(B) For a dark fringe at point $P$,the path difference $\Delta$ is given by the condition:$\Delta = S_1P - S_2P = (2n - 1) \frac{\lambda}{2}$where $n$

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(B) For a dark fringe at point $P$,the path difference $\Delta$ is given by the condition:$\Delta = S_1P - S_2P = (2n - 1) \frac{\lambda}{2}$where $n$ is the order of the dark fringe and $\lambda$ is the wavelength of light.Given $n = 3$ and $\lambda = 6000 \ \mathring A = 6000 \times 10^{-10} \ m = 0.6 \ \mu m$.Substituting the values:$\Delta = (2(3) - 1) \frac{0.6 \ \mu m}{2} = 5 \times 0.3 \ \mu m = 1.5 \ \mu m$.Thus,the path difference is $1.5 \ \mu m$.

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

Medium

In Young's double-slit experiment,an interference pattern is obtained on a screen by light of wavelength $6000 \ \mathring A$,coming from coherent sources $S_1$ and $S_2$. At a certain point $P$ on the screen,the third dark fringe is formed. Then the path difference $S_1P - S_2P$ in microns is:

A
$0.75$
B
$1.5$
C
$3$
D
$4.5$

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Class 12 Questions

Class 12

Physics Questions

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Wave Optics

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Young's Double Slit Experiment (YDSE)

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

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In Young's double-slit experiment,the distance between the two slits is $1 \, mm$ and the distance between the slits and the screen is $1 \, m$. If the wavelength of the light used is $500 \, nm$,then the distance of the third minimum from the central maximum is ... $mm$.

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In a Young's double-slit experiment,light of wavelength $5890 \ \mathring A$ is used,and the angular fringe width on the screen is $0.20^\circ$. If the entire apparatus is immersed in water,find the new angular fringe width. (Refractive index of water $\mu = 4/3$) (in $^\circ$)

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The maximum intensity of fringes in Young's experiment is $I$. If one of the slits is closed,then the intensity at that place becomes $I_o$. Which of the following relations is true?

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If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)

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In Young's double-slit experiment,an interference pattern is obtained on a screen by light of wavelength $6000 \ \mathring A$,coming from coherent sources $S_1$ and $S_2$. At a certain point $P$ on the screen,the third dark fringe is formed. Then the path difference $S_1P - S_2P$ in microns is:

Similar Questions

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

In Young's double-slit experiment,the distance between the two slits is $1 \, mm$ and the distance between the slits and the screen is $1 \, m$. If the wavelength of the light used is $500 \, nm$,then the distance of the third minimum from the central maximum is ... $mm$.

In a Young's double-slit experiment,light of wavelength $5890 \ \mathring A$ is used,and the angular fringe width on the screen is $0.20^\circ$. If the entire apparatus is immersed in water,find the new angular fringe width. (Refractive index of water $\mu = 4/3$) (in $^\circ$)

The maximum intensity of fringes in Young's experiment is $I$. If one of the slits is closed,then the intensity at that place becomes $I_o$. Which of the following relations is true?

If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)

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