Question

A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generated a wave of intensity $l_0$ which is equally divided into two parts : One part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed. If a maxima is formed at the detector then, the magnitude of wavelength $\lambda$ of the wave produced is given by :

• A. $\pi R$
• B. $\frac{\pi R}{2}$
• C. $\frac{\pi R}{4}$
• D. $\frac{2\pi R}{3}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\pi R$
B $\frac{\pi R}{2}$
C $\frac{\pi R}{4}$

Net path difference between the waves travelling through path $I$ and

$II$, $x=\frac{3}{4}2\pi R-\frac{1}{4}2\pi R$

$⟹x=\pi R$

Let $\lambda$ be the wavelength of the waves.

Now for maxima to occur at D, path difference $=n\lambda$

where $n=1,2,.........$

Thus $n\lambda =\pi R⟹\lambda =\frac{\pi R}{n}$

Now $n=1,2,$ and $4$ $⟹\lambda =\pi R,\frac{\pi R}{2}$ and $\frac{\pi R}{4}$ respectively.