In a Young's double slit experiment,the angul | vedclass

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

Question

(B) The angular width of a fringe in a Young's double slit experiment is given by $	heta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength and $

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The angular width of a fringe in a Young's double slit experiment is given by $	heta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit separation.When the system is immersed in a medium of refractive index $\mu$,the wavelength changes to $\lambda' = \frac{\lambda}{\mu}$.Therefore,the new angular width $	heta'$ is given by $	heta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{	heta}{\mu}$.Given $	heta = 0.35^{\circ}$ and $\mu = 7/5 = 1.4$.Thus,$	heta' = \frac{0.35^{\circ}}{1.4} = \frac{0.35}{1.4} = \frac{35}{140} = \frac{1}{4}^{\circ}$.Comparing this with $\frac{1}{\alpha}$,we get $\alpha = 4$.

Similar Questions

Two coherent sources $O_1$ and $O_2$ in Young's double slit experiment are illuminated with monochromatic light of wavelength $5000 \text{ Å}$. If a second order dark fringe is formed at a point $R$ on the screen,the path difference $O_1 R \sim O_2 R$ is (in $\mu m$)

In Young's experiment,the distance between two slits is $d/3$ and the distance between the screen and the slits is $3D$. The number of fringes in $1/3 \ m$ on the screen,formed by monochromatic light of wavelength $3\lambda$,will be

In Young's double-slit experiment,the intensity at a point is $1/4^{th}$ of the maximum intensity. The angular position of this point is

In the interference pattern of white light, what is the color of the bright fringe closest to the center?

Vedclass Test Series

Exam Paper Generator

Online Exam Module