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The ratio of maximum intensity to minimu...

The ratio of maximum intensity to minimum intensity due to interference of two light waves of amplitudes `a_(1) and a_(2)` is

A

`(I_("max"))/(I_("min")) =((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))`

B

`(I_("max"))/(I_("min")) =(a_(1)+a_(2))/(a_(1)-a_(2))`

C

`(I_("max"))/(I_("min")) =(a_(1)^(2)+a_(2)^(2))/(a_(1)^(2)-a_(2)^(2))`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`I= a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2) cos (alpha_(1)-alpha_(2))`
Now, `(I_("max"))/(I_("min"))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))`
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Knowledge Check

  • If ratio of maximum to minimum intensity in an interference experiment is 16:1 then ratio of amplitudes of individual waves is-

    A
    `4:1`
    B
    `16:1`
    C
    `5:3`
    D
    `25:9`

  • Assertion: Ratio of maximum intensity and minimum intensity in interference is 25:1 . The amplitudes ratio of two waves should be 3:2 . Reason: I_(max)/I_(min) = ((A_1+A_2)/(A_1-A_2))^2 .

    A
    If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
    B
    If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    C
    If Assertion is true, but the Reason is false.
    D
    If Assertion is false but the Reason is true.

  • The phenomenon of producing oalternate points of maximum and minimum intensity due to the superposition of two light waves is

    A
    refraction of light
    B
    reflection of light
    C
    interference of light
    D
    diffraction of light

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