Question

Young's double slit experiment is performed in a medium of refractive index of 1.33. The maximum intensity is $I_0$. The intensity at a point on the screen, where path difference between the light coming out from slits is $\frac{\lambda}{4}$, is

• A. 0
• B. $\frac{I_0}{2}$
• C. $\frac{3I_0}{8}$
• D. $\frac{2I_0}{3}$

Answer 1

Answer: (B)

$\frac{I_0}{2}$

Answer 2

The intensity $I$ at a point in Young's double-slit experiment is given by $I = I_{max} \cos^2(\frac{\phi}{2})$, where $I_{max}$ is the maximum intensity and $\phi$ is the phase difference between the waves from the two slits. The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.

Given maximum intensity $I_{max} = I_0$ and path difference $\Delta x = \frac{\lambda}{4}$.

First, calculate the phase difference: $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.

Next, substitute the phase difference into the intensity formula: $I = I_0 \cos^2\left(\frac{\phi}{2}\right) = I_0 \cos^2\left(\frac{\pi/2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right)$.

We know that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

So, $\cos^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.

Therefore, the intensity at the point is $I = I_0 \times \frac{1}{2} = \frac{I_0}{2}$.

The refractive index information is not needed as the path difference is given in terms of the wavelength $\lambda$ in the medium.