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In a two-slit interference pattern, the intensity at the peak of the central maximum is I_0. a....

Question:

In a two-slit interference pattern, the intensity at the peak of the central maximum is {eq}I_0 {/eq}.

a. At a point in the pattern where the phase difference between the waves from the two slits is {eq}62^o {/eq}, what is the intensity? Give your answer as a fraction of {eq}I_0. {/eq}

b. What is the path difference for light with a wavelength of {eq}473 \ nm {/eq} from the two slits at a point where the phase angle is {eq}62^o {/eq}?

Phase of Waves

It is a state of oscillation of a particle which gives magnitude and direction of displacement of the particle performing oscillations. S. I. unit of phase is radian (rad).Phase of a wave is related to different physical quantities associated with wave.When 2 particles are separated by a distance equal to wave length, phase difference between the 2 particles is 2 X 180 = 360 degrees.

Answer and Explanation:

{eq}\displaystyle{ \\ }{/eq}

Symbols Used :-

1) {eq}\space I_0,\space I,\space \phi {/eq} are the intensity of central maximum, intensity at the given point and the phase difference respectively.

2) {eq}\space x, \space \lambda{/eq} are the path difference and the wave length of light respectively.


Given Data :-

  • {eq}\lambda = 473 \ \rm nm = \space 473 \times 10^{-9} \ \rm m,\\ \phi = 62^{0} {/eq}


Required

a) The intensity at a point in the pattern where the phase difference between the waves from the two slits is 62 degrees =?

b) The path difference for light with a wavelength of 473 nm from the two slits at a point where the phase angle is 62 degrees =?

Solution:-

The intensity at a point in the pattern where the phase difference between the waves from the two slits is {eq}\phi \space {/eq} given by

$$\begin{align} \space I &= 4 \times I_0 \times \cos^{2} \phi \\[0. 3 cm] I &= 4 \times I_0 \times \cos^{2} 62^{0} \\[0. 3 cm] I &= 4 \times I_0 \times 0.6735^{2} \\[0. 3 cm] I &= 4 \times I_0 \times 0.4536 \\[0. 3 cm] I &= 1.8144 \times I_0 \\[0. 3 cm] I &\cong 1.81 I_0 \\[0. 3 cm] \end{align} $$

b) The relation between path difference(x) and phase difference {eq}(\phi) \space {/eq} is given by

$$\begin{align} \phi &= \frac{ 2 \times \pi \times x}{\lambda} \\[0.3 cm] x &= \frac{ \phi \times \lambda} { 2 \times \pi} \\[0.3 cm] x &= \frac{ 62^{0} \times 473 \times 10^{-9} \ \rm m} { 2 \times 180^{0}} \\[0.3 cm] x &= \frac{ 62^{0} \times 473 \times 10^{-9} \ \rm m} { 360^{0}} \\[0.3 cm] x &= \frac{ 29236^{0} \times 10^{-9} \ \rm m} { 360^{0}} \\[0.3 cm] x &= 81.46 \times 10^{-9} \ \rm m \\[0.3 cm] x &= 8.146 \times 10^{-8} \ \rm m \\[0.3 cm] x &\cong 8.2 \times 10^{-8} \ \rm m \\[0.3 cm] \end{align} $$

Answers:-

a) The intensity at a point in the pattern where the phase difference between the waves from the two slits is 62 degrees is 1.81 {eq}I_0 {/eq}.

b) The path difference for light with a wavelength of 473 nm from the two slits at a point where the phase angle is 62 degrees is

{eq}8.2 \times 10^{-8} \ \rm m {/eq}.


Learn more about this topic:

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Constructive and Destructive Interference

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16
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