Question

In a single slit diffraction experiment, slit width 'a' is illuminated by wavelength '$\lambda$' and the width of central maxima is 'y'. When half the slit is covered and illuminated by $(1.5)\lambda$, the width of the central maximum becomes ______.

• A. $\frac{3}{2}y$
• B. $\frac{2}{3}y$
• C. $3y$
• D. $\frac{y}{3}$

Hint:

Remember the proportionality trick: $W \propto \frac{\lambda}{a}$. If $\lambda$ increases by a factor of 1.5, the width increases by 1.5. If $a$ is halved, the width doubles. Total scaling factor = $1.5 \times 2 = 3$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The angular width of the central maximum in single-slit diffraction is $2\theta \approx \frac{2\lambda}{a}$. The linear width $y$ is proportional to this angular width.

Step 2: Formula Application:
Initial width $y \propto \frac{\lambda}{a}$. New conditions: wavelength $\lambda' = 1.5\lambda$ and slit width $a' = a/2$ (since half is covered).

Step 3: Explanation:
New width $y' \propto \frac{\lambda'}{a'} = \frac{1.5\lambda}{a/2} = 3 \left( \frac{\lambda}{a} \right)$. Therefore, $y' = 3y$.

Step 4: Final Answer:
The width of the central maximum becomes $3y$.