(B) In Young's double slit experiment,the position of the $n^{th}$ bright fringe (maximum) from the central maximum is given by the formula: $x_n = \f

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

(B) In Young's double slit experiment,the position of the $n^{th}$ bright fringe (maximum) from the central maximum is given by the formula: $x_n = \frac{n \lambda D}{d}$.Here,$n = 4$ (for the fourth maximum),$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.Since $n$,$D$,and $d$ are constant for both experiments,the position $x$ is directly proportional to the wavelength $\lambda$ $(x \propto \lambda)$.Given $\lambda_{blue} = 4360 \, \mathring{A}$ and $\lambda_{green} = 5460 \, \mathring{A}$.Since $\lambda_{blue} < \lambda_{green}$,it follows that $X_{blue} < X_{green}$.

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

Medium

The Young's double slit experiment is performed in succession using blue light of wavelength $4360 \, \mathring{A}$ and green light of wavelength $5460 \, \mathring{A}$. If the distance of the fourth maximum from the central maximum is $x$,then:

A
$X_{blue} > X_{green}$
B
$X_{blue} < X_{green}$
C
$X_{blue} = X_{green}$
D
$\frac{X_{blue}}{X_{green}} = \frac{5460}{4360}$

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 Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?

Medium

View Solution

In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

Easy

View Solution

In Young's double-slit experiment,the slits are separated by $0.2 \, cm$ and illuminated by light of wavelength $\lambda = 5896 \, \mathring{A}$. What is the fringe width on a screen placed $1 \, m$ away from the plane of the slits? What will be the fringe width if the entire system is immersed in water (in $, mm$)? (Refractive index of water $\mu_w = 4/3$)

Medium

View Solution

In a double slit experiment with monochromatic light,fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5 \times 10^{-2} \,m$ towards the slits,the change in fringe width is $3 \times 10^{-3} \,cm$. If the distance between the slits is $1 \,mm$,then the wavelength of the light will be . . . . . . $nm$.

DifficultJEE MAIN 2022

View Solution

The fringe width in an interference pattern is $X$. The distance between the sixth dark fringe from one side of the central bright band to the fourth bright fringe on the other side is: (in $X$)

MediumMHT CET 2024

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

The Young's double slit experiment is performed in succession using blue light of wavelength $4360 \, \mathring{A}$ and green light of wavelength $5460 \, \mathring{A}$. If the distance of the fourth maximum from the central maximum is $x$,then:

Similar Questions

In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

The fringe width in an interference pattern is $X$. The distance between the sixth dark fringe from one side of the central bright band to the fourth bright fringe on the other side is: (in $X$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module