Question

The intensity at the maximum in a Young’s double slit experiment is $I_0$. Distance between two slits is $d=5\lambda$, where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d ?

• A. $\frac{I_0}{4}$
• B. $\frac{3}{4}{I}_0$
• C. $\frac{I_0}{2}$
• D. $I_0$

Answer 1

Answer: (C)

Path difference
$=S_{2}P-S_{1}P$
$=\sqrt{D^2+d^2}-D$
$=D\left(1+\frac{1}{2}\frac{d^2}{D^2}\right)-D$
$=D\left[1+\frac{d^2}{2D^2}-1\right]=\frac{d^2}{2D}$
$\Delta x=\frac{d^2}{2\times 10d}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}$
$\Delta \varphi =\frac{2\pi }{\lambda }.\frac{\lambda }{4}=\frac{\pi }{2}$
So, intensity at the desired point is
$I=I_{0}co{s}^2\frac{\varphi }{2}=I_{0}co{s}^2\frac{\pi }{4}=\frac{I_0}{2}$