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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Answer in brief: Explain what is optical path length. How is it different from actual path length?

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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? 

थोडक्यात उत्तर
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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.

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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.  

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Wave Optics - Exercises [पृष्ठ १८४]

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Answer in brief:

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`[sin 45^circ = cos 45^circ = 1/sqrt 2]`


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