Wave Optics Concept Page - 7

Definition

Energy redistribution in diffraction and interference

In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe.
There is no gain or loss of energy, which is consistent with the principle of conservation of energy.

Result

Differentiate between fresnel's and fraunhofer's diffraction

	Fraunhofer Diffraction	Fresnel Diffraction
Diffraction patterns	Shape and intensity of a Fraunhofer diffraction pattern stay constant.	Change as we propagate them further downstream of the source of scattering.
Wave fronts	Planar wave fronts	Cylindrical wave fronts
Observation distance	Observation distance is infinite. In practice, often at focal point of lens.	Source of screen at finite distance from the obstacle.
Movement of diffraction pattern	Fixed in position	Move in a way that directly corresponds with any shift in the object.
Surface of calculation	Fraunhofer diffraction patterns on spherical surfaces.	Fresnel diffraction patterns on flat surfaces.

Definition

Interference and diffraction

(i) The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side.
(ii) We calculate the interference pattern by superposing two waves originating from the two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(iii) For a single slit of width a$a$, the first null of the interference pattern occurs at an angle of λ/a$\lambda /a$. At the same angle of λ/a$\lambda /a$, we get a maximum (not a null) for two narrow slits separated by a distance a$a$.

Definition

Fresnel distance

A beam of width a$a$ travels a distance of a2λ${\displaystyle \frac{a^2}{\lambda }}$, called the Fresnel distance, before it starts to spread out due to diffraction.
Example: The source is at some distance from an obstacle. Distance between obstacle and the point of observation is b$b$ and wavelength of light is λ$\lambda$. Find the distance of nth$n^{th}$ Fresnel Zone from the point of observation.

Solution:
If the distance between the obstacle and the point of observation is b$b$, then the distance of the nth$n^{th}$ Fresnel Zone = b+nλ2$b+{\displaystyle \frac{n\lambda }{2}}$ from the point of observation.

Definition

Construction and Working of Fresnel's Biprism Experiment

A Fresnel Biprism is a variation of the Youngs Slits experiment. Here we use a biprism to get interference pattern by the division of wavefront method. 

Construction :

• The biprism consists of two right-angled prisms with their bases in contact. 
• Here two sources S1 and S2 act as virtual images of the fine slit S as shown in Figure. 
• The experimental arrangement consists of a slit S, the biprism ABC and the microscope M. 
• All are mounted on an optical bench. 
• These are adjusted at the same height and can move and rotate as required. 

Working : 

• The light emerging from the slit fall on the biprism. 
• The edge A of the biprism divides the incident wavefront into two parts.
• One is refracted through upper half AB of biprism and appears to come from virtual source S1. Other is from lower half AC of biprism and appears to come virtual S2. 
• The interference fringes can be seen in the overlapping region XY and can be seen the eyepiece.

Example

Calculate the refractive index of the material of the biprism

Example: On placing a thin sheet of mica of 1.2 μm$1.2\mu m$ thickness in the path of one of the interfering beams in a biprism experiment, it is found that the central bright band shifts to a distance equal to the width of a bright fringe. The refractive index of mica is x2${\displaystyle \frac{x}{2}}$. Find x$x$. (λ=600 nm$\lambda =600nm$).

Solution:
The width of the bright fringe is given by : λDd${\displaystyle \frac{\lambda D}{d}}$
The shift in central maxima due to placing mica is given by : (μ−1)tDd${\displaystyle \frac{(\mu -1)tD}{d}}$
It is given that the shift is equal to the length of a bright fringe.
So, λDd${\displaystyle \frac{\lambda D}{d}}$ = (μ−1)tDd${\displaystyle \frac{(\mu -1)tD}{d}}$
or, 600Dd${\displaystyle \frac{600D}{d}}$ = (μ−1)1200Dd${\displaystyle \frac{(\mu -1)1200D}{d}}$
or, μ$\mu$ = 1 + 12=32${\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{2}}$
∴$\therefore$ x=3$x=3$

Example

Calculate the seperation between the slits in the Fresnel's biprism

Example: In a Fresnel biprism experiment, the two positions of lens give separation between the slits as 16 cm$16cm$ and 9 cm$9cm$ respectively. What is the actual distance of separation?

Solution:
Actual separation (d$d$) can not be determined directly since the two slits are purely virtual.
d$d$ is determined by placing a converging lens between the bi-prism and the screen and forming real images of the virtual slits on the screen.
So, d=d1d2−−−−√=16×9−−−−−√=144−−−√=12cm$d=\sqrt{d_1{d}_2}=\sqrt{16\times 9}=\sqrt{144}=12cm$

Example

Position of fringes in biprism experiment

In a biprism experiment, the micrometer readings for the zero order and 10th${10}^{th}$ order fringe are 1.25 mm$1.25mm$ and 2.37 mm$2.37mm$ respectively, when light of 600 nm$600nm$ is used. If the wavelength is changed to 750 nm$750nm$, what will be the respective position of zero and 10th${10}^{th}$ order fringes ?

Since the reading of micrometer for zero order is non-zero, there is a zero error equal to 1.25 mm$1.25mm$.
Hence this needs to be subtracted to find the true positions of fringes.
Distance between zero order and 10th${10}^{th}$ order=2.37 mm−1.25 mm=1.12 mm$=2.37mm-1.25mm=1.12mm$.
Now since position of nth$n^{th}$ order fringe is yn=nDλd$y_n={\displaystyle \frac{nD\lambda }{d}}$,Δynyn=Δλλ=150600=0.25${\displaystyle \frac{\mathrm{\Delta }{y}_n}{y_n}}={\displaystyle \frac{\mathrm{\Delta }\lambda }{\lambda }}={\displaystyle \frac{150}{600}}=0.25$
Therefore new y′n=yn+Δyn=1.25yn=1.25×1.12 mm=1.4 mm$y_n^{\prime }=y_n+\mathrm{\Delta }{y}_n=1.25y_n=1.25\times 1.12mm=1.4mm$
Adding the zero error of the micrometer gives the new position of 10th${10}^{th}$ fringe at 2.65 mm$2.65mm$
Obviously the position of the zero order fringe remains unchanged.

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