In the figure shown, a parallel beam of light (of wavelength `lambda_1` in
medium `mu_1`) is incident at an angle `theta`. Distance `S_1O = S_2O`. Distance between the
slits is d.
`
Using d = 1mm, D = 1m, `theta = 30^@` , `lambda_1 = 0.3mm, mu_1 = 4//3` and `mu_2= 10//9`, find
(a) the y-coordinate of the point where the total phase difference between the interfering
waves is zero.
(b) If the intensity due to each light wave at point O is `I_0`, then find the resultant intensity
at O.
Find y-coordinate of the nearest maxima above O.

Answer

A, B

Explanation

(a)Wavelength of light in air,
` lambda_0 = mu_1lambda_1 = 0.4mm`
Wavelength of light in medium-2,
`lambda_2= lambda_0/mu_2 = 0.4/((10//9)) = 0.36mm `
Let net path difference at some angle alpha is zero, then
` Delta phi_1+ Delta phi_2 = 0 `
` or `2pi/lambda_1 (Deltax_1) + 2pi/lambda_2 (Deltax_2) = 0 `
or ` Deltax_1/lambda_1+Deltax_2/lambda_2=0`
`or ((d sin theta)/lambda_1) + (d sin alpha)/lambda_2 = 0 `
` sin alpha = - lambda_2/lambda_1sin theta `
` = - 0.36/0.4 xx 1/2 = - 0.45`
` :. alpha ~~ -26.74^@ `
` y= D tan alpha `
` = (1m) tan (-26.74^@)`
` ~~ -1/2m.`
(b) At O, net phase difference,
` Delta phi or phi = ((2pi)/lambda_1) Deltax_1 `
`=(2pi)/lambda_1 (d sin theta) `
` = (2pi)/0.3 (1) sin 30^@ `
`=(10pi)/3 = 600^@`
Using the equation ` I=4 I_0 cos^(2) phi/2` , we have
`I = 4 (I_0) cos^(2)(300^@)`
`=I_0`
(c)At O, phase difference `Deltaphi_1` is `(10pi)/3` which is greater than` 2pi` but less than `4pi`. So, to make it `4pi`,
we must have
`Delta phi_1 + Deltaphi_2 = 4pi`
`Deltaphi_2 =4pi - (10pi)/3=(2pi)/3`
` Deltaphi_2 = (2pi)/3 = (2pi)/lambda_2 (Deltax_2) = (2pi)/lambda_2(d sin alpha)`
or `sinalpha = lambda_2/(3d)=0.36/(3xx1) = 0.12`
`:. alpha ~~ 6.89^(@)`
Now, `y=D tan alpha`
`=(1m)tan(6.89^@)`
=0.12m.