(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

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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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Wave Optics

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Young's Double Slit Experiment (YDSE)

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In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is

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$A$ source simultaneously emitting light at two wavelengths $400 \,nm$ and $800 \,nm$ is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is $I_{0}$,then the maximum intensity that can be observed at any point on the screen is

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In a Young's double-slit experiment,two slits act as coherent sources of waves of equal wavelength $\lambda$ and amplitude $A$. In another experiment,the two slits are arranged to act as incoherent sources of waves of the same amplitude and wavelength. If the intensity at the central point of the screen in the first case is $I_1$ and in the second case is $I_2$,then the ratio $I_1/I_2$ is:

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The maximum intensity of fringes in Young's experiment is $I$. If one of the slits is closed,then the intensity at that place becomes $I_o$. Which of the following relations is true?

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In an interference experiment,the third bright fringe is obtained using light of wavelength $700 \, nm$. What should be the wavelength of light to obtain the fifth bright fringe at the same point?

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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

Similar Questions

In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is

$A$ source simultaneously emitting light at two wavelengths $400 \,nm$ and $800 \,nm$ is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is $I_{0}$,then the maximum intensity that can be observed at any point on the screen is

The maximum intensity of fringes in Young's experiment is $I$. If one of the slits is closed,then the intensity at that place becomes $I_o$. Which of the following relations is true?

In an interference experiment,the third bright fringe is obtained using light of wavelength $700 \, nm$. What should be the wavelength of light to obtain the fifth bright fringe at the same point?

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