Chapter 13 - Optics and interference

Question 1

Why does red light refract less than violet light?

Answer 1

Red light has a longer wavelength and so travels faster in a material than violet light. The refractive index of red light in the material is less than that of violet light.

Question 2

Why do diamonds sparkle so much?

Answer 2

Diamond has a very high refractive index and so has a small critical angle. Light is more likely to totally internally reflect many times before leaving the substance.

Question 3

Why do optical fibres need to be highly transparent?

Answer 3

To minimise absorption of light.

Question 4

Why is cladding needed around an optical fibre?

Answer 4

Fibres would be in direct contact and so light could cross over from one to another. This means data is not secure.

Question 5

What is multipath dispersion?

Answer 5

If a core of the optical fibre is too wide, then light travelling along the core will travel less distance than light totally internally reflected. The pulse of light would become longer than it should, merging with the next pulse.

Question 6

Why is monochromatic light used in optical fibres?

Answer 6

To prevent spectral dispersion. This is pulse merging of many wavelengths of light travelling at different speeds through the core. Red is faster than violet.

Question 7

What does coherent fibre bundle mean?

Answer 7

Fibre ends at each end of the endoscope are in the same relative position.

Question 8

Give the conditions needed for total internal reflection to occur.

Answer 8

Light needs to be moving from one material with a higher refractive index than the material it is travelling towards.
The angle of incidence must be greater than the critical angle.

Question 9

Describe the set up to see Young’s fringes.

Answer 9

Either a monochromatic light source (laser) or a white light source and a filter.
A single slit to create maximum diffraction.
A double slit to produce coherent sources of light.
A screen to observe fringes on.

Question 10

Why does a bright fringe form in Young’s double slit experiment?

Answer 10

Constructive interference occurs due to waves having a path difference of a whole number of wavelengths and so arriving at the screen in phase.

Question 11

Why does a dark fringe appear in Young’s Double slit experiment?

Answer 11

The two coherent waves destructively interfere as they have a path difference of half, one and a half two and a half etc. wavelengths. The waves arrive at the screen antiphase.

Question 12

Give safety precautions taken when using a laser.

Answer 12

Never look directly at the laser beam or even a reflection.
Wear laser safety glasses.
Put a sign up on the door when using the laser.

Question 13

How would you accurately measure the fringe spacing in a double slit experiment?

Answer 13

Measure from the Middle of a dark fringe over several dark fringes then divide your answer by the number of fringes measured across.
Centre of dark fringes are easier to locate than bright fringes.

Question 14

Describe the difference in interference patterns obtained in a double slit or diffraction grating experiment when changing a green laser to a red laser.

Answer 14

The bright dots (diffraction grating) or fringes for double slit will be further apart with a red laser as the wavelength is larger.

Question 15

Give two differences between using a white light source for a double slit experiment and a red laser.

Answer 15

With white light you get a central white maxima, not red. The subsidiary maximas with white light are a spectrum of colours, blue nearest the Centre and red furthest away.
The subsidiary maximas are wider than the ones with red light.
The spacing between maximas is smaller with white light.

Question 16

What is different about the fringes seen through single slit diffraction of red light compared to blue light?

Answer 16

The fringes get wider the longer the wavelength, so red fringes are wider than blue.

Question 17

Give the characteristics of the diffraction pattern seen when red light passes through a single slit.

Answer 17

The central fringe is twice as wide as each of the outer fringes.
The peak intensity of each fringe decreases with distance from the centre.
Each outer fringe is the same width.
The outer fringes are much less intense than the central fringe.

Question 18

State a use for a diffraction grating in a spectrometer.

Answer 18

Study light from stars to find their composition from absorption or emission spectra.
Obtain spectra showing size of red shift and so position of a galaxy relative to Earth.

Question 19

How would you find d, the spacing between lines in a diffraction grating with 600 lines per mm?

Answer 19

Calculate 1 x 10 ^ -3 divide by 600. This will give you the value in metres.

Question 20

How do y determine the maximum number of orders observed from a light source using a diffraction grating?

Answer 20

Calculate d/wavelength. The maximum angle of 90 is set so sin 90 = 1. Then round down to the nearest whole number.

Question 21

What is the relationship between distance between slits and angle of diffraction?

Answer 21

The smaller the distance between slits, the bigger the angle of diffraction.

Question 22

List the equipment needed to calculate the wavelength of light emitted from a laser.

Answer 22

A laser :-)
A diffraction grating
A screen
(A telescope - identify maxima and determine the angle of the 1st order etc.)

Question 23

Explain how to directly find the critical angle for the boundary between two materials.

Answer 23

From the normal, increase the incidence angle of light at the boundary until the angle of refraction is 90 degrees. Measure the angle of incidence at this point and that is the critical angle.

Question 24

Explain how to calculate the critical angle from a boundary between two materials knowing the refractive index of these materials.

Answer 24

Sin c = refractive index of boundary light is moving towards divided by refractive index of material light is travelling in.

The refractive index of the material light is in must be bigger than the other so sin c is less than 1!

Question 25

Give two reasons why optical fibres used for communication have a cladding.

Answer 25

Protection of the core from scratches
Prevent crossover of signal or loss of information
Increase rate of data transfer
Reduce pulse broadening or dispersion.

Question 26

Give a use for optical fibres in society and why they have been beneficial.

Answer 26

Endoscope, communications.

Improved medical diagnosis, better transmission of data, high speed internet.

Question 27

What did Young’s double slit experiment suggest about the nature of light?

Answer 27

That light was a wave rather than a particle.

Question 28

A single slit diffraction pattern is obtained from a monochromatic light source. How will the pattern change if the slit was made narrower?

Answer 28

The fringes will be wider or fringes will have an increased separation.
The fringes will have a lower intensity (less light getting through)

Question 29

How does the frequency, wavelength and speed of light change when it moves from a material with a greater optical density or refractive index to a material with a smaller one?

Answer 29

Increase
Constant
Increase

Question 30

How does wave theory account for areas of minimum on the screen of a double slit experiment?

Answer 30

Waves cancel or destructively interfere due to the light from one slit meets light from the other in antiphase or with a path difference of 1/2 wavelength.

Question 31

Give two differences between laser light and light emitted by a filament bulb.

Answer 31

The light is monochromatic
The light is coherent
The light is polarised

Question 32

What is the name given to light spreading out through double slits?

Answer 32

Diffraction.