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प्रश्न
A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)?
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उत्तर
Given: θ1 = 0.20°, nw = 1.33
In the first approximation,
D sin θ1 = y1 and D sin θ2 = y2
`therefore (sin theta_2)/(sin theta_1) = "y"_2/"y"_1` ...(1)
Now, `y prop (lambdaD)/d`
For given d and D,
y ∝ λ
∴ `y_2/y_1 = lambda_2/lambda_1` ...(2)
Now, `n_w = lambda_1/lambda_2` ...(3)
From Eqs. (1), (2) and (3), we get
`(sin theta_2)/(sin theta_1) = lambda_2/lambda_1 = 1/n_w`
∴ `sin theta_2 = sin theta_1/n_w`
`sin theta_2 = (sin 0.2)/1.33`
`sin theta_2 = 0.0035/1.33`
sinθ2 = 0.0026
θ2 = sin−1 0.0026
θ2 = 9'
θ2 = 0.15°
This is the required angular fringe separation.
संबंधित प्रश्न
In Young’s double slit experiment, the slits are separated by 0.5 mm and screen is placed 1.0 m away from the slit. It is found that the 5th bright fringe is at a distance of 4.13 mm from the 2nd dark fringe. Find the wavelength of light used.
Answer in brief:
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