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. The maximum value of $\lambda$ to produce a maxima at D is given by :

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

Answer 1

The correct option is A $\pi R$

Net path difference between the waves travelling through path $I$ and $II$,
$\Delta x=\frac{3}{4}2\pi R-\frac{1}{4}2\pi R$
$⟹\Delta x=\pi R$
Let $\lambda$ be the wavelength of the waves.
Now for maxima to occur at D, path difference $=n\lambda$
where $n=1,2,.........$
Thus $n\lambda =\pi R⟹\lambda =\frac{\pi R}{n}$
For maximum value of wavelength, $n=1$
$⟹{\lambda }_{max}=\pi R$