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प्रश्न
Obtain the equation for resultant intensity due to interference of light.
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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.
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संबंधित प्रश्न
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)
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In Young's double-slit experiment, if the two sources of light are very wide, then ______.
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A double slit experiment is immersed in water of refractive index 1.33. The slit separation is 1 mm, distance between slit and screen is 1.33 m. The slits are illuminated by a light of wavelength 6300 Å. The fringe width is ______.
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(cos 60° = 0.5, `lambda` = wavelength of light, d = slit width)
With a neat labelled ray diagram explain the use of Fresnel's biprism to obtain two coherent sources.
