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 Question:
We are dealing with Fraunhofer single slit diffraction. We need to find how the width of the central maximum changes when both the slit width and the incident wavelength are altered.

Step 2: Key Formula or Approach:
The linear width of the central maximum in a single slit diffraction pattern is given by:
$$W = \frac{2\lambda D}{a}$$
Where $D$ is the distance to the screen, $\lambda$ is the wavelength, and $a$ is the slit width.
From this, we see that Width $\propto \frac{\lambda}{a}$.

Step 3: Detailed Explanation:
Initial conditions: Width $= y$, Wavelength $= \lambda$, Slit width $= a$.
$$y = \frac{2\lambda D}{a}$$
New conditions:
The slit is half-covered, so the effective new slit width is $a' = a/2$.
The new wavelength is $\lambda' = 1.5\lambda = \frac{3}{2}\lambda$.
Substitute these into the width formula:
$$y' = \frac{2\lambda' D}{a'} = \frac{2 \left(\frac{3}{2}\lambda\right) D}{\frac{a}{2}}$$
$$y' = \frac{3\lambda D}{a / 2} = 6 \frac{\lambda D}{a}$$
Rewrite this in terms of the original width $y$:
Since $y = \frac{2\lambda D}{a}$, we can substitute $\frac{\lambda D}{a} = \frac{y}{2}$:
$$y' = 6 \left(\frac{y}{2}\right) = 3y$$

Step 4: Final Answer:
The new width is $3y$, matching option (c).