(D) The formula for fringe width in Young's double-slit experiment is $\beta = \frac{\lambda D}{d}$.Since $D$ (distance between slits and screen) and

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

(D) The formula for fringe width in Young's double-slit experiment is $\beta = \frac{\lambda D}{d}$.Since $D$ (distance between slits and screen) and $d$ (distance between slits) are constant,the fringe width is directly proportional to the wavelength,i.e.,$\beta \propto \lambda$.The wavelengths of the colors follow the order: $\lambda_R > \lambda_G > \lambda_B$.Therefore,the fringe widths follow the same order: $\beta_R > \beta_G > \beta_B$.

Class 12 Physics Wave Optics Young's Double Slit Experiment (YDSE)

Medium

Young's double-slit experiment is carried out by using green,red,and blue light,one color at a time. The fringe widths recorded are $\beta_G$,$\beta_R$,and $\beta_B$ respectively. Then:

A
$\beta_G > \beta_B > \beta_R$
B
$\beta_B > \beta_G > \beta_R$
C
$\beta_R > \beta_B > \beta_G$
D
$\beta_R > \beta_G > \beta_B$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Wave Optics

Subtopic

Young's Double Slit Experiment (YDSE)

In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is

DifficultMHT CET 2026

View Solution

In a Young's double slit experiment,$I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the central maximum will be

Difficult

View Solution

In a double-slit experiment,the distance between the slits is $d$. The screen is at a distance $D$ from the slits. If a bright fringe is formed opposite to one of the slits,its order $n$ is:

Difficult

View Solution

In Young's double-slit experiment with a source of light of wavelength $6320 \ \mathring{A}$,the first maxima will occur when:

Easy

View Solution

Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

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

Young's double-slit experiment is carried out by using green,red,and blue light,one color at a time. The fringe widths recorded are $\beta_G$,$\beta_R$,and $\beta_B$ respectively. Then:

Similar Questions

In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is

In a Young's double slit experiment,$I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the central maximum will be

In a double-slit experiment,the distance between the slits is $d$. The screen is at a distance $D$ from the slits. If a bright fringe is formed opposite to one of the slits,its order $n$ is:

In Young's double-slit experiment with a source of light of wavelength $6320 \ \mathring{A}$,the first maxima will occur when:

Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module