Advertisements
Advertisements
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?
Advertisements
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
State any one difference between interference of light and diffraction of light
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 long narrow horizontal slit is paced 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?
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.
What is meant by coherent sources?
Describe Young's double-slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.
What is interference of light?
What is phase of a wave?
How does wavefront division provide coherent sources?
How do source and images behave as coherent sources?
In Young’s double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0 = 750 nm and λ = 900 nm. What is the minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other?
A thin transparent sheet is placed in front of a slit in Young's double slit experiment. The fringe width will ____________.
The distance between the first and ninth bright fringes formed in a biprism experiment is ______.
(`lambda` = 6000 A, D = 1.0 m, d = 1.2 mm)
The phenomenon of interference is based on ______.
Two sources of light 0.5 mm apart are placed at a distance of 2.4 m and wavelength of light is 5000 Å. The phase difference between the two light waves interfering on the screen at a point at a distance 3 mm from central bright band is ____________.
If two waves represented by `"y"_1 = 3 "sin" omega "t"` and `"y"_2 = 5 "sin" (omega "t" + pi/3)` interfere at a point, then the amplitude of the resulting wave will be about ____________.
In a double slit experiment, the separation between the slits is d and distance of screen from slits is D. If the wavelength of light used is `lambda` and I is the intensity of central bright fringe, then intensity at distance x from central maximum is given by ____________.
Two identical light sources s1 and s2 emit light of same wavelength `lambda`. These light rays will exhibit interference if their ______.
In biprism experiment, if the 5th bright band with wavelength 'λ1' coincides with the 6th dark band with wavelength 'λ2' then the ratio `(lambda_2/lambda_1)` is ______
The graph shows the variation of fringe width (β) versus distance of the screen from the plane of the slits (D) in Young's double-slit experiment Keeping other parameters the same. The wavelength of light used can be calculated as d = distance between the slits ______
Waves from two coherent sources of light having an intensity ratio I1 : I2 equal to 'x' interfere. Then in the interference pattern obtained on the screen, the value of (Imax - Imin)/(Imax + Imin) is ______
Two waves with same amplitude and frequency superpose at a point. The ratio of resultant intensities when they arrive in phase to that when they arrive 90° out of phase is ______.
`[cos pi/2=0]`
If we have two coherent sources S1 and S2 vibrating in phase, then for an arbitrary point P constructive interference is observed whenever the path difference is ______.
What is meant by Constructive interference?
How will the interference pattern of Young's double slit change if one of the two slits is covered by a paper which transmits only half of the light intensity?
Two coherent sources P and Q produce interference at point A on the screen where there is a dark band which is formed between 4th bright band and 5th bright band. Wavelength of light used is 6000 Å. The path difference between PA and QA is ______.
In biprism experiment, the 3rd dark band is formed opposite to one of the slits. The wavelength of light used is ______.
(D = distance between source and screen, d = distance between the two narrow slits)
