(C) The distance of the $n^{th}$ bright fringe from the central maximum is given by the formula: $y_n = \frac{n \lambda D}{d}$.Since $n$,$D$,and $d$ a

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$x(\text{Blue}) < x(\text{Green})$

(C) The distance of the $n^{th}$ bright fringe from the central maximum is given by the formula: $y_n = \frac{n \lambda D}{d}$.Since $n$,$D$,and $d$ are constant for both cases,we have $y_n \propto \lambda$.Therefore,the ratio of the distances for the $4^{th}$ fringe is: $\frac{x(\text{Blue})}{x(\text{Green})} = \frac{\lambda(\text{Blue})}{\lambda(\text{Green})} = \frac{4360}{5460}$.Since $4360 < 5460$,it follows that $x(\text{Blue}) < x(\text{Green})$.

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

Medium

In a Young's double-slit experiment,the experiment is performed with blue light of wavelength $4360 \; \mathring{A}$ and green light of wavelength $5460 \; \mathring{A}$. If the distance of the $4^{th}$ bright fringe from the central maximum is $x$,then:

AIPMT 1990 All AIPMT

A
$x(\text{Blue}) = x(\text{Green})$
B
$x(\text{Blue}) > x(\text{Green})$
C
$x(\text{Blue}) < x(\text{Green})$
D
$\frac{x(\text{Blue})}{x(\text{Green})} = \frac{5460}{4360}$

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Young's Double Slit Experiment (YDSE)

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AIPMT 1990 Paper All AIPMT 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 ....... $mm$.

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In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

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In a Young's double-slit experiment,the wavelength of light used is $5000 \, \mathring{A}$ and the fringe width obtained is $1 \, mm$. If the wavelength of light is changed to $6000 \, \mathring{A}$ while keeping the experimental setup unchanged,the new fringe width will be ........ $mm$.

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

Difficult

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In Young's double slit experiment, carried out with light of wavelength $\lambda = 5000\;\mathring{A}$, the distance between the slits is $d = 0.2\;mm$ and the screen is at $D = 200\;cm$ from the slits. The central maximum is at $x = 0$. The third maximum (taking the central maximum as the zeroth maximum) will be at $x$ equal to......$cm$.

MediumAIPMT 1992

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In a Young's double-slit experiment,the experiment is performed with blue light of wavelength $4360 \; \mathring{A}$ and green light of wavelength $5460 \; \mathring{A}$. If the distance of the $4^{th}$ bright fringe from the central maximum is $x$,then:

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 wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

In a Young's double-slit experiment,the wavelength of light used is $5000 \, \mathring{A}$ and the fringe width obtained is $1 \, mm$. If the wavelength of light is changed to $6000 \, \mathring{A}$ while keeping the experimental setup unchanged,the new fringe width will be ........ $mm$.

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