Advertisements
Advertisements
प्रश्न
Obtain the equation for resultant intensity due to interference of light.
Advertisements
उत्तर
- The phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and a decrease in intensity at some other points is called interference of light.
- Let us consider two light waves from the two sources S1 and S2 meeting at a point P as shown
- The wave from S1 at an instant t and P is, y1 = a1 sin ωt
The wave from S2 an instant t at P is
y2 = a2 sin(ωt + Φ)
Superposition principle - The two waves have different amplitudes a1 and a2, same angular frequency ω’ and a phase difference of Φ between them. The resultant displacement will be given by.
y = y1 + y2 = a1 sin ωt + a1 sin2 (ωt + Φ) y = A sin (ωt + Φ)
Where, A = `sqrt("a"_1^2 + "a"_2^2 + 2"a"_1"a"_2 cos phi)` .....(1)
`theta = tan^-1 ("a"_2 sin phi)/("a"_1 + "a"_2 cos phi)` ......(2) - The resultant amplitude is maximum.
Amax = `sqrt(("a"_1 + "a"_2)^2)`;
when Φ = 0, ±2π, ± 4π …….(3) - The resultant amplitude is minimum.
Amin = `sqrt(("a"_1 - "a"_2)^2)`;
when Φ = 0, ±π, ± 3π ± 5π …..(4) - The intensity of light is proportional to the square of amplitude.
- I α A2 ……(5)
Now equation (1) becomes
I α I1 + I2 + 2`sqrt("I"_1"I"_2)` cos Φ .(6) - 9. If the phase difference, Φ = 0, ± 2π, ± 4π., it corresponds to the condition for maximum intensity of light called as constructive interference.
- The resultant maximum intensity is,
Imax α (a1 + a2)2 …….(7) - If the phase difference, Φ = + π, ± 3π, ± 5π …., it corresponds to the condition for the minimum intensity of light called destructive interference.
- The resultant minimum intensity is Imin α
(a1 – a2)2 α I1 + I2 – 2`sqrt("I"_1"I"_2)` ......(8)
As a special case, if a1 = a2 = a, then equation (1) becomes,
A = `sqrt(2"a"^2 + 2"a"^2 cos phi)`
= `sqrt(2"a"^2 (1 + cos phi))`
= `sqrt(2"a"^2 2cos^2 (phi//2))` -
A = 2 a cos(Φ/2) ….(9)
I α 4a2 cos2 (Φ/2) [∴ I α A2] ……(10)
I α 4 I0 cos2 (Φ/ 2) [ΦI0 α a2] …….(11)
IMax = 4I0 when, Φ = o, ± 2π, 4π …..(12)
Imin = 0 when, Φ = ± π, ± 3π, ± 5π …..(13)
Conclusion:
The phase difference between the two waves decides the intensity of light meet at a point.
APPEARS IN
संबंधित 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)?
Draw a neat labelled ray diagram of the Fresnel Biprism experiment showing the region of interference.
Obtain the relation between phase difference and path difference.
What is intensity (or) amplitude division?
How do source and images behave as coherent sources?
Discuss the interference in thin films and obtain the equations for constructive and destructive interference for transmitted and reflected light.
The ratio of maximum and minimum intensities in an interference pattern is 36 : 1. What is the ratio of the amplitudes of the two interfering waves?
A metal rod has length, cross-sectional area and Young's modulus as L, A and Y, respectively. If the elongation in the rod produced is l, then work done is proportional to ______.
On a rainy day, a small oil film on water shows brilliant colours. This is due to ____________.
Band width for red light of wavelength 6600 Å is 0.42 mm. If red Light is replaced by blue light of wavelength 4400 Å, then the change m bandwidth will be ____________.
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)
Two identical light waves having phase difference 'Φ' propagate in same direction. When they superpose, the intensity of the resultant wave is proportional to ______.
In Young's double slit experiment with a source of light of wavelength 5860 Å, the first maxima will occur when ____________.
In interference experiment, intensity at a point is `(1/4)^"th"` of the maximum intensity. The angular position of this point is at (sin30° = cos60° = 0.5, `lambda` = wavelength of light, d = slit width) ____________.
In a biprism experiment, red light of wavelength 6500 Å was used. It was then replaced by green light of wavelength 5200 Å. The value of n for which (n + 1)th green bright band would coincide with nth red bright band for the same setting 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 ______
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 ______
Describe Young's double-slit interference experiment.
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]`
