(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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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,${I_o}$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ distant $x$ from the centre will be

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In the Young's double slit experiment,if the phase difference between the two waves interfering at a point is $\phi$,the intensity at that point can be expressed by the expression:

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The wavelength of light $500 \, nm$ is used in a Young's double slit experiment. The distance between the slits and the screen is $100 \, cm$ and the slits are separated by $1 \, mm$. Find the distance between the fifth $(5^{th})$ and third $(3^{rd})$ bright fringes. (in $mm$)

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In a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

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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 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,${I_o}$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ distant $x$ from the centre will be

In the Young's double slit experiment,if the phase difference between the two waves interfering at a point is $\phi$,the intensity at that point can be expressed by the expression:

The wavelength of light $500 \, nm$ is used in a Young's double slit experiment. The distance between the slits and the screen is $100 \, cm$ and the slits are separated by $1 \, mm$. Find the distance between the fifth $(5^{th})$ and third $(3^{rd})$ bright fringes. (in $mm$)

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