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The intensity of the light coming from one of the slits in Young's experiment is twice the intensity of the light coming from the other slit. - Physics

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प्रश्न

The intensity of the light coming from one of the slits in Young's experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern?

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उत्तर

Data: I1 : I2 = 2 : 1

If E10 and E20 are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is

`("I"_"max")/("I"_"min") = (("E"_10 + "E"_20)/("E"_10 - "E"_20))^2 = (("r + 1")/("r - 1"))^2`

where r = `"E"_10/"E"_20`

∴ `"I"_1/"I"_2 = (("E"_10)/("E"_20))^2 = "r"^2`

∴ r = `sqrt("I"_1/"I"_2) = sqrt2`

∴ `("I"_"max")/("I"_"min") = ((sqrt2 + 1)/(sqrt2 - 1))^2 = (2.414/0.414)^2 = (5.83)^2`

= 33.99 ≈ 34.

∴ The ratio of the intensities of the bright and dark fringes in the resulting interference pattern is 34: 1.

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पाठ 7: Wave Optics - Exercises [पृष्ठ १८५]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
पाठ 7 Wave Optics
Exercises | Q 20. | पृष्ठ १८५

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)


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