(C) The position of the $n^{\text{th}}$ bright fringe for wavelength $\lambda_1$ is given by $y_n = \frac{n \lambda_1 D}{d}$.The position of the $m^{\

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

(C) The position of the $n^{\text{th}}$ bright fringe for wavelength $\lambda_1$ is given by $y_n = \frac{n \lambda_1 D}{d}$.The position of the $m^{\text{th}}$ dark fringe for wavelength $\lambda_2$ is given by $y_m = \frac{(2m-1) \lambda_2 D}{2d}$.Since the fringes coincide,we equate their positions:$\frac{n \lambda_1 D}{d} = \frac{(2m-1) \lambda_2 D}{2d}$.Canceling $D$ and $d$ from both sides,we get:$n \lambda_1 = \frac{(2m-1) \lambda_2}{2}$.Rearranging to find the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{2m-1}{2n}$.

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

Easy

In an interference experiment,the $n^{\text{th}}$ bright fringe for light of wavelength $\lambda_1$ $(n=0, 1, 2, 3, \ldots)$ coincides with the $m^{\text{th}}$ dark fringe for light of wavelength $\lambda_2$ $(m=1, 2, 3, \ldots)$. The ratio $\frac{\lambda_1}{\lambda_2}$ is

MHT CET 2024 All MHT CET

A
$\frac{m-1}{n}$
B
$\frac{2m-1}{n}$
C
$\frac{2m-1}{2n}$
D
$\frac{2m+1}{2n}$

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)

Previous Year

MHT CET 2024 Paper All MHT CET years →

In Young's double-slit experiment,the fringe width on the screen is $0.2 \, mm$. If the wavelength of the light used is increased by $10\%$ and the distance between the two slits $S_1$ and $S_2$ is also increased by $10\%$,the new fringe width will be ....... $mm$.

Easy

View Solution

In Young's double slit experiment,the $y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $\mu = 1.5$,what will be the corresponding $y$-coordinates?

Difficult

View Solution

In a Young's double slit experiment,light of $500 \ nm$ is used to produce an interference pattern. When the distance between the slits is $0.05 \ mm$,the angular width (in degrees) of the fringes formed on the distant screen is close to $........^o$.

MediumJEE MAIN 2020

View Solution

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

DifficultNEET 2016

View Solution

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

MediumMHT CET 2016

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

In an interference experiment,the $n^{\text{th}}$ bright fringe for light of wavelength $\lambda_1$ $(n=0, 1, 2, 3, \ldots)$ coincides with the $m^{\text{th}}$ dark fringe for light of wavelength $\lambda_2$ $(m=1, 2, 3, \ldots)$. The ratio $\frac{\lambda_1}{\lambda_2}$ is

Similar Questions

In Young's double-slit experiment,the fringe width on the screen is $0.2 \, mm$. If the wavelength of the light used is increased by $10\%$ and the distance between the two slits $S_1$ and $S_2$ is also increased by $10\%$,the new fringe width will be ....... $mm$.

In Young's double slit experiment,the $y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $\mu = 1.5$,what will be the corresponding $y$-coordinates?

In a Young's double slit experiment,light of $500 \ nm$ is used to produce an interference pattern. When the distance between the slits is $0.05 \ mm$,the angular width (in degrees) of the fringes formed on the distant screen is close to $........^o$.

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module