(D) For the $m^{th}$ dark fringe,the distance from the central bright fringe is given by $x_m' = (2m - 1) \frac{D\lambda}{2d}$.Given: $D = 1 \, m = 10

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(D) For the $m^{th}$ dark fringe,the distance from the central bright fringe is given by $x_m' = (2m - 1) \frac{D\lambda}{2d}$.Given: $D = 1 \, m = 100 \, cm$,$d = 0.12 \, mm = 0.012 \, cm$,$\lambda = 6000 \, \mathring{A} = 6 \times 10^{-5} \, cm$,and $m = 3$.Substituting the values:$x_3' = (2 \times 3 - 1) \times \frac{100 \times 6 \times 10^{-5}}{2 \times 0.012}$$x_3' = 5 \times \frac{6 \times 10^{-3}}{0.024}$$x_3' = 5 \times 0.25 = 1.25 \, cm$.

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

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In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

A
$1$
B
$0.75$
C
$0.60$
D
$1.25$

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Wave Optics

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Young's Double Slit Experiment (YDSE)

Young's double slit experiment is carried out by using green,red,and blue light,one color at a time. The fringe widths recorded are $\beta_G, \beta_R$,and $\beta_B$,respectively. Then:

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Sodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is:

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Two coherent light sources $S_1$ and $S_2$ $(\lambda = 6000 \ \mathring{A})$ are $1 \, mm$ apart from each other. The screen is placed at a distance of $25 \, cm$ from the sources. The width of the fringes on the screen should be (in $, cm$)

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In a Young's double-slit experiment,sodium light of wavelength $5890 \ \mathring A$ is used,and the angular width of the fringe is found to be $0.20^\circ$. If the angular width is to be increased by $10\%$,what is the required change in the wavelength?

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In the Young's double slit experiment,for which colour is the fringe width the least?

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In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

Similar Questions

Young's double slit experiment is carried out by using green,red,and blue light,one color at a time. The fringe widths recorded are $\beta_G, \beta_R$,and $\beta_B$,respectively. Then:

Sodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is:

Two coherent light sources $S_1$ and $S_2$ $(\lambda = 6000 \ \mathring{A})$ are $1 \, mm$ apart from each other. The screen is placed at a distance of $25 \, cm$ from the sources. The width of the fringes on the screen should be (in $, cm$)

In a Young's double-slit experiment,sodium light of wavelength $5890 \ \mathring A$ is used,and the angular width of the fringe is found to be $0.20^\circ$. If the angular width is to be increased by $10\%$,what is the required change in the wavelength?

In the Young's double slit experiment,for which colour is the fringe width the least?

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