(B) The fringe width $\beta$ in a medium with refractive index $\mu$ is given by the formula: $\beta = \frac{\lambda' D}{d}$,where $\lambda' = \frac{\

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(B) The fringe width $\beta$ in a medium with refractive index $\mu$ is given by the formula: $\beta = \frac{\lambda' D}{d}$,where $\lambda' = \frac{\lambda}{\mu}$.Thus,$\beta = \frac{\lambda D}{\mu d}$.Given values:$\lambda = 6300 \ \mathring{A} = 6300 \times 10^{-10} \ m$$D = 1.33 \ m$$d = 1 \ mm = 10^{-3} \ m$$\mu = 1.33$Substituting these values into the formula:$\beta = \frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 10^{-3}}$$\beta = 6300 \times 10^{-7} \ m$$\beta = 6.3 \times 10^{-4} \ m = 0.63 \ mm$.

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

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$A$ Young's double-slit experimental setup is immersed in water of refractive index $1.33$. It has a slit separation of $1 \ mm$ and the distance between the slits and the screen is $1.33 \ m$. If the wavelength of incident light on the slits is $6300 \ \mathring{A}$,then the fringe width on the screen is:

AP EAMCET 2019 All AP EAMCET

A
$6.3 \ mm$
B
$0.63 \ mm$
C
$0.63 \ m$
D
$6.3 \ m$

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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 intensity of light at each slit is $I_0$. An interference pattern is observed along a direction parallel to the line $S_1 S_2$ on screen $S$. Find the minimum,maximum,and the intensity averaged over the entire screen,respectively.

MediumKVPY 2012

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In a double slit interference experiment,the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement,if the slit separation is increased by one-and-half times the previous value,then the fringe width is (in $ \text{ mm}$)

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In a Young's double slit experiment,the slit separation is $1 \, mm$ and the screen is $1 \, m$ from the slit. For a monochromatic light of wavelength $500 \, nm$,the distance of the $3rd$ minima from the central maxima is.....$mm$

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In Young's double-slit experiment,one of the slits is painted such that its intensity is half that of the other slit. Then:

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In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

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In a Young's double slit experiment,the intensity of light at each slit is $I_0$. An interference pattern is observed along a direction parallel to the line $S_1 S_2$ on screen $S$. Find the minimum,maximum,and the intensity averaged over the entire screen,respectively.

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