(C) The path difference $\Delta x$ between the two waves reaching point $X$ is given by:$\Delta x = |QX - PX| = |(n + 2)\lambda - n\lambda| = 2\lambda

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(C) The path difference $\Delta x$ between the two waves reaching point $X$ is given by:$\Delta x = |QX - PX| = |(n + 2)\lambda - n\lambda| = 2\lambda$.For constructive interference (bright fringe),the condition is $\Delta x = m\lambda$,where $m = 0, 1, 2, ...$.Here,$\Delta x = 2\lambda$,which corresponds to $m = 2$.Since the central fringe $(m = 0)$ is at the center,$m = 1$ corresponds to the first bright fringe,and $m = 2$ corresponds to the second bright fringe.Therefore,the second bright fringe is formed at $X$.

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

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The figure shows a double slit experiment where $P$ and $Q$ are the slits. The path lengths $PX$ and $QX$ are $n\lambda$ and $(n + 2)\lambda$ respectively,where $n$ is a whole number and $\lambda$ is the wavelength. Taking the central fringe as zero,what is formed at $X$?

A
First bright
B
First dark
C
Second bright
D
Second dark

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Wave Optics

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Young's Double Slit Experiment (YDSE)

The fringe width in Young's double-slit experiment can be increased if we decrease

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In a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

MediumJEE MAIN 2020

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Statement-$1$: If white light is used in $YDSE$,then the central bright fringe will be white.
Statement-$2$: In the case of white light used in $YDSE$,all the wavelengths produce their zero-order maxima at the same position.

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In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is

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In two separate set-ups of the Young's double slit experiment,fringes of equal width are observed when lights of wavelengths in the ratio $1:2$ are used. If the ratio of the slit separation in the two cases is $2:1$,the ratio of the distances between the plane of the slits and the screen in the two set-ups is

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The figure shows a double slit experiment where $P$ and $Q$ are the slits. The path lengths $PX$ and $QX$ are $n\lambda$ and $(n + 2)\lambda$ respectively,where $n$ is a whole number and $\lambda$ is the wavelength. Taking the central fringe as zero,what is formed at $X$?

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The fringe width in Young's double-slit experiment can be increased if we decrease

Statement-$1$: If white light is used in $YDSE$,then the central bright fringe will be white.Statement-$2$: In the case of white light used in $YDSE$,all the wavelengths produce their zero-order maxima at the same position.

In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is

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Statement-$1$: If white light is used in $YDSE$,then the central bright fringe will be white.
Statement-$2$: In the case of white light used in $YDSE$,all the wavelengths produce their zero-order maxima at the same position.