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 l0 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 λ to produce a minima at D is given by :

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 l0​ 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 λ to produce a minima at D is given by :

Filo · Accepted Answer

Net path difference between the waves travelling through path

I

and

II

,

x = \frac{3}{4} 2 π R - \frac{1}{4} 2 π R

⟹ x = π R

Let

λ

be the wavelength of the waves.

Now for minima to occur at D, path difference

= (n + \frac{1}{2}) λ

where

n = 0, 1, 2, .........

Thus

\frac{( 2 n + 1 ) λ}{2} = π R ⟹ λ = \frac{2 π R}{2 n + 1}

Now wavelength is maximum for

n = 0

Thus

λ_{ma x} = 2 π R

Question Text	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 $λ$ to produce a minima at D is given by :
Answer Type	Text solution:1
Upvotes	150

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