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The Intensity at the Central Maxima in Young’S Double Slit Experiment is I0. - Physics

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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.`

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Solution

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`

Concept: Interference of Light Waves and Young’S Experiment
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