(B) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.Given: $\lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, m$, $D

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(B) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.Given: $\lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, m$, $D = 0.5 \, m$, and $d = 0.5 \, mm = 0.5 \times 10^{-3} \, m$.Substituting the values: $\beta = \frac{5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}} = 5890 \times 10^{-7} \, m = 589 \times 10^{-6} \, m$.The distance between the $n^{th}$ and $m^{th}$ bright fringe is $(n-m) \beta$.For the first and third bright fringe, the distance is $(3-1) \beta = 2 \beta$.Distance $= 2 \times 589 \times 10^{-6} \, m = 1178 \times 10^{-6} \, m$.

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

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In Young's double slit arrangement, slits are separated by a gap of $0.5 \, mm$, and the screen is placed at a distance of $0.5 \, m$ from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of $5890 \, \text{Å}$ is

JEE MAIN 2021 All JEE MAIN

A
$1178 \times 10^{-9} \, m$
B
$1178 \times 10^{-6} \, m$
C
$1178 \times 10^{-12} \, m$
D
$5890 \times 10^{-7} \, m$

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Two sources of light are $0.6 \, mm$ apart and the screen is placed at a distance of $1.2 \, m$ from them. $A$ light of wavelength $6000 \, \mathring{A}$ is used. The phase difference between the two light waves interfering on the screen at a point at a distance of $3 \, mm$ from the central bright band is:

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Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

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In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made

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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 an interference experiment,let $D$ be the distance between the slit and the eyepiece. When the distance between two virtual sources is changed from $d_{A}$ to $d_{B}$,the fringe width changes from $Z_{A}$ to $Z_{B}$. The ratio $Z_{A} / Z_{B}$ is:

MediumMHT CET 2020

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In Young's double slit arrangement, slits are separated by a gap of $0.5 \, mm$, and the screen is placed at a distance of $0.5 \, m$ from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of $5890 \, \text{Å}$ is

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Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made

In an interference experiment,let $D$ be the distance between the slit and the eyepiece. When the distance between two virtual sources is changed from $d_{A}$ to $d_{B}$,the fringe width changes from $Z_{A}$ to $Z_{B}$. The ratio $Z_{A} / Z_{B}$ is:

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