Question

When light of wavelength $\lambda$ incidents on a single slit of width 'a', then the angular width between the third order diffraction maxima on either side of the central maximum is:

• A. $\frac{9\lambda}{a}$
• B. $\frac{7\lambda}{a}$
• C. $\frac{5\lambda}{a}$
• D. $\frac{3\lambda}{a}$

Hint:

The angular width between any symmetric $n^{\text{th}}$ secondary maxima on either side of the central peak is always:
$\Delta \theta = (2n + 1)\frac{\lambda}{a}$.
For $n = 3$, substituting gives $\Delta \theta = (2(3) + 1)\frac{\lambda}{a} = \frac{7\lambda}{a}$.

Answer 1

The Correct Option is B

Step 1: Recall the total-width shortcut for the nth maxima.
Secondary maxima sit at $\theta_n = \left(n+\frac{1}{2}\right)\frac{\lambda}{a}$ on each side of the centre, so the full angular gap between the nth maxima on the two sides is simply double this value: \[ W_n = 2\theta_n = (2n+1)\frac{\lambda}{a} \]
Step 2: Plug in the order asked for.
Here the third order maxima are wanted, so $n = 3$: \[ W_3 = (2(3)+1)\frac{\lambda}{a} \]
Step 3: Simplify. \[ W_3 = 7\frac{\lambda}{a} \] \[ \boxed{W_3 = \dfrac{7\lambda}{a}} \]