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Karnataka Board PUCPUC Science Class 11

Two Coherent Point Sources S1 and S2, Vibrating in Phase, Emit Light of Wavelength λ . the Separation Between the Sources is 2 λ . - Physics

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Question

Two coherent point sources S1 and S2, vibrating in phase, emit light of wavelength \[\lambda.\] The separation between the sources is \[2\lambda.\] Consider a line passing through S2 and perpendicular to the line S1 S2. What is the smallest distance from S2 where a minimum intensity occurs?

Sum

Solution

Let P be the point of minimum intensity.

For minimum intensity at point P,

\[S_1 P - S_2 P = x = \left( 2n + 1 \right) \frac{\lambda}{2}\]

Thus, we get

\[\sqrt{Z^2 + \left( 2\lambda \right)^2} - Z = \left( 2n + 1 \right)  \frac{\lambda}{2}\]

\[ \Rightarrow  Z^2  + 4 \lambda^2  =  Z^2    \left( 2n + 1 \right)^2   \frac{\lambda^2}{4} + 2Z  \left( 2n + 1 \right)  \frac{\lambda}{2}\]

\[ \Rightarrow Z = \frac{4 \lambda^2 - \left( 2n + 1 \right)  2 \lambda^2 /4}{\left( 2n + 1 \right)  \lambda}\]

\[             = \frac{16 \lambda^2 - \left( 2n + 1 \right)}{4  \left( 2n + 1 \right)  \lambda}............(1)\]

\[\text{When }n = 0, Z = \frac{15\lambda}{4}\]

\[ n = - 1, Z = \frac{- 15\lambda}{4}\]

\[ n = 1, Z = \frac{7\lambda}{12}\]

\[ n = 2 , Z = \frac{- 9\lambda}{20}\]

Thus, \[Z = \frac{7\lambda}{12}\] is the smallest distance for which there will be minimum intensity.

 

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Chapter 17: Light Waves - Exercise [Page 382]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 17 Light Waves
Exercise | Q 27 | Page 382
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