In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

  • A
    $2\lambda$
  • B
    $\frac{2\lambda}{3}$
  • C
    $\frac{\lambda}{3}$
  • D
    $\lambda$

Explore More

Similar Questions

In a double slit experiment, the distance between the slits is $0.1 \ cm$ and the screen is placed at $50 \ cm$ from the slits plane. When one slit is covered with a transparent sheet having thickness $t$ and refractive index $n = 1.5$, the central fringe shifts by $0.2 \ cm$. The value of $t$ is . . . . . . $cm$.

Calculate the wavelength of light used in an interference experiment from the following data: Fringe width $\beta = 0.03 \, cm$. The distance between the slits and the eyepiece is $D = 1 \, m$. The distance between the images of the virtual source,when a convex lens of focal length $f = 16 \, cm$ is used at a distance of $v = 80 \, cm$ from the eyepiece,is $d' = 0.8 \, cm$.

Difficult
View Solution

Consider the figure (not drawn to scale) in which a converging lens of radius $R = 1 \ cm$ and focal length $f = 20 \ cm$ is cut in the middle. The upper part is lifted up by $d = 1 \ mm$ and the lower part is pulled down by the same distance. The gap between them is blocked by an opaque sheet. $A$ point light source with wavelength $\lambda = 500 \ nm$ is placed on the optical axis at a distance of $2f$ from the split lens. $A$ large screen is placed at $L = 1 \ m$ from the right focus of the lens. Find the approximate number of interference fringes on the screen.

In a $YDSE$,light of wavelength $\lambda = 5000 \; \mathring{A}$ is used,which emerges in phase from two slits separated by a distance $d = 3 \times 10^{-7} \; m$. $A$ transparent sheet of thickness $t = 1.5 \times 10^{-7} \; m$ and refractive index $\mu = 1.17$ is placed over one of the slits. What is the new angular position of the central maxima of the interference pattern,and what is its linear position $y$ from the center of the screen?

Difficult
View Solution

$A$ monochromatic beam of light falls on a $YDSE$ apparatus at an angle $\theta$ as shown in the figure. $A$ thin sheet of glass of thickness $t$ and refractive index $\mu$ is inserted in front of the lower slit $S_2$. The central bright fringe (path difference $= 0$) will be obtained:

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