Question

Draw the variation of intensity with angle of diffraction in single slit diffraction pattern. Write the expression for value of angle corresponding to zero intensity locations.

Hint:

The central peak is the tallest; zeros occur where \(a\sin\theta = n\lambda\).

Answer 1

Step 1: Set up the single slit pattern:
When light of wavelength \(\lambda\) falls on a single slit of width \(a\), a diffraction pattern forms on the screen. It has a bright central maximum with weaker bright fringes on both sides, separated by dark fringes.
Let \(\theta\) be the angle of diffraction measured from the central axis.

Step 2: Describe the graph of intensity against angle:
Draw the angle \(\theta\) on the horizontal axis with \(\theta = 0\) at the middle, and the intensity \(I\) on the vertical axis.
At \(\theta = 0\) the curve has a tall central peak of intensity \(I_0\).
The curve falls to zero at \(\theta = \pm \lambda/a\), \(\pm 2\lambda/a\), \(\pm 3\lambda/a\) and so on.
Between these zeros there are small secondary peaks, each much lower than the central one (the first is only about 4.5 percent of \(I_0\)). The curve is symmetric about \(\theta = 0\), and the central peak is twice as wide as each secondary peak.

Step 3: Condition for zero intensity:
Divide the slit into two halves. Rays from a point in the upper half and the matching point in the lower half differ in path by \(\dfrac{a}{2}\sin\theta\).
They cancel in pairs when this path difference is \(\dfrac{\lambda}{2}\), which gives the first minimum \(a\sin\theta = \lambda\).
Dividing the slit into 4, 6, ... equal parts in the same way gives the general result.
\[ a \sin\theta = n\lambda, \quad n = \pm 1, \pm 2, \pm 3, \ldots \]
For small angles, \(\sin\theta \approx \theta\), so \(\theta = \dfrac{n\lambda}{a}\). Here \(n = 0\) is excluded because it is the central maximum.

Final Answer:
The intensity graph has a tall central maximum at zero angle with weak secondary maxima on both sides. The intensity is zero at angles given by \(\sin\theta = n\lambda/a\) (\(n = \pm 1, \pm 2, \ldots\)), that is \(\theta \approx n\lambda/a\). \[ \boxed{a\sin\theta = n\lambda,\ n = \pm1, \pm2, \ldots} \]