The minimum thickness of a mica sheet having refractive index $\mu = 3/2$,which should be placed in front of one of the slits in $YDSE$ to reduce the intensity at the center of the screen to half of the maximum intensity,is:

  • A
    $\lambda /4$
  • B
    $\lambda /8$
  • C
    $\lambda /2$
  • D
    $\lambda /3$

Explore More

Similar Questions

On placing a thin film of mica of thickness $12 \times 10^{-5} \text{ cm}$ in the path of one of the interfering waves in Young's double slit experiment using monochromatic light,the fringe pattern shifts through a distance equal to the width of a bright fringe. If the wavelength used is $\lambda = 6 \times 10^{-5} \text{ cm}$,the refractive index of mica is:

In a Young's double slit experiment,each of the two slits $A$ and $B$,as shown in the figure,are oscillating about their fixed center with a mean separation of $0.8 \ mm$. The distance between the slits at time $t$ is given by $d = (0.8 + 0.04 \sin \omega t) \ mm$,where $\omega = 0.08 \ rad \ s^{-1}$. The distance of the screen from the slits is $1 \ m$ and the wavelength of the light used to illuminate the slits is $6000 \ \mathring A$. The interference pattern on the screen changes with time,while the central bright fringe (zeroth fringe) remains fixed at point $O$.
$(1)$ The $8^{\text{th}}$ bright fringe above the point $O$ oscillates with time between two extreme positions. The separation between these two extreme positions,in micrometer $(\mu m)$,is. . . . .
$(2)$ The maximum speed in $\mu m/s$ at which the $8^{\text{th}}$ bright fringe will move is. . . . .

The central fringe of the interference pattern produced by light of wavelength $6000 \mathring A$ is found to shift to the position of the $4^{\text {th}}$ bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit in Young's experiment. The thickness of the glass plate will be ......... $\mu m$.

In Young's double-slit experiment performed using a monochromatic light of wavelength $\lambda$,when a glass plate $(\mu=1.5)$ of thickness $t = x \lambda$ is introduced in the path of one of the interfering beams,the intensity at the position where the central maximum occurred previously remains unchanged. The value of $x$ will be..........

In a double-slit experiment,the wavelength $\lambda$ of the light source is $400 \ nm$,the slit separation $d$ is $20 \ \mu m$,and the individual slit width $a$ is $4 \ \mu m$. Considering the interference of light from the two slits and the diffraction of light through each slit,find the number of bright interference fringes that lie within the central peak of the diffraction envelope.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo