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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)
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`.
What are the two methods for obtaining coherent sources in the laboratory?
What is interference?
What are coherent sources of light?
One of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will,
What is phase of a wave?
A graph is plotted between the fringe-width Z and the distance D between the slit and eye-piece, keeping other adjustment same. The correct graph is
A. |
B. |
C. |
D. |
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`[cos pi/2=0]`
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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 ______.
