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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
Write the important characteristic features by which the interference can be distinguished from the observed diffraction pattern.
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)
Explain constructive and destructive interference with the help of a diagram?
The ratio of maximum and minimum intensities in an interference pattern is 36 : 1. What is the ratio of the amplitudes of the two interfering waves?
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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.
A thin transparent sheet is placed in front of a slit in Young's double slit experiment. The fringe width will ____________.
On a rainy day, a small oil film on water shows brilliant colours. This is due to ____________.
A wire of length 'L' and area of cross-section · A' is made of material of Young's modulus 'Y'. It is stretched by an amount 'x'. The work done in stretching the wire is ______.
In a double slit experiment, the two slits are 2 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?
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).
If the two slits in Young's double slit experiment have width ratio 9 : 1, the ratio of maximum to minimum intensity in the interference pattern is ______.
Two identical light sources s1 and s2 emit light of same wavelength `lambda`. These light rays will exhibit interference if their ______.
If two light waves reaching a point produce destructive interference, then the condition of phase difference is ______
In Young's double-slit experiment, if the two sources of light are very wide, then ______.
In Young's double slit experiment, for wavelength λ1 the nth bright fringe is obtained at a point P on the screen. Keeping the same setting, source of light is replaced by wavelength λ2 and now (n + 1)th bright fringe is obtained at the same point P on the screen. The value of n is ______.
In Young's double-slit experiment, the distance between the slits is 3 mm and the slits are 2 m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength 480 nm and 600 run respectively. The separation on the screen between the 5th order bright fringes on the two interference patterns is ______
In Young's double-slit experiment, the distance between the slits is 3 mm and the slits are 2 m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength 480 nm and 600 run respectively. The separation on the screen between the 5th order bright fringes on the two interference patterns is ______
Two coherent sources of intensities I1 and I2 produce an interference pattern on the screen. The maximum intensity in the interference pattern is ______
In a double-slit experiment, the optical path difference between the waves coming from two coherent sources at a point P on one side of the central bright is 7.5 µm and that at a point Q on the other side of the central bright fringe and 1.8 µm. How many bright and dark fringes are observed between points P and Q if the wavelength of light used is 600 nm?
