1.

In Young's double-slit experiment, one of the slit is wider than other so that the amplitude of the light from one slit is double of that from other slit. If I_m be the maximum intensity, the resultant intensity I when they interfere at phase difference phi is given by:

Answer»

`(I_m)/(9)(1 + 8 COS^2 (theta)/2)`
`(I_m)/(9)(4 + 5 cos^2 phi)`
`(I_m)/(3)(1 + 2 cos^2 (phi)/2)`
`(I_m)/(5)(1 + 3 cos^2 (phi)/2)`

Solution :LET `I_1 and I_2` be intensities which are coming from SLITS `S_1 and S_2`
The resultant intensity
`:. I = I_1 + I_2 + 2sqrt(I_1I_2) cos theta`
`I_2 = K(2A)^2`
`= 4KA^2`
`:. I_2 = 4I_1`

`I_m = I_1 + I_2 + 2sqrt(I_1I_2) cos 0^@`
`= I_1 + 4I_1 + 2sqrt(4I_1^2)`
`= I_1 + 4I_1 + 2 xx 2I_1`
`I_m = 9I_1 implies I_1 = (I_m)/9 implies I_2 = (4I_m)/(9)`
`I = (I_m)/9 + (4I_m)/9 + 2 SQRT((I_m)/(9) xx (4I_m)/(9)) cos theta`
`= (I_m)/(9) + (4I_m)/9 + 2sqrt((4I_m^2)/(81)) cos theta`
`= (I_m)/(9) (5 + 4 cos theta) = (I_m)/9 (1 + 8 cos^2 theta//2)`


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