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प्रश्न
Answer in brief:
Explain what is the optical path length. How is it different from actual path length?
What is Optical Path length? How is it different from the actual Path length?
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उत्तर १
Consider, a light wave with an angular frequency of w and a wave vector of k travelling in the x-direction through a vacuum. The phase of this wave is (kx - ωt). In a vacuum, light speed at c, but in a medium, it speeds at v.
k = `(2pi)/lambda = (2pi v)/(v lambda) = omega/v` as ω = 2πv and v = vλ, where v is the frequency of light.
If the wave travels a distance Δ x, its phase changes by Δ Φ = kΔx = ω Δx/v.
Similarly, if the wave is travelling in vacuum,
k = ω/c and Δ Φ = ω Δ x/c
Now, consider a wave travelling a distance Δ x in the medium, the phase difference generated is,
Δ Φ' = k' Δ x = ωn Δ x/c = ω Δ x'/c ...(1)
where Δ x' = n Δ x .....(2)
The distance nΔ x is called the optical path length of the light in the medium; it is the distance the light would have travelled in the same time t in vacuum (with the speed c).
The optical path length in a medium is the corresponding path in a vacuum that light traverses at the same time as it does in the medium.
Now, speed = `"distance"/"time"`
∴ time = `"distance"/"speed"`
∴ t = `"d"_"medium"/"v"_"medium" = "d"_"vaccum"/"v"_"vaccum"`
Hence, the optical path = `"d"_"vacuum"`
`= "v"_"vaccum"/"v"_"medium" xx "d"_"medium"`
`= "n" xx "d"_"medium"`
Thus, a distance d travelled in a medium of refractive index n introduces a path difference = nd - d = d (n - 1) over a ray travelling equal distance through vacuum.
उत्तर २
i. When a wave travels a distance Δx through a medium having a refractive index of n, its phase changes by the same amount as it would if the wave had travelled a distance nΔx in a vacuum.
ii. Thus, a path length of Δx in a medium of refractive index n is equivalent to a path length of nΔx in a vacuum.
iii. nΔx is called the optical path travelled by a wave.
iv. This means, the optical path through a medium is the effective path travelled by light in a vacuum to generate the same phase difference.
v. Optical path in a medium can also be defined as the corresponding path in a vacuum that the light travels at the same time as it takes in the given medium.
i.e., time = `"d"_"medium"/"v"_"medium" = "d"_"vacuum"/"v"_"vacuum"`
∴ `"d"_"vacuum" = "v"_"vacuum"/"v"_"medium" xx "d"_"medium" = "n" xx "d"_"medium"`
But `"d"_"vacuum"` = Optical path
∴ Optical path = n × `"d"_"medium"`
Thus, a distance d travelled in a medium of refractive index n introduces a path difference = nd - d = d(n - 1) over a ray travelling an equal distance through the vacuum.
संबंधित प्रश्न
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`.
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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