Maxima in tube

Classical Mechanics Level 3

A narrow tube is bent in the form of a circle of radius R R as shown Two small holes S a n d D S~and~D are made in the tube at positions right angle to each other. A source placed at S S generated a wave of intensity I o I_{o} which is equally divided into two parts. One part travels along the longer path while the other travels along the shorter path. Both the waves meet the point D D where a detector is placed.

If maxima is found at the detector then, the magnitude of wavelength λ \lambda cannot be

4 π R 6 \frac{4 \pi R}{6} 2 π R 8 \frac{2 \pi R}{8} 2 π R 16 \frac{2 \pi R}{16} π R 2 \frac{ \pi R}{2}

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Nishant Rai
May 17, 2015

For a maxima at D D , path difference Δ x = n λ \Delta x = n\lambda

3 π R 2 − π R 2 = n λ \frac{3\pi R}{2} - \frac{\pi R}{2} = n\lambda

⇒ λ = π R n \Rightarrow ~ \lambda = \frac{\pi R}{n}

λ = π R , π R 2 , π R 3 , π R 4 , e t c . \lambda = {\pi R} , \frac{\pi R}{2} , \frac{\pi R}{3} , \frac{\pi R}{4} , etc.

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