In a Young's double-slit experiment with wave | vedclass

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

Question

(A) The shift in the fringe pattern due to a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{\beta}{\lambda} (\m

Vedclass · Accepted Answer

$\frac{\beta (\mu - 1)}{\lambda} (t_1 - t_2)$

Vedclass · Answer

(A) The shift in the fringe pattern due to a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{\beta}{\lambda} (\mu - 1) t$.When two plates of thicknesses $t_1$ and $t_2$ are placed in the paths of the two beams,the shift produced by the first plate is $\Delta x_1 = \frac{\beta}{\lambda} (\mu - 1) t_1$ and the shift produced by the second plate is $\Delta x_2 = \frac{\beta}{\lambda} (\mu - 1) t_2$.Since the plates are placed in the paths of the two different beams,the net shift is the difference between the individual shifts:$\Delta x_{net} = |\Delta x_1 - \Delta x_2| = \left| \frac{\beta}{\lambda} (\mu - 1) t_1 - \frac{\beta}{\lambda} (\mu - 1) t_2 \right|$.Simplifying this,we get $\Delta x_{net} = \frac{\beta}{\lambda} (\mu - 1) (t_1 - t_2)$.

In a Young's double-slit experiment with wavelength $\lambda$,the fringe width is $\beta$. When two glass plates of thicknesses $t_1$ and $t_2$ $(t_1 > t_2)$ and refractive index $\mu$ are placed in the paths of the two light beams respectively,by what distance will the fringe pattern shift?

Similar Questions

The width of the fringe is $2\,mm$ on the screen in a double-slit experiment for light of wavelength $400\,nm$. The width of the fringe for light of wavelength $600\,nm$ will be $..............\,mm$.

$A$ beam of electrons is used in $Y.D.S.E.$ The slit width is $d$. When the velocity of the electrons is increased,then:

Obtain the formula for the path difference at a point on the screen in Young's double-slit experiment in terms of $x$,$d$,and $D$.

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

On using red light $(\lambda = 6600 \ \text{Å})$ in Young's double slit experiment,$60$ fringes are observed in the field of view. If violet light $(\lambda = 4400 \ \text{Å})$ is used,the number of fringes observed will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module