Question

In a single-slit diffraction experiment, light of wavelength $\lambda$ illuminates the slit of width 'a'. The diffraction pattern is observed on a screen kept at a distance $D$ from the slits.

(i) Depict variation of intensity in the fringe pattern with the angular position of the fringes.

(ii) How is the linear width of central maximum affected when separation between the slit and the screen is decreased ?

Hint:

For single-slit diffraction, \[ W=\frac{2D\lambda}{a} \] Increase \(D\) or \(\lambda\) to obtain a wider central maximum.

Answer 1

Step 1: Recall the single-slit pattern:
Light of wavelength \(\lambda\) falls on a slit of width \(a\). The pattern forms on a screen at distance \(D\).
Minima occur where \(a\sin\theta = n\lambda\), with \(n = \pm 1, \pm 2, \ldots\)
The central maximum lies between the first minima on the two sides.

Step 2: Part (i), draw the intensity graph:
Take the angular position \(\theta\) on the horizontal axis and the intensity \(I\) on the vertical axis.
Draw a tall, broad peak at \(\theta = 0\). This is the central maximum with intensity \(I_0\).
The curve falls to zero at \(\theta = \pm\dfrac{\lambda}{a}\). These are the first minima.
Beyond them draw small side peaks between \(\pm\dfrac{\lambda}{a}\) and \(\pm\dfrac{2\lambda}{a}\), between \(\pm\dfrac{2\lambda}{a}\) and \(\pm\dfrac{3\lambda}{a}\), and so on.
Zeros lie at \(\pm\dfrac{\lambda}{a}, \pm\dfrac{2\lambda}{a}, \pm\dfrac{3\lambda}{a}\).
The central peak is twice as wide as each side peak. The first side peak is only about \(4.5\%\) of \(I_0\). The graph is symmetric about \(\theta = 0\).

Step 3: Part (ii), width of the central maximum:
The first minimum is at angle \(\theta \approx \dfrac{\lambda}{a}\).
Its distance from the centre of the screen is \(x = D\tan\theta \approx \dfrac{D\lambda}{a}\).
The linear width of the central maximum is the distance between the two first minima.
\[ W = 2x = \frac{2D\lambda}{a} \]

Step 4: Effect of decreasing the screen distance:
Here \(\lambda\) and \(a\) stay the same, so \(W \propto D\).
When \(D\) decreases, \(W\) decreases in the same ratio.
For example, if \(D\) is halved, \(W\) is also halved.
The angular width \(\dfrac{2\lambda}{a}\) does not change.

Final Answer:
(i) The graph has one broad central peak with weak, narrower side maxima and zeros at \(\theta = \pm n\lambda/a\). (ii) \(W = \dfrac{2D\lambda}{a}\), so the linear width of the central maximum decreases when the slit-screen distance decreases.