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Question
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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Solution
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.
RELATED QUESTIONS
Laser light of wavelength 630 nm is incident on a pair of slits which are separated by 1.8 mm. If the screen is kept 80 cm away from the two slits, calculate:
1) fringe separation i.e. fringe width.
2) distance of 10th bright fringe from the centre of the interference pattern
Four light waves are represented by
(i) \[y = a_1 \sin \omega t\]
(ii) \[y = a_2 \sin \left( \omega t + \epsilon \right)\]
(iii) \[y = a_1 \sin 2\omega t\]
(iv) \[y = a_2 \sin 2\left( \omega t + \epsilon \right).\]
Interference fringes may be observed due to superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
The intensity at the central maximum (O) in a Young’s double slit experimental set-up shown in the figure is IO. If the distance OP equals one-third of the fringe width of the pattern, show that the intensity at point P, would equal `(I_0)/4`.
Why are multiple colours observed over a thin film of oil floating on water? Explain with the help of a diagram.
Answer in brief:
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