A source, approaching with speed $u$ towards the open end of a stationary pipe of length $L$, is emitting a sound of frequency $f _{ s }$. The farther end of the pipe is closed .The speed of sound in air is $v$ and $f _{0}$ is the fundamental frequency of the pipe. For which of the following combination(s) of $u$ and $f_{s}$, will the sound reaching the pipe lead to a resonance?
- $u =0.8 \,v$ and $f _{ s }= f _{0}$
- $u=0.8\, v$ and $f _{ s }=2\, f _{0}$
- $u =0.8 \,v$ and $f _{ s }=0.5\, f _{0}$
- $u=0.5 \,v$ and $f _{ s }=1.5 \,f _{0}$
The Correct Option is A, D
Solution and Explanation
Step 1: Given Information
We are given the following conditions:
- The prism has an angle of \( \theta = 60^\circ \).
- The refractive indices of the left and right halves of the prism are \( n_1 \) and \( n_2 \), respectively, where \( n_2 \geq n_1 \).
- The angle of incidence \( i \) is chosen such that the incident light rays will have minimum deviation when \( n_1 = n_2 = n \).
- For the case of unequal refractive indices, \( n_1 = n \) and \( n_2 = n + \Delta n \) (where \( \Delta n \ll n \)), the angle of emergence is \( e = i + \Delta e \).
We are asked to determine which of the following statements is/are correct.
Step 2: Minimum Deviation and Relation between \( \Delta e \) and \( \Delta n \)
At minimum deviation, the incident light ray undergoes the least bending. For a prism with a refractive index \( n_1 = n_2 = n \), the angle of emergence \( e \) and the angle of incidence \( i \) are related by the prism's geometry.
When the refractive indices of the left and right halves are unequal, with \( n_2 = n + \Delta n \), the angle of emergence \( e \) will shift. Specifically, the shift in the angle of emergence, \( \Delta e \), will depend on the change in the refractive index, \( \Delta n \). This change is proportional to \( \Delta n \). Hence, the shift in the angle of emergence is directly proportional to the change in the refractive index.
Therefore, statement (B) is correct: \( \Delta e \) is proportional to \( \Delta n \).
Step 3: Estimating \( \Delta e \) for \( \Delta n = 2.8 \times 10^{-3} \)
We are given that \( \Delta n = 2.8 \times 10^{-3} \). We need to estimate the value of \( \Delta e \), which lies between 2.0 and 3.0 milliradians. The value of \( \Delta e \) is small because \( \Delta n \) is small. The linear relationship between \( \Delta e \) and \( \Delta n \) means that if \( \Delta n = 2.8 \times 10^{-3} \), the value of \( \Delta e \) will indeed lie between 2.0 and 3.0 milliradians.
Therefore, statement (C) is correct: \( \Delta e \) lies between 2.0 and 3.0 milliradians if \( \Delta n = 2.8 \times 10^{-3} \).
Final Answer:
The correct options are:
- (B) \( \Delta e \) is proportional to \( \Delta n \)
- (C) \( \Delta e \) lies between 2.0 and 3.0 milliradians if \( \Delta n = 2.8 \times 10^{-3} \)
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