Advertisements
Advertisements
Questions
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
Solution 1
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.
Solution 2
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.
RELATED QUESTIONS
Write the important characteristic features by which the interference can be distinguished from the observed diffraction pattern.
Write the necessary conditions to obtain sustained interference fringes.
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
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.
Why two light sources must be of equal intensity to obtain a well-defined interference pattern?
Draw a neat labelled ray diagram of the Fresnel Biprism experiment showing the region of interference.
What is interference?
Obtain the relation between phase difference and path difference.
How do source and images behave as coherent sources?
Obtain the equation for resultant intensity due to interference of light.
Explain Young’s double-slit experimental setup and obtain the equation for path difference.
Discuss the interference in thin films and obtain the equations for constructive and destructive interference for transmitted and reflected light.
Two independent monochromatic sources cannot act as coherent sources, why?
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio `("I"_"max" - "I"_"min")/("I"_"max" + "I"_"min")` will be ______
On a rainy day, a small oil film on water shows brilliant colours. This is due to ____________.
In Young's experiment for the interference of light, the separation between the silts is d and the distance of the screen from the slits is D. If D is increased by 0.6% and d is decreased by 0.2%, then for the light of a given wavelength, which one of the following is true?
"The fringe width ____________."
In Young's double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength `lambda`. In another experiment with the same set up the two slits are of equal amplitude A and wavelength `lambda`. but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is ____________.
In Young's double slit experiment fifth dark fringe is formed opposite to one of the slits. If D is the distance between the slits and the screen and d is the separation between the slits, then the wavelength of light used is ______.
Two coherent light sources of intensity ratio 'n' are employed in an interference experiment. The ratio of the intensities of the maxima and minima in the interference pattern is (I1 > I2).
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 ______
In Young's double slit experiment, for wavelength λ1 the nth bright fringe is obtained at a point P on the screen. Keeping the same setting, source of light is replaced by wavelength λ2 and now (n + 1)th bright fringe is obtained at the same point P on the screen. The value of n is ______.
In Young's double-slit experiment, the distance between the slits is 3 mm and the slits are 2 m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength 480 nm and 600 run respectively. The separation on the screen between the 5th order bright fringes on the two interference patterns is ______
Two coherent sources of intensities I1 and I2 produce an interference pattern on the screen. The maximum intensity in the interference pattern is ______
White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. The difference in wavelengths of red and violet light is ______ nm.
In biprism experiment the maximum intensity is ‘I0’. If the path difference between the two interfering waves is ‘λ/4’ then intensity at the point on the screen is ______.
`[sin 45^circ = cos 45^circ = 1/sqrt 2]`
