To make the central fringe appear at the centre $O,$ a mica sheet of refractive index $\mu = 1.5$ is introduced. Choose the correct statement:

  • A
    The thickness of the sheet is $2(\sqrt{2}-1)d$ in front of $S_1$.
  • B
    The thickness of the sheet is $(\sqrt{2}-1)d$ in front of $S_2$.
  • C
    The thickness of the sheet is $2\sqrt{2}d$ in front of $S_1$.
  • D
    The thickness of the sheet is $(2\sqrt{2}-1)d$ in front of $S_1$.

Explore More

Similar Questions

Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between them is $2 \lambda$ as shown in the figure. The first bright fringe is formed at $P$ due to interference on a screen placed at a distance $D$ from $S_1$ $(D >> \lambda)$. Find the distance $OP$.

$A$ small transparent slab of refractive index $\mu = 1.5$ and thickness $L = d/4$ is placed along the path $AS_2$ (see figure). What will be the distance from $O$ of the principal maxima and the first minima on either side of the principal maxima obtained in the absence of the glass slab? Given $AC = CD = D$ and $S_1C = S_2C = d$ (where $d << D$).

Difficult
View Solution

Monochromatic light of wavelength $500 \, nm$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $= 1.5$), the central maximum is shifted to a position previously occupied by the $4^{th}$ bright fringe. The thickness of the glass plate is ..................... $\mu m$.

In a Young's double-slit experiment,let $A$ and $B$ be the two slits. Thin films of thicknesses $t_A$ and $t_B$ and refractive indices $\mu_A$ and $\mu_B$ are placed in front of slits $A$ and $B$ respectively. If $\mu_A t_A = \mu_B t_B$,the central maximum will:

$A$ monochromatic light source $S$ of wavelength $440 \,nm$ is placed slightly above a plane mirror $M$ as shown below. The image of $S$ in $M$ can be used as a virtual source to produce interference fringes on the screen. The distance of source $S$ from $O$ is $20.0 \,cm$ and the distance of the screen from $O$ is $100.0 \,cm$ (figure is not to scale). If the angle $\theta = 0.50 \times 10^{-3} \,radians$, then the width of the interference fringes observed on the screen is ............... $mm$.

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