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प्रश्न
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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उत्तर
`"I"_"max"/"I"_"min" = 36/1`
`"I"_"max"/"I"_"min" = ("a"_1 + "a"_2)^2/("a"_1 - "a"_2)^2`
`("a"_1 + "a"_2)^2/("a"_1 - "a"_2)^2 = 36/1 = 6^2/1^2`
`("a"_1 + "a"_2)/("a"_1 - "a"_2) = 6/1`
a1 + a2 = 6a1 - 6a2
a2 + 6a2 = 6a1 - a1
7a2 = 5a1
`7/5 = "a"_1/"a"_2`
a1 : a2 = 7 : 5
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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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Answer in brief:
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