(C) In Young's double slit experiment,the position of the $n^{th}$ bright fringe is given by $y_n = n \frac{\lambda D}{d}$.For the first case,$n_1 = 5

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(C) In Young's double slit experiment,the position of the $n^{th}$ bright fringe is given by $y_n = n \frac{\lambda D}{d}$.For the first case,$n_1 = 5$,$\lambda_1 = 600 nm$,and $y_1 = 6 mm$.So,$6 = 5 \times \frac{600 D}{d} \Rightarrow \frac{D}{d} = \frac{6}{5 \times 600} = \frac{1}{500} mm/nm$.For the second case,$n_2 = 3$,$\lambda_2 = 400 nm$.The position of the third bright fringe is $y_2 = n_2 \frac{\lambda_2 D}{d}$.Substituting the values: $y_2 = 3 \times 400 \times \frac{1}{500} = \frac{1200}{500} = 2.4 mm$.

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

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When monochromatic light of wavelength $600 nm$ is used in Young's double slit experiment,the fifth order bright fringe is formed at $6 mm$ from the central bright fringe on the screen. If the experiment is conducted with light of wavelength $400 nm$,the third order bright fringe will be located at: (in $mm$)

AP EAMCET 2022 All AP EAMCET

A
$1.6$
B
$2$
C
$2.4$
D
$3$

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Wave Optics

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Young's Double Slit Experiment (YDSE)

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In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.

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Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the fringe width will be:

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In an interference pattern,the $(n + 4)^{th}$ order blue bright fringe and the $n^{th}$ order red bright fringe coincide at a point. If the wavelengths of red and blue light are $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$ respectively,then the value of $n$ is . . . . . .

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In a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

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In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

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In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.

Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the fringe width will be:

In an interference pattern,the $(n + 4)^{th}$ order blue bright fringe and the $n^{th}$ order red bright fringe coincide at a point. If the wavelengths of red and blue light are $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$ respectively,then the value of $n$ is . . . . . .

In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

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