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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

State any one difference between interference of light and diffraction of light


In Young’s double slit experiment, the slits are separated by 0.5 mm and screen is placed 1.0 m away  from the slit. It is found that the 5th bright fringe is at a distance of 4.13 mm from the 2nd dark fringe.  Find the wavelength of light used.  


Answer in brief:

Explain what is the optical path length. How is it different from actual path length?


Why two light sources must be of equal intensity to obtain a well-defined interference pattern?


In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to ______.


What is phase of a wave?


The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio `("I"_"max" - "I"_"min")/("I"_"max" + "I"_"min")` will be ______


In Young's double-slit experiment, if the width of the 2nd bright fringe is 4 x 10-2 cm, then the width of the 4th bright fringe will be ______ cm.


In Young's double-slit experiment, in an interference pattern, a second minimum is observed exactly in front of one slit. The distance between the two coherent sources is 'd' and the distance between source and screen is 'D'. The wavelength of the light source used is ______


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 ______.


In a Young's double-slit experiment, the intensity at a point where the path difference is `lambda/3` (`lambda` being the wavelength of the light used) is I. If I0 denotes the maximum intensity, then `"I"/"I"_0` is equal to ______.


In Young's double slit experiment fifth dark fringe is formed opposite to one of the slits. If D is the distance between the slits and the screen and d is the separation between the slits, then the wavelength of light used is ______.


Two coherent light sources of intensity ratio 'n' are employed in an interference experiment. The ratio of the intensities of the maxima and minima in the interference pattern is (I1 > I2).


In biprism experiment, the 4th dark band is formed opposite to one of the slits. The wavelength of light used is ______.


Two identical light sources s1 and s2 emit light of same wavelength `lambda`. These light rays will exhibit interference if their ______.


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 ______ 

 


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 ______.


With a neat labelled ray diagram explain the use of Fresnel's biprism to obtain two coherent sources.


In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index ‘μ’ is placed in one of the interfering paths. Now 7th bright fringe is obtained at the same point. If ‘A’ is the wavelength of light used, the thickness of film is equal to ______.


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