(B) In $YDSE$,the position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.For the fringes to coincide,their positions must be

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

(B) In $YDSE$,the position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.For the fringes to coincide,their positions must be equal: $y_n = y_m$.Therefore,$n \lambda_1 = m \lambda_2$.Substituting the given values: $n(6000) = m(5500)$.This simplifies to $\frac{n}{m} = \frac{5500}{6000} = \frac{11}{12}$.Since $n$ and $m$ must be integers,the smallest values are $n = 11$ and $m = 12$.

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

Medium

Light consisting of wavelengths $6000\,\mathring{A}$ and $5500\,\mathring{A}$ falls on the double slits in $YDSE$. The $n^{th}$ order bright fringe of $\lambda_1 = 6000\,\mathring{A}$ is found to coincide with the $m^{th}$ order bright fringe of $\lambda_2 = 5500\,\mathring{A}$. The smallest values of $n$ and $m$ are respectively:

A
$22, 23$
B
$11, 12$
C
$7, 9$
D
$15, 17$

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)

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

KVPY 2015

View Solution

In a Young's double slit experiment,two slits are illuminated with light of wavelength $\lambda = 800 \, nm$. The line joining $A_1 P$ is perpendicular to $A_1 A_2$ as shown in the figure. If the first minimum is detected at $P$,the value of slit separation '$a$' will be $......... \, mm$. The distance of the screen from the slits is $D = 5 \, cm$.

MediumJEE MAIN 2023

View Solution

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

Medium

View Solution

In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

Easy

View Solution

Obtain the formula for the path difference at a point on the screen in Young's double-slit experiment in terms of $x$,$d$,and $D$.

Medium

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

Similar Questions

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

Obtain the formula for the path difference at a point on the screen in Young's double-slit experiment in terms of $x$,$d$,and $D$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module