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# Why We Get Maximum Wavelength for Two Narrow Slits Separated by a Distance "A" - Physics

ConceptFraunhofer Diffraction Due to a Single Slit

#### Question

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.

#### Solution

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/awe get a maximum for two narrow slits separated by a distance "a".

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Solution Why We Get Maximum Wavelength for Two Narrow Slits Separated by a Distance "A" Concept: Fraunhofer Diffraction Due to a Single Slit.
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