Interference

Physics 202Professor Lee

CarknerLecture 22

Interference

Due to a phase difference between the incoming waves the amplitude of the resultant wave can be larger or smaller than the original

Light can experience such interference as well

The interference of light demonstrates the wave nature of light

Speed of Light Why does light bend when entering a new

medium?

n = c/v Consider a wavefront half in air and half in

glass

Note that:

Light also changes wavelength in a new medium (v = f

Frequency stays the same

Phase Change

If light travels through a medium of length L with index of refraction n, the number of wavelengths in that medium is:

N = L/new = Ln/

The difference in wavelengths between two different paths is:

N2 - N1 = (L/)(n2-n1)

Phase

We can represent phase in different ways

Phase differences are seen as brightness variations

Destructive interference ( = ½) produces a dark spot

Intermediate interferences produces intermediate brightness

Diffraction

When a planar wavefront passes through a slit the wavefront flares out

Diffraction can be produced by any sharp edge e.g. a circular aperture or a thin edge In order to see light it has to pass through a

circular aperture (e.g. your eye), so diffraction affects all images

Diffraction

Basic Interference

In Young’s experiment sunlight passes through a double slit with a screen on the other side

The two resultant wavefronts overlap to produce constructive and destructive interference

The interference patterns appear on the screen and show bright and dark maxima and minima, or fringes

Interference Patterns

The two rays travel different distances and so will be in or out of phase depending on if the difference is a multiple of 1 or an odd multiple of 0.5 wavelengths L = 1, L = ½,

What is the path length difference (L) for a given set-up?

Path Length Difference Consider a double slit

system where d is the distance between the slits and is the angle between the normal and the point on the screen we are interested in

The path length difference for rays emerging from the two slits is

L = d sin This is strictly true only when

the distance to the screen D is much larger than d

Maxima and Minima

d sin = m For minima (dark spots) the path

length must be equal to a odd multiple of half wavelengths:

d sin = (m+½)

Location of Fringes

For example: m=1max = sin-1 (/d)

min = sin-1 (1.5/d) Zeroth order maxima is straight in front of the

slits and order numbers increase to each side

Interference Patterns

You can also find the location of maxima in terms of the linear distance from the center of the interference pattern (y):

For small angles (or large D) tan = sin

where m is the order number The same equation holds for for minima if

you replace m with (m+½)

Wavelength of Light

If you measure d and (or y and D) you can solve the above equations for the wavelength of light

Young found 570 nm, fairly close to the true value of 555 nm

Next Time

Read: 35.6-35.8

For a diverging lens what is the sign of f, the type of image and the orientation of the image?

A) +, real, uprightB) +, virtual, uprightC) -, virtual, invertedD) -, virtual, uprightE) + real, inverted

For a converging lens (with object close to the mirror) what is the sign of f, the type of image and the orientation of the image?

A) +, real, uprightB) +, virtual, uprightC) -, virtual, invertedD) -, virtual, uprightE) +, real, inverted

For a converging lens (with object far from the mirror) what is the sign of f, the type of image and the orientation of the image?

A) +, real, uprightB) +, virtual, uprightC) -, virtual, invertedD) -, virtual, uprightE) +, real, inverted