_2}{)}^2=(\sqrt{I}+2\sqrt{I}{)}^2=9I$
$I=I_1+I_2+2\sqrt{I_1{I}_1}cos\Phi$
$=I+4I+2\sqrt{I\left(4I\right)}cos\phi =5I+4I[2co{s}^2\left(\frac{\phi }{2}\right)-1]=I[1+8co{s}^2\left(\frac{\Phi }{2}\right)]$
$\frac{I_m}{9}[1+8co{s}^2\left(\frac{\Phi }{2}\right)]$