Flashcards in 9.3 Interference Deck (16):

1

## What happens during double slit diffraction?

### Diffraction takes place at each slit so waves spread out from each slit and far from these slits rays from one slit meet rays from the other and interfere. In this way regions of maximum and minimum intensity are formed.

2

## Where will maximum intensity occur?

### Constructive interference at points where the path differences is an integral multiple of the wavelength.

3

## Where will minimum intensity occur>

### Destructive interference where the path difference is a half interval multiple of the wavelength.

4

## What is the path difference for constructive interference where d is the separation of the slits?

### dsin@ = nlambda

5

## What is the path difference for destructive interference where d is the separation of the slits?

### dsin@ = (n + 1/2)lambda

6

## What happens to the diffraction pattern if you increase the number of slits?

### There will become secondary maximum in between the primary maxima, the primary maximum have narrowed and the central maximum intensity has increased.

7

## What happens to the diffraction pattern if you use a diffraction grating?

### The secondary maxima shrine in size and the primary maxima grow very intense and very narrows, the diffraction grating produces well separated, very narrows, very bright maxima of essential equal intensity.

8

## How is the position of a maxima found on a diffraction grating?

### dsin@ = nlambda - this formula gives the primary maxima for any number of slits.

9

## How else can a diffraction grating be used?

### To measure the wavelength of light from its diffraction pattern using dsin@ = nlambda

10

## What is the resolving power of a diffraction grating?

###
Its ability to distinguish between a particular pair of wavelengths

defined as

_

R = lambda/(delta lambda)

_

where lambda is the average of the two wavelengths and delta lambda is their difference.

11

## What does R = mN show?

###
R = mN where m is the order of the intensity maximum at which the observation is made and N is the total number of slits in the diffraction grating. So the smallest difference in wavelengths that can be resolved with a particular diffraction grating is

_

delta lambda = lambda/mN

12

## Why will a thin transparent film appear coloured from above?

### Because of interference of rats reflected from its upper and lower boundaries.

13

## What happens during thin film interference?

### The ray reflected at the entrance of the film will interfere with the ray reflected from the bottom of the film. The path difference is approximately 2d where d is the thickness of the film assuming we are looking directly form above. The conditions of interference can be applied.

14

## Why are interference conditions in thin film interference complicated?

### Because the rays may undergo a phase change upon reflection. There is no phase change when a ray is reflected at a boundary into a medium of LOWER refractive index but a phase change of pi (half cycle) when it is reflected at a boundary into a medium of HIGHER refractive index.

15

## What are the conditions for constructive and destructive interference when there is no phase change or a phase change at both boundaries?

###
2d = m(lambda/n) CONSTRUCTIVE

2d = (m + 1/2) lambda/n DESTRUCTIVE

16