Q. In young’s double slit experiment one of the slits is wider than other, so that amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I, when they interfere at phase difference φ is given by

(a) I_{m}/9(4 + 5 cosφ )

(b) I_{m}/3 (1+2 cos^{2} (φ/2 ))

(c) I_{m}/5 (1+4 cos^{2} (φ/2))

(d) I_{m}/9 (1+8 cos^{2} (φ/2 ))

Ans: (d)

Sol: a_{1} = 2 a_{2}

I_{1} = 4I’ & I_{2} = I’

$ \displaystyle I_m = (\sqrt {I_1} + \sqrt {I_2} )^2 $

$ \displaystyle I_m = (\sqrt {4I’} + \sqrt {I’} )^2 = 9I’ $

$\displaystyle I’ = \frac{I_m}{9} $ ….(i)

$\displaystyle I = I_1 + I_2 + 2\sqrt{I_1 I_2} cos\phi $

$ \displaystyle I = 4I’ + I’ + 2\sqrt{4I’ \times I’} cos\phi $

$ \displaystyle = 5I’ + 4I’ cos\phi $

$ \displaystyle = 5I’ + 4I'(2 cos^2 \phi/2 – 1 ) $

$ \displaystyle = I’ + 8I’cos^2 \phi/2 $

$ \displaystyle = I'(1 + 8 cos^2 \phi/2 ) $

*From(i)*

$\displaystyle I = \frac{I_m}{9}(1 + 8 cos^2 \phi/2 ) $