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प्रश्न
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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उत्तर
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`
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