Question

The intensity at the central maxima in Young’s double slit experiment is I₀. Find out the intensity at a point where the path difference is` lambda/6,lambda/4 and lambda/3.`

Answer 1

The intensity of central maxima is I₀. Let I₁ and I₂ be the intensity emitted by the two slits S₁ and S₂, respectively.

The expression for resultant intensity is

`I=I_1+I_2+2sqrt(I_1I_2)cosphi`

For central maxima, I = I₀ and Φ = 0

We assume I₁ =  I₂

∴ I₀=2I₁+2I₁ cos0=4I₁

∴ I₁ = I₂= `I_0/4`

Now, when the path difference is  `lambda/6`we get

`phi=(2pi)/lambdaxxp.d=(2pi)/lambdaxxlambda/6=pi/3`

`:.I'=I_1+I_2+2sqrt(I_1I_2)cos`

`:.I'=2I_0/4+2I_0/4xx1/2`

`:.I'=I_0/2+I_0/4=(3I_0)/4`

Similarly, when the path difference is `lambda/4`we get

`phi=(2pi)/lambdaxxp.d=(2pi)/lambdaxxlambda/4=pi/2`

`:.I'=I_1+I_2+2sqrt(I_1I_2)cos""pi/2`

 `:.I'=2I_0/4+0`

`:.I'=I_0/2`

 Finally, when the path difference is `lambda/3`we get

`phi=(2pi)/lambdaxxp.d=(2pi)/lambdaxxlambda/3=(2pi)/3`

`:.I'=I_1+I_2+2sqrt(I_1I_2)cos ""(2pi)/3`

`:.I'=2I_0/4+2I_0/4xx(-1/2)`

`:.I'=I_0/2-I_0/4=I_0/4`