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In Young's double slit experiment the ra...

In Young's double slit experiment the ratio of intensitities of bright and dark fringes is 9 this means that

A

the intensities at the screen due to the two slits are 5 units and 4 units, respectively

B

the intensities at the screen due to the two slits are 4 unit and 1 units, respectively

C

the amplitude ratio is 3

D

the amplitude ratio is 2

Text Solution

Verified by Experts

The correct Answer is:
b.,d

We know that
`(I_(max))/(I_(min)) = ((a + b)^(2))/((a - b)^(2))`
Where `I_(1) prop a^(2)` (a is amplitude of 1 wave) and `I_(2) prop b^(2)` (b is amplitude of 2 wave). Hence,
`(I_(max))/(I_(min)) = (9)/(1) implies (a + b)/(a - b) = (3)/(1) implies (a)/(b) = (1)/(2)`
`:. (I_(1)^(2))/(I_(2)^(2)) = (a^(2))/(b^(2)) = (1)/(4)`
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Knowledge Check

  • In Young’s double slit experiment, the intensity of light coming from the first slit is double the intensity from the second slit. The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is

    A
    `2:1`
    B
    `34:1`
    C
    `9:1`
    D
    `8:1`
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