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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
How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit screens is doubled?
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)
A narrow slit S transmitting light of wavelength λ is placed a distance d above a large plane mirror, as shown in the following figure. The light coming directly from the slit and that coming after the reflection interfere at a screen ∑ placed at a distance D from the slit. (a) What will be the intensity at a point just above the mirror, i.e. just above O? (b) At what distance from O does the first maximum occur?
Answer in brief:
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