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

Hint:

Distinguish between:
Linear width $\beta_0 = \frac{2\lambda D}{a}$ (depends on $D$, decreases when $D$ decreases).
Angular width $2\theta = \frac{2\lambda}{a}$ (independent of $D$, stays constant).

Answer 1

Concept:
• The angular half-width of central maximum is $\theta = \frac{\lambda}{a}$.

• The linear width $\beta_0$ of the central maximum on a screen placed at distance $D$ is $\beta_0 = 2 y_1 = \frac{2 \lambda D}{a}$.

Step 1: Formula for linear width
The distance of the first minimum from the center of the screen is $y_1 = \frac{\lambda D}{a}$.
The total linear width $\beta_0$ of the central maximum extends between first minima on either side:
\[ \beta_0 = 2 y_1 = \frac{2 \lambda D}{a} \]

Step 2: Analyze effect of decreasing D
From the formula $\beta_0 = \frac{2 \lambda D}{a}$, linear width $\beta_0$ is directly proportional to slit-to-screen distance $D$ ($\beta_0 \propto D$).
When the separation $D$ between the slit and the screen is decreased, the linear width $\beta_0$ of the central maximum decreases proportionally.

Step 3: Conclusion
Decreasing the distance $D$ causes the central maximum on the screen to become narrower linearly. Note that angular width ($2\theta = \frac{2\lambda}{a}$) remains unchanged as it depends only on wavelength $\lambda$ and slit width $a$.