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प्रश्न
For a single slit of width "a", the first minimum of the interference pattern of a monochromatic light of wavelength λ occurs at an angle of λa. At the same angle of λa, we get a maximum for two narrow slits separated by a distance "a". Explain.
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उत्तर
The path difference between two secondary wavelets is given by nλ = asinθ. Since, θ is very small sinθ = θ. So, for the first order diffraction n = 1, the angle is λ/a. Now we know that θ must be very small θ = 0 (nearly) because of which the diffraction pattern is minimum.
Now for interference case, for two interfering waves of intensity I1 and I2 we must have two slits separated by a distance. We have the resultant intensity
`I=I_1+I_2+2sqrt(I_1I_2)costheta`
Since, θ = 0 (nearly) corresponding to angle λ/a so cosθ = 1 (nearly)
So,
`I=I_1+I_2+2sqrt(I_1I_2)costheta`
`=>I=I_1+I_2+2sqrt(I_1I_2)cos(0)`
`=>I=I_1+I_2+2sqrt(I_1I_2)`
We see the resultant intensity is sum of the two intensities, so there is a maxima corresponding to the angle λ/a.
This is why, at the same angle of `lambda/a`we get a maximum for two narrow slits separated by a distance "a".
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