Two coherent point sources S_1 and S_2 vibrat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The path difference between the waves reaching point $P$ from $S_1$ and $S_2$ is given by $\Delta x = S_1 P - S_2 P$.For the first bright fringe,

Vedclass · Accepted Answer

$\sqrt{3} D$

Vedclass · Answer

(C) The path difference between the waves reaching point $P$ from $S_1$ and $S_2$ is given by $\Delta x = S_1 P - S_2 P$.For the first bright fringe, the path difference must be equal to the wavelength, so $\Delta x = \lambda$.From the geometry of the setup, the path difference can also be expressed as $\Delta x = d \cos 	heta$, where $d = 2 \lambda$ is the separation between the sources.Thus, $2 \lambda \cos 	heta = \lambda$.This simplifies to $\cos 	heta = \frac{1}{2}$, which means $	heta = 60^{\circ}$.From the right-angled triangle formed by the screen, we have $	an 	heta = \frac{OP}{D}$.Substituting $	heta = 60^{\circ}$, we get $	an 60^{\circ} = \frac{OP}{D}$.Since $	an 60^{\circ} = \sqrt{3}$, we have $\sqrt{3} = \frac{OP}{D}$.Therefore, $OP = \sqrt{3} D$.

Similar Questions

When one of the slits in Young's experiment is covered with a transparent sheet of thickness $3.6 \times 10^{-3} \ cm$,the central fringe shifts to a position originally occupied by the $30^{th}$ bright fringe. If $\lambda = 6000 \ \mathring{A}$,then what is the refractive index of the sheet?

In the figure shown,if a parallel beam of white light is incident on the plane of the slits,then the distance of the white spot on the screen from $O$ is [Assume $d << D, \lambda << d$].

Young's double slit experiment is conducted in a liquid of refractive index $\mu_1$ as shown in the figure. $A$ thin transparent slab of refractive index $\mu_2$ and thickness $t$ is placed in front of the slit $S_2$. The magnitude of the optical path difference at point $O$ is

Each plate reflects $25\%$ of the incident light intensity. When $AB$ and $A'B'$ are taken as the two slits in Young's double-slit experiment,what is the ratio of maximum to minimum intensity ${I_{\max }}/{I_{\min }}$ (in $: 1$)?

The fringe widths are found to be $\omega_1$ and $\omega_2$ respectively if a Young's double slit experiment is performed in media of refractive indices $n_1$ and $n_2$ respectively. The correct statement is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module