Wave Optics Concept Page - 3

Example

Applications of Doppler effect in light

The use of the Doppler effect for light is important in astronomy. If a far away cluster of stars are moving away from the Earth, then they would appear to be shifted downward in the frequency of the emitted radiation (a red shift). Whereas the frequency is shifted upward (a blue shift) if the star is moving towards the Earth.This phenomenon helps in understanding the theory of expanding universe.

Example

Describe the applications of Doppler Effect in Light

The applications of Doppler effect in light are as follows:
1. The Doppler effect for electromagnetic waves such as light is of great use in astronomy and results in either a so-called redshift or blueshift.
2. It has been used to measure the speed at which stars and galaxies are approaching or receding from us; that is, their radial velocities.
3. This may be used to detect if an apparently single star is, in reality, a close binary, to measure the rotational speed of stars and galaxies, or to detect exoplanets

Definition

Red shift and Blue shift

1. If the light source is moving away from the observer (positive velocity) then the observed frequency is lower and the observed wavelength is greater (red shifted).

2. If the source is moving towards the observer (negative velocity), the observed frequency is higher and the wavelength is shorter (blue shifted).

Example

Radial velocity for distant galaxy

What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm$589.0nm$ is observed at 589.6 nm$589.6nm$?
Since νλ=c$\nu \lambda =c$, Δνν=−Δλλ${\displaystyle \frac{\mathrm{\Delta }\nu }{\nu }}=-{\displaystyle \frac{\mathrm{\Delta }\lambda }{\lambda }}$ (for small changes in ν$\nu$ and λ$\lambda$). For Δν=589.6−589.0=0.6 nm$\mathrm{\Delta }\nu =589.6-589.0=0.6nm$ we get,
Δνν=−Δλλ=−νradialc${\displaystyle \frac{\mathrm{\Delta }\nu }{\nu }}=-{\displaystyle \frac{\mathrm{\Delta }\lambda }{\lambda }}=-{\displaystyle \frac{{\nu }_{radial}}{c}}$
or, νradial≃+c(0.6589.0)=3.06×105 m/s=306 km/s${\nu }_{radial}\simeq +c\left({\displaystyle \frac{0.6}{589.0}}\right)=3.06\times {10}^{5}m/s=306km/s$
Therefore, the galaxy is moving away from us.

Example

Problem on Intensity of unpolarized light

Example: When an unpolarized light of intensity I0$I_0$ is incident on a polarizing sheet, find the intensity of the light which does not get transmitted.

Solution:
Since the incident light is unpolarized, the intensity of transmitted light at an angle θ$\theta$ with respect to the polarizing axis is: I=I0cos2θ$I=I_0{\cos }^2\theta$
Integrating and then averaging over all angles between 00$0^0$ and 3600${360}^0$ we get: I0(cos2θ)avg=I02$I_0({\cos }^2\theta {)}_{avg}={\displaystyle \frac{I_0}{2}}$
Hence, the intensity of light not transmitted is I0−I02=I02$I_0-{\displaystyle \frac{I_0}{2}}={\displaystyle \frac{I_0}{2}}$

Definition

Doppler effect

The apparent change in frequency of wave due to motion of the source or the observer is called Doppler effect.
Example: Imagine a light bulb giving off pure yellow light; when it moves towards you the light that reaches you eye will be bluer, when the bulb moves away form you the light reaching your eye will be reddish.

Definition

Constructive and Destructive interference

Two sources are said to interfere constructively if they are in phase with each other. In destructive interference, the sources are out of phase and they cancel out each other's effect.

Example

Condition for constructive interference

Example: For constructive interference between two waves of equal wavelength, find the phase angle δ$\delta$.

Solution:
For constructive interference between two waves of equal wavelength the path difference should be a multiple of λ$\lambda$ .
Let it be nλ$n\lambda$
It path difference is nλ$n\lambda$ then phase difference is 2nπ$2n\pi$ because λ$\lambda$ corresponds to 2π$2\pi$.
Thus δ=2nπ$\delta =2n\pi$
cos2δ2=1$co{s}^2\frac{\delta }{2}=1$ in this case
 

Example

Use of relation between path & phase difference

Example: Light of wavelength λ$\lambda$ in the air enters a medium of refractive index μ$\mu$. Two points in this medium, lying along the path of this light are at a distance x$x$ apart. Find the phase difference between these points.

Solution:
λ$\lambda$ is the wavelength of light in air.Let t$t$ be the distance between points S and A.Optical path length travelled by light via SB =μ(t+x)$=\mu (t+x)$
Optical path length travelled by light via SA =μt$=\mu t$Path difference between points  A and B: Δx=SB−SA=μx$\mathrm{\Delta }x=SB-SA=\mu x$
Now phase difference, ϕ=Δx×2πλ$\varphi =\mathrm{\Delta }x\times {\displaystyle \frac{2\pi }{\lambda }}$
Hence,  ϕ=μ(2πλ)x$\varphi =\mu ({\displaystyle \frac{2\pi }{\lambda }})x$

Example

Write intensity as a function of phase difference for two coherent light sources

Example: Two coherent waves each of amplitude a$a$ travelling with a phase difference δ$\delta$ superpose with each other. Find the resultant intensity at a given point on the screen.

Solution:
Let the waves be acosωt$a\cos \omega t$ and acos(ωt+δ)$a\cos (\omega t+\delta )$
So, resultant wave is acosωt+acos(ωt+δ)$a\cos \omega t+a\cos (\omega t+\delta )$
Simplifying: acosωt+acosωtcosδ−asinωtsinδ$a\cos \omega t+a\cos \omega t\cos \delta -a\sin \omega t\sin \delta$ =a(1+ cosδ)cosωt−asinδsinωt$=a(1+cos\delta )\cos \omega t-a\sin \delta \sin \omega t$
The amplitude of this wave =a2(1+cosδ)2+a2sin2δ−−−−−−−−−−−−−−−−−−−√$=\sqrt{a^2(1+\cos \delta {)}^2+a^{2}si{n}^2\delta }$
=a2+a2cos2δ+2a2cosδ+a2sin2δ−−−−−−−−−−−−−−−−−−−−−−−−−−√$=\sqrt{a^2+a^{2}co{s}^2\delta +2a^{2}cos\delta +a^{2}si{n}^2\delta }$
=2a2+2a2cosδ−−−−−−−−−−−√$=\sqrt{2a^2+2a^{2}cos\delta }$
Intensity is square of amplitude.
So, I=2a2(1+cosδ)$I=2a^2(1+cos\delta )$

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