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

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प्रश्न

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

संख्यात्मक
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उत्तर

  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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पाठ 7: Wave Optics - Evaluation [पृष्ठ १०४]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 7 Wave Optics
Evaluation | Q 3. | पृष्ठ १०४

संबंधित प्रश्‍न

How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit screens is doubled?


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Answer the following question.
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Answer in brief:

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A.
B.
C.
D.

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