Question

For a single slit of width "a", the first minimum of the interference pattem of a monochromatic light of wavelength i occurs at an angle of ;. At the same angle of )" [, we get a maximum for two narrow slits separated by a distance "a". Expla

Answer 1

Width of the slit = a

The path difference between two secondary wavelets is given by,

N λ = a sin θ.

Since, θ is very small sin θ = 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 = straight I subscript 1 space plus space straight I subscript 2 space plus 2 square root of straight I subscript 1 straight I subscript 2 cosθ end root

Since, θ = 0 (nearly) corresponding to angle λ/a so cos θ = 1 (nearly)

So,

space space space space space space space space space straight I space equals straight I subscript 1 space plus space straight I subscript 2 space plus 2 square root of straight I subscript 1 straight I subscript 2 cosθ end root

rightwards double arrow space space space italic space straight I space equals space straight I subscript 1 space plus space straight I subscript 2 space plus 2 space square root of straight I subscript 1 straight I subscript 2 end root cos left parenthesis 0 right parenthesis

rightwards double arrow space space space space straight I space equals space straight I subscript 1 space plus space straight I subscript 2 space plus 2 space square root of straight I subscript 1 straight I subscript 2 end root italic space

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