In a Young's double-slit experiment,sources of equal intensities are used. The distance between the slits is $d$ and the wavelength of light used is $\lambda$ (where $\lambda \ll d$). Find the angular separation of the nearest points on either side of the central maximum where the intensity becomes half of the maximum value.

  • A
    $\frac{\lambda}{d}$
  • B
    $\frac{\lambda}{2d}$
  • C
    $\frac{\lambda}{4d}$
  • D
    $\frac{\lambda}{6d}$

Explore More

Similar Questions

Two slits are $1 \, mm$ apart from each other and the distance of the screen is $1 \, m$. If illuminated with light of wavelength $5 \times 10^{-7} \, m$,the distance between the $3^{rd}$ dark fringe and the $5^{th}$ bright fringe is......$mm$.

In a $YDSE$ apparatus,if we use white light,then:

Difficult
View Solution

In Young's double slit experiment,if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled,then the fringe width becomes:

In Young's double-slit experiment,monochromatic light is used to illuminate the two slits $A$ and $B$. Interference fringes are observed on a screen placed in front of the slits. If a thin glass plate is placed normally in the path of the beam coming from slit $A$,then:

In Young's double-slit experiment,which of the following phenomena occur?

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