In a Young's double slit experiment,{I_o} is | vedclass

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

Question

(C) The path difference at point $P$ is given by $\Delta = \frac{xd}{D}$.Since the fringe width $\beta = \frac{\lambda D}{d}$,we can write the path di

Vedclass · Accepted Answer

${I_o}{\cos ^2}\frac{{\pi x}}{\beta }$

Vedclass · Answer

(C) The path difference at point $P$ is given by $\Delta = \frac{xd}{D}$.Since the fringe width $\beta = \frac{\lambda D}{d}$,we can write the path difference as $\Delta = \frac{x\lambda}{\beta}$.The phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta = \frac{2\pi}{\lambda} \cdot \frac{x\lambda}{\beta} = \frac{2\pi x}{\beta}$.The intensity at any point is given by $I = I_{max} \cos^2(\frac{\phi}{2})$.Given $I_{max} = I_o$,the intensity at point $P$ is $I = I_o \cos^2(\frac{2\pi x}{2\beta}) = I_o \cos^2(\frac{\pi x}{\beta})$.

In a Young's double slit experiment,${I_o}$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ distant $x$ from the centre will be

Similar Questions

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is ......... $\mathring{A}$.

Two parallel slits $0.6 \, mm$ apart are illuminated by a light source of wavelength $6000 \, \mathring{A}$. The distance between two consecutive dark fringes on a screen $1 \, m$ away from the slits is ........ $mm$.

What happens to the fringe width in Young's double-slit experiment if it is performed in glycerine instead of air?

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module