Question

In a single slit experiment, the width of the slit is doubled. Which one of the following statements is correct?

• A. The intensity and width of the central maximum are unaffected.
• B. The intensity and angular width both are doubled.
• C. The intensity increases by a factor 4 and the angular width decreases by a factor of $\frac{1}{2}$.
• D.

Hint:

Logic Tip: A wider slit results in a narrower and much brighter central diffraction peak.

Answer 1

The Correct Option is D

Concept:
For Fraunhofer diffraction due to a single slit of width $d$: \[ W=\frac{2\lambda D}{d} \] where:

• $W$ = linear width of central maximum
• $\lambda$ = wavelength of light
• $D$ = distance of screen from slit
• $d$ = slit width

Hence, width of central maximum is inversely proportional to slit width: \[ W \propto \frac{1}{d} \] Also, the amplitude at the center is proportional to slit width, therefore intensity is: \[ I \propto d^2 \]
Step 1: Given change in slit width
Original slit width: \[ d \] New slit width after doubling: \[ d' = 2d \]
Step 2: Find new angular / linear width of central maximum
Since: \[ W \propto \frac{1}{d} \] Therefore: \[ W'=\frac{W}{2} \] So, the width of the central maximum becomes half of its original value.
Step 3: Find new intensity
Since: \[ I \propto d^2 \] Then: \[ I' \propto (2d)^2 = 4d^2 \] Thus: \[ I' = 4I \] So, the intensity becomes four times the original intensity.
Step 4: Final Result
When the slit width is doubled:

• Width of central maximum becomes half.
• Intensity of central maximum becomes four times.

\[ \boxed{W'=\frac{W}{2}, \qquad I'=4I} \] Quick Tip:
In diffraction, increasing slit width makes the pattern narrower but brighter.