Question

In a Young's double slit experiment the intensity at a point where the path difference is $\frac{\lambda }{6}$ ($\lambda$ being the wavelength of the light used) is $I$. if $I_0$ denotes the maximum intensity, is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D $\frac{3}{4}$

For maximum intensity $\varphi =0$

Assume that $I_1=I_2=I_3$(source intensity)

$\therefore I_0=I_3+I_2+2\sqrt{I_2{I}_3}{I}_0={4I}_3⟶1\Delta x=\frac{\lambda }{6}\therefore \Delta \varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2x}{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}\therefore I=I_s+I_s+2\sqrt{I_s{I}_s}\cos \frac{\pi }{3}=3I_{3}I=\frac{3}{4}{I}_0$