At two points P and Q on a screen in Young`s double-slit experiment, waves from slit S_1 and S_2...

Question:

At two points P and Q on a screen in Young`s double-slit experiment, waves from slit S_1 and S_2 have a path difference of 0 and {eq}\lambda {/eq}/4, respectively. Then the ratio of intensities at P and Q will be _____.

Intensity Distribution In Young's Double Slit Experiment

The Young's double-slit arrangement ensures that light intensities add up coherently. In the arrangement the two interfering waves have the same amplitude, the same frequency, and a fixed phase difference. Therefore the resultant intensity on the screen is no longer the sum of the individual intensities. There is an additional interference term. This leads to a redistribution of the incident energy in the form of alternating high and low intensity regions.

Answer and Explanation:

Since the electric field of the electromagnetic wave is an oscillating quantity we may represent it by a phasor. The amplitude is represented by the length of the phasor and the relative phase difference by the angle between the phasors. Then by using the parallelogram law of addition, the resultant amplitude may be found. The intensity (the energy falling on unit perpendicular area per second) is proportional to the square of the amplitude.

If {eq}\displaystyle {a_1} {/eq} is the amplitude of one phasor and {eq}\displaystyle {a_2} {/eq} that of the other and if the relative phase between the two is {eq}\displaystyle {\phi} {/eq} then the amplitude of the resultant wave is,

{eq}\displaystyle {a_r= \sqrt{a_1^2+a_2^2+2a_1a_2 \cos \phi}}---------(1) {/eq}

Here {eq}\displaystyle {a_1=a_2=a} {/eq}.

Therefore the resultant intensity at a point on the screen is,

{eq}\displaystyle { I=k(4a^2cos^2{\frac{\phi}{2})}}---------(2) {/eq}.

Here k is a proportionality constant.

Now a path difference of {eq}\displaystyle {\lambda} {/eq} corresponds to a phase difference of {eq}\displaystyle {2 \pi} {/eq}.

At a point P on the screen where there is no path diffference, the phase difference is zero. Hence from (2) we get the intensity as,

{eq}\displaystyle { I_P=4ka^2} {/eq}

At a point Q where the path difference is {eq}\displaystyle {\frac{\lambda}{4}} {/eq}, the phase difference is {eq}\displaystyle {\pi/2} {/eq}

Hence (2) gives,

{eq}\displaystyle { I_Q=4ka^2\cos^2{\frac{\pi}{4}}=2ka^2} {/eq}

Therefore,

{eq}\displaystyle {\frac{I_P}{I_Q}=2} {/eq}.


Learn more about this topic:

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The Resultant Amplitude of Two Superposed Waves

from MEGA Physics: Practice & Study Guide

Chapter 15 / Lesson 10
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