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Obtain the equation for resultant intensity due to interference of light. - Physics

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Question

Obtain the equation for resultant intensity due to interference of light.

Numerical

Solution

  1. 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.
  2. Let us consider two light waves from the two sources S1 and S2 meeting at a point P as shown
  3. 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
  4. 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)
  5. The resultant amplitude is maximum.
    Amax = `sqrt(("a"_1 + "a"_2)^2)`;
    when Φ = 0, ±2π, ± 4π …….(3)
  6. The resultant amplitude is minimum.
    Amin = `sqrt(("a"_1 - "a"_2)^2)`;
    when Φ = 0, ±π, ± 3π ± 5π …..(4)
  7. The intensity of light is proportional to the square of amplitude.
  8. I α A2 ……(5)
    Now equation (1) becomes
    I α I1 + I2 + 2`sqrt("I"_1"I"_2)` cos Φ .(6)
  9. 9. If the phase difference, Φ = 0, ± 2π, ± 4π., it corresponds to the condition for maximum intensity of light called as constructive interference.
  10. The resultant maximum intensity is,
    Imax α (a1 + a2)2 …….(7)
  11. If the phase difference, Φ = + π, ± 3π, ± 5π …., it corresponds to the condition for the minimum intensity of light called destructive interference.
  12. 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))`
  13. 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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Chapter 7: Wave Optics - Evaluation [Page 104]

APPEARS IN

Samacheer Kalvi Physics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Wave Optics
Evaluation | Q 3. | Page 104

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