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Slit 1 of a double slit is wider than sl...

Slit 1 of a double slit is wider than slit 2, so that the light from slit 1 has an amplitudes three times that of the light from slit 2. Show that equation `I=I_(max) "cos"^2 phi/2` is replaced by the equation,
`I = (I_(max)/4)1+(3cos^2(phi/2))` .

Text Solution

Verified by Experts

`A_1 =3A_2`
`So, I_1 = 9I_2`
Let `A_2 =A_0 and I_2 = I_0`
Then, `A_1 = 3A_0` and `I_1 = 9I_0`
`A_(max) = (A_1+A_2) = 4A_0`
and `I_(max) = 16I_0`
Now, `I= I_1 +I_2 + 2sqrtI_1I_2 cos phi`
`=9I_0 + I_0 + 2sqrt(9I_0 xx I_0) cos phi`
`=10I_(0)+6I_(0) cosphi`
`=10xxI_(max)/16+6((I_max)/16) cosphi`
`=5/8I_(max)+3/8I_(max) cos(phi)`
` =5/8I_(max)+3/8I_(max) (2"cos"^2 phi/2 -1)
`=1/4I_(max)+3/4I_(max).cos^2phi/2`
`I_(max)/(4)(1+3cos^2phi/2)`.
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