Question

If wavelength of light and separation between slit and screen are fixed. If the slit width is halved, then the angular width become

Hint:

Narrower slit $\implies$ broader diffraction pattern. The angular spread is purely a consequence of the ratio of wavelength to slit size ($\lambda/a$).

Answer 1

Step 1: Understanding the Concept:
This refers to Fraunhofer diffraction due to a single slit. The light passing through the slit spreads out, forming a central bright maximum flanked by secondary maxima and minima. The angular width describes how much the central maximum spreads out angularly.

Step 2: Key Formula or Approach:
The angular width ($2\theta$) of the central maximum in a single slit diffraction pattern is given by:
\[ \text{Angular width } (2\theta) = \frac{2\lambda}{a} \]
Where:
$\lambda$ = wavelength of light
$a$ = width of the slit

Step 3: Detailed Explanation:
The problem states that the wavelength ($\lambda$) is fixed. The separation between slit and screen ($D$) is also fixed, though it only affects linear width, not angular width.
From the formula, the angular width is inversely proportional to the slit width ($a$):
\[ \text{Angular width} \propto \frac{1}{a} \]
Let initial slit width be $a_1 = a$. Initial angular width is $\theta_1 = \frac{2\lambda}{a}$.
The new slit width is halved: $a_2 = a/2$.
The new angular width $\theta_2$ will be:
\[ \theta_2 = \frac{2\lambda}{a_2} \]
\[ \theta_2 = \frac{2\lambda}{a/2} \]
\[ \theta_2 = 2 \times \left(\frac{2\lambda}{a}\right) \]
\[ \theta_2 = 2 \times \theta_1 \]
The new angular width is twice the original angular width.

Step 4: Final Answer:
The angular width becomes doubled.