(C) The width of the segment on the screen is given by $y = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Thus,$y = n \lambda \left

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(C) The width of the segment on the screen is given by $y = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Thus,$y = n \lambda \left(\frac{D}{d}\right)$.Since the segment length $y$ and the experimental setup parameters $D$ and $d$ remain constant,we have $n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 8$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$.Substituting the values: $8 \times 600 = n_2 \times 400$.$n_2 = \frac{8 \times 600}{400} = \frac{4800}{400} = 12$.Therefore,the student will observe $12$ fringes.

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

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In a Young's double slit experiment,a student observes $8$ fringes in a certain segment of the screen when a monochromatic light of $600 \ nm$ wavelength is used. If the wavelength of light is changed to $400 \ nm$,then the number of fringes he would observe in the same region of the screen is:

NEET 2022 All NEET

A
$8$
B
$9$
C
$12$
D
$6$

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Wave Optics

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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 $X$?

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The path difference between two identical light waves at a point $Q$ on the screen is $3 \mu m$. If the wavelength of the waves is $5000 \text{ Å}$,then at point $Q$ there is:

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In Young's double slit experiment,if the source of light changes from orange to blue,then

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In a double slit experiment shown in the figure,when light of wavelength $400 \ nm$ is used,a dark fringe is observed at $P$. If $D=0.2 \ m$,the minimum distance between the slits $S_1$ and $S_2$ is . . . . . . $mm$.

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In a Young's double slit experiment,a student observes $8$ fringes in a certain segment of the screen when a monochromatic light of $600 \ nm$ wavelength is used. If the wavelength of light is changed to $400 \ nm$,then the number of fringes he would observe in the same region of the screen 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 $X$?

In a Young's double-slit experiment,the wavelength of light used is $6000 \, \mathring A$. The phase difference between the central bright fringe and the third bright fringe is .......

The path difference between two identical light waves at a point $Q$ on the screen is $3 \mu m$. If the wavelength of the waves is $5000 \text{ Å}$,then at point $Q$ there is:

In Young's double slit experiment,if the source of light changes from orange to blue,then

In a double slit experiment shown in the figure,when light of wavelength $400 \ nm$ is used,a dark fringe is observed at $P$. If $D=0.2 \ m$,the minimum distance between the slits $S_1$ and $S_2$ is . . . . . . $mm$.

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