Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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The Following Figure Shows Three Equidistant Slits Being Illuminated by a Monochromatic Parallel Beam of Light. - Physics

Sum

The following figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let \[B P_0  - A P_0  = \lambda/3\text{ and }D >  > \lambda.\] (a) Show that in this case \[d = \sqrt{2\lambda D/3}.\] (b) Show that the intensity at P0 is three times the intensity due to any of the three slits individually.

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Solution

(a) Given:-

Wavelength of light = \[\lambda\]

Path difference of wave fronts reaching from A and B is given by

\[∆  x_B  =  {BP}_0  -  {AP}_0  = \frac{\lambda}{3}\]

\[ \Rightarrow \sqrt{D^2 + d^2} - D = \frac{\lambda}{3}\]

\[ \Rightarrow  D^2  +  d^2  =    D^2  + \frac{\lambda^2}{9}   + \frac{2\lambda D}{3}\]

We will neglect the term \[\frac{\lambda^2}{9},\] as it has a very small value.

\[\therefore d = \sqrt{\frac{\left( 2\lambda D \right)}{3}}\]

(b) To calculating the intensity at P0consider the interference of light waves coming from all the three slits.

Path difference of the wave fronts reaching from A and C is given by

\[{CP}_0 - {AP}_0 = \sqrt{D^2 + \left( 2d \right)^2} - D\]

\[ = \sqrt{D^2 + \frac{8\lambda D}{3}} - D ........\left(\text{Using the value of d from part a} \right)\]

\[ = D \left\{ 1 + \frac{8\lambda}{3D} \right\}^\frac{1}{2} - D\]

Expanding the value using binomial theorem and neglectingthe higher order terms, we get

\[ D\left\{ 1 + \frac{1}{2} \times \frac{8\lambda}{3D} + . . . \right\} - D\]

\[{CP}_0 - {AP}_0 = \frac{4\lambda}{3}\]

So, the corresponding phase difference between the wave fronts from A and C is given by

\[\phi_c = \frac{2\pi ∆ x_C}{\lambda} = \frac{2\pi \times 4\lambda}{3\lambda}\]

\[ \Rightarrow \phi_c = \frac{8\pi}{3}\text{ or }\left( 2\pi + \frac{2\pi}{3} \right)\]

\[ \Rightarrow \phi_c = \frac{2\pi}{3}...........(1)\]

Again, \[\phi_B = \frac{2\pi ∆ x_B}{\lambda}\]

\[ \Rightarrow \phi_B = \frac{2\pi\lambda}{3\lambda} = \frac{2\pi}{3}..........(2)\]

So, it can be said that light from B and C are in the same phase, as they have the same phase difference with respect to A.

Amplitude of wave reaching P0 is given by

\[A = \sqrt{\left( 2a \right)^2 + a^2 + 2a \times a\cos\left( \frac{2\pi}{3} \right)}\]

\[               = \sqrt{4 a^2 + a^2 + 2 a^2 \sqrt{3}}\]

\[ \therefore  l_{po}  = K   \left( \sqrt{3}  r \right)^2  = 3  K r^2  = 3l\]

Here, I is the intensity due to the individual slits and Ipo is the total intensity at P0.

Thus, the resulting amplitude is three times the intensity due to the individual slits.

Concept: Refraction of Monochromatic Light
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 17 Light Waves
Q 28 | Page 382
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