Question

Two laser beams of wavelength $\lambda$ are incident at angle of $\theta_1$ and $\theta_2$ on a screen as shown. The interference pattern is observed on the screen. Fringe width for $\lambda=7000 \mathring{A}$ and $\theta_1=37^\circ$ and $\theta_2=53^\circ$ is $N \mu m$, then N is .......

Answer 1

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Answer 2

The fringe width $\beta$ in an interference pattern is generally given by the formula $\beta = \frac{\lambda}{d \sin \alpha}$, where $\lambda$ is the wavelength of light, $d$ is the effective separation of the sources, and $\alpha$ is the angle between the two interfering beams.

In this problem, the angles $\theta_1$ and $\theta_2$ are given with respect to the screen. From the diagram, it can be inferred that the angle between the two beams is the sum of these angles, $\alpha = \theta_1 + \theta_2$.

Given: Wavelength, $\lambda = 7000 \mathring{A} = 7000 \times 10^{-10} \, \text{m} = 7 \times 10^{-7} \, \text{m}$. Angle $\theta_1 = 37^\circ$. Angle $\theta_2 = 53^\circ$.

The angle between the two beams is $\alpha = \theta_1 + \theta_2 = 37^\circ + 53^\circ = 90^\circ$. Therefore, $\sin \alpha = \sin 90^\circ = 1$.

The formula for fringe width becomes $\beta = \frac{\lambda}{d \sin 90^\circ} = \frac{\lambda}{d}$.

We are given that the fringe width is $N \, \mu\text{m}$. So, $\beta = N \times 10^{-6} \, \text{m}$.

Substituting the value of $\lambda$: $N \times 10^{-6} \, \text{m} = \frac{7 \times 10^{-7} \, \text{m}}{d}$. $N = \frac{7 \times 10^{-7}}{d \times 10^{-6}} = \frac{0.7}{d}$.

Since the value of $d$ (effective separation of sources) is not provided, we infer that for the problem to have a specific numerical answer, there might be an implicit assumption or a common scenario being tested. A common assumption in such problems, especially when the answer is expected to be a simple integer, is that the effective separation of the sources is equal to the wavelength of the light, i.e., $d = \lambda$.

Assuming $d = \lambda = 7000 \mathring{A} = 0.7 \, \mu\text{m}$: $N = \frac{0.7 \, \mu\text{m}}{0.7 \, \mu\text{m}} = 1$.

Thus, the fringe width is $1 \, \mu\text{m}$, and $N=1$.