Question

2 identical beams A and B of plane coherent waves of same intensity and wavelength 6 x 10^-7m fall on a plane screen. the direction of beam propagation makes 37 deg and 53 deg respectively with normal and lie in same plane. find distance between adjacent. interference fringes

Answer 1

Approx. $3.05 \times 10^{-6}$ m

Answer 2

The path difference at a point $x$ on the screen is due to the different projections of $x$ along the directions of the two beams. If beam A makes an angle $\theta_A=37^\circ$ and beam B makes $\theta_B=53^\circ$ with the normal, the phase difference is given by:

$$\Delta \phi = kx \left(\sin\theta_B - \sin\theta_A\right), \quad \text{where } k=\frac{2\pi}{\lambda}.$$

For constructive interference (bright fringes), we need:

$$kx \left(\sin\theta_B-\sin\theta_A\right)=2\pi m.$$

The fringe spacing $\Delta x$ (distance between adjacent bright fringes) is:

$$k\,\Delta x \left(\sin\theta_B-\sin\theta_A\right)=2\pi \quad \Rightarrow \quad \Delta x = \frac{2\pi}{k\left(\sin\theta_B-\sin\theta_A\right)} = \frac{\lambda}{\sin\theta_B-\sin\theta_A}.$$

Given:

$$\lambda = 6 \times 10^{-7}\, \text{m},$$ $$\sin 53^\circ \approx 0.7986,\quad \sin 37^\circ \approx 0.6018.$$

Thus,

$$\sin\theta_B-\sin\theta_A \approx 0.7986 - 0.6018 = 0.1968.$$ $$\Delta x \approx \frac{6 \times 10^{-7}}{0.1968} \approx 3.05 \times 10^{-6}\,\text{m}.$$

Core Explanation:

• Fringe spacing is given by $\Delta x=\frac{\lambda}{\sin53^\circ-\sin37^\circ}$.
• Substitute $\lambda=6\times10^{-7}\,\text{m}$, using $\sin53^\circ\approx 0.7986$ and $\sin37^\circ\approx 0.6018$.
• Compute $\Delta x\approx3.05\times10^{-6}\,\text{m}$.