Question

In Young's double slit experiment using monochromatic light of wavelength '$\lambda$', the maximum intensity of light at a point on the screen is 'K' units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ is $(cos 60^\circ = sin 30^\circ = 0.5, sin 60^\circ = cos 30^\circ = \sqrt{3}/2)$

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

Answer 1

Answer: (A)

$\frac{3K}{4}$

Answer 2

1. The phase difference corresponding to a path difference of $\lambda/6$ is:

   $$\phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{6} = \frac{\pi}{3}.$$

2. The intensity in Young's double slit experiment is given by:

   $$I = I_{\max}\cos^2\left(\frac{\phi}{2}\right).$$

   For maximum intensity $K$, we have $I_{\max} = K$.

3. Therefore, for $\phi = \frac{\pi}{3}$:

   $$I = K\cos^2\left(\frac{\pi}{6}\right) = K\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3K}{4}.$$