Question 13

For maximum amplitude at the detector $D$, the path difference $\Delta x$ between the different routes must be an integer multiple of the wavelength $\lambda$. For the smallest value of $l$, we set $\Delta x = \lambda = 0.29$ m.

(P) The two paths are the semi-circle (arc length $= \pi(l/2)$) and the straight line segment (length $= l$).

$\Delta x = \frac{\pi l}{2} - l = l\left(\frac{\pi}{2} - 1\right) \approx 0.57l$

$0.57l = 0.29 \implies l = \frac{0.29}{0.57} \approx 0.51$ m. So, P $\to$ 3.

(Q) The top path has length $0.5l + l + 0.5l = 2l$ and the bottom path has length $l$.

$\Delta x = 2l - l = l$

$l = 0.29$ m. So, Q $\to$ 4.

(R) The two paths are the 'corner' of a square ($l+l=2l$) and the semi-circular arc whose diameter is the diagonal $\sqrt{2}l$.

Path length of arc $= \frac{\pi(\sqrt{2}l)}{2} \approx 1.57 \times 1.414l \approx 2.22l$

$\Delta x = 2.22l +l -l= 2.22l$

$2.22l = 0.29 \implies l = \frac{0.29}{2.22} \approx 0.13$ m. So, R $\to$ 5.

(S) The two paths are the triangle sides ($a+b$) and the base $l$. The angles are $45^\circ$ and $30^\circ$ ($180^\circ - 105^\circ - 45^\circ = 30^\circ$).

Using the Sine Rule: $\frac{a}{\sin 30^\circ} = \frac{b}{\sin 45^\circ} = \frac{l}{\sin 105^\circ}$

Top path $L = a + b = l \frac{\sin 30^\circ + \sin 45^\circ}{\sin 105^\circ} = l \frac{0.5 + 0.707}{0.97} \approx 1.244l$

$\Delta x = 1.244l - l = 0.244l$

$0.244l = 0.29 \implies l = \frac{0.29}{0.244} \approx 1.19$ m. So, S $\to$ 2.

Final match: P-3, Q-4, R-1, S-2. Correct Option: D