(B) The distance of the $n^{th}$ bright fringe from the central maxima is given by the formula: $x_n = \frac{n \lambda D}{d}$.Here,$n = 3$,$\lambda =

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(B) The distance of the $n^{th}$ bright fringe from the central maxima is given by the formula: $x_n = \frac{n \lambda D}{d}$.Here,$n = 3$,$\lambda = 6000\ \mathring A = 6000 \times 10^{-10}\ m$,$D = 1.0\ m$,and $d = 0.5\ mm = 0.5 \times 10^{-3}\ m$.Substituting these values into the formula:$x_3 = \frac{3 \times 6000 \times 10^{-10} \times 1.0}{0.5 \times 10^{-3}}$$x_3 = \frac{18000 \times 10^{-10}}{0.5 \times 10^{-3}}$$x_3 = 36000 \times 10^{-7}\ m = 3.6 \times 10^{-3}\ m = 3.6\ mm$.Therefore,the separation is $3.6\ mm$.

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

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In Young's double slit experiment,the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1.0\, m$ from the plane containing the slits. If the wavelength of the incident light is $6000\ \mathring A$,then the separation between the third bright fringe and the central maxima is......$mm$.

A
$4$
B
$3.6$
C
$3$
D
$2.5$

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Wave Optics

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Young's Double Slit Experiment (YDSE)

Light consisting of wavelengths $6000\,\mathring{A}$ and $5500\,\mathring{A}$ falls on the double slits in $YDSE$. The $n^{th}$ order bright fringe of $\lambda_1 = 6000\,\mathring{A}$ is found to coincide with the $m^{th}$ order bright fringe of $\lambda_2 = 5500\,\mathring{A}$. The smallest values of $n$ and $m$ are respectively:

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The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

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In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\theta$ is the angle $S_1CS_2$ and $\lambda$ is the wavelength,the fringe width will be

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In the interference pattern of white light, what is the color of the bright fringe closest to the center?

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Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.
Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.

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The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\theta$ is the angle $S_1CS_2$ and $\lambda$ is the wavelength,the fringe width will be

In the interference pattern of white light, what is the color of the bright fringe closest to the center?

Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.

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Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.
Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.