Question:
In Young's double slit experiment, the distance between two sources is $0.1 \,mm$. The distance of the screen from the source is $20\, cm$. Wavelength of light used is $5460\, ?$. The angular position of the first dark fringe is
In Young's double slit experiment, the distance between two sources is $0.1 \,mm$. The distance of the screen from the source is $20\, cm$. Wavelength of light used is $5460\, ?$. The angular position of the first dark fringe is
Updated On: Jul 5, 2022
- $0.08^{\circ}$
- $0.16^{\circ}$
- $0.20^{\circ}$
- $0.32^{\circ}$
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The Correct Option is B
Solution and Explanation
For first dark fringe
$x=\left(2n-1\right)\frac{\lambda}{2} \frac{D}{d}=\frac{\lambda}{2} \frac{D}{d} \left(\because\,n=1\right)$
Angular position, $\theta=\frac{x}{D}=\frac{\lambda}{2d}$
$=\frac{5460\times10^{-10}}{2\times10^{-4}}$ radian
$=2730\times10^{-6}\times\frac{180}{\pi } $
degree $= 0.16^{\circ}$
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Concepts Used:
Young’s Double Slit Experiment
- Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
- The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
- Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.
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