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 $D=10d$?

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

Answer 1

Answer: (A)

Path difference $=S_{2}P-S_{1}P$

$=\sqrt{D^2+d^2}-D$

$=D(1+\frac{1}{2}\frac{d^2}{D^2}-1)$

$=\frac{d^2}{2D}$

$\Delta x=\frac{d^2}{2\times 10d}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}$

$\Delta \varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

Therefore, intensity at the desired point will be:

$I=I_0{cos}^2\frac{\varphi }{2}$$=I_0{cos}^2\frac{\pi }{4}=\frac{I_0}{2}$