14 — light

Question 1

Light

Answer 1

Light is a form of energy which consists of waves
-> forms part of electromagnetic spectrum.

Question 2

Spd of light

Answer 2

C = 3.0 x 10^8m/s

Question 3

SPD of sound

Answer 3

330m/s

Question 4

Refraction

Answer 4

Bending of light at the boundary of 2 optical mediums due to different optical densities as it travels from one optical medium into another.

Question 5

Cause of refraction

Answer 5

Change in speed of light in diff optical mediums

Question 6

Refractive index

Answer 6

Refractive index, n, of a medium is
1. Defined as the ratio of the spd of light in a vacuum to the spd of light in that medium.
2. the ratio of sin i/sin r for a light ray passing from vacuum into given medium

The higher the refractive index n, the more the bending of light at the boundaries
The higher the refractive index n, the higher the chance that TIR occurs
(Got 2 defs)

Question 7

Critical angle

Answer 7

The angle of incidence in an optically denser medium for which the angle of refraction of the less dense medium is 90dg.

Question 8

Total internal reflection

Answer 8

Complete reflection of a light ray in an optically denser medium at the boundary with an optically less dense medium when angle of incidence is greater than critical angle.

Incident ray must travel from an optically denser medium to an optically less dense medium + angle incidence > critical angle for TIR to occur

Question 9

Focal length

Answer 9

Distance betw the optical centre C and the principal focus point F.

Question 10

Applications of TIR

Answer 10

1. Optical fibres
2. Periscope
3. Binoculars

Question 11

The laws of reflection

Answer 11

The first law of reflection states that the incident ray, reflected ray and the normal at the point of incidence lie in the same plane.

The second law of reflection states that the angle of incidence, theta, is equal to the angle of reflection, theta r.

Question 12

The laws of refraction

Answer 12

The first law of refraction states that the incident ray, reflected ray and the normal lies in the same plane.

The 2nd law of refraction states that for 2 given media, the ratio of the sine of the angle of incidence to the sine of angle of refraction is a constant. Thus, sin i/ sin r is a constant.

Question 13

Characteristics of image formed by plane mirror

Answer 13

1. Virtual
2. Image formed by plane mirror and obj r of the same size
3. Laterally inverted
4. Upright
5. Image formed and obj has an equal distance from mirror

Question 14

Characteristics of image formed when obj is at >2F distance from lens C

Answer 14

Real
Diminished
Inverted

Rays meet at F<x<2F
Applications: Camera

Question 15

Characteristics of image formed when obj is at 2F distance from lens C

Answer 15

Real
Inverted
Same size

Rays meet at 2F

Applications: photocopier

Question 16

Characteristics of image formed when obj is at F<x<2F distance from lens C

Answer 16

Real
Inverted
Magnified

Rays meet at >2F

Applications: projector

Question 17

Characteristics of image formed when obj is at <F1 distance from lens C

Answer 17

Upright
Magnified
Virtual

Image forms behind object, rays form >F1

Applications: magnifying glass

Question 18

Characteristics of image formed when obj is at F distance from lens C (usually from very far distance)

Answer 18

Inverted
Diminished
Real

Rays converge at F, parallel before lens

Applications: telescope

Question 19

Define Real image and virtual image

Answer 19

Real image is an image formed by a lens that can be captured on a screen and is formed by the intersection of light rays.

Virtual image is an image formed by a lens that cannot be captured on a screen and is not formed by the actual intersection of light rays.

Question 20

Answ tech for TIR

Answer 20

Since light is travelling from an optically denser medium to an optically less dense medium, and angle of incidence at Q is greater than the critical angle of medium x, total internal reflection occurs.

Question 21

Answ tech for refraction

Answer 21

When light hits surface x, angle i < angle c, refraction occurs, light emerge from x.

Question 22

Answ tech for motion of light in medium

Answer 22

Light enters optically denser medium from a less dense medium at an angle i -> decreases in speed -> bends towards the normal -> change in direction of light rays

Question 23

What happens when a thicker converging lens is used?

Answer 23

Image formed is closer to lens, is inverted and real, and diminished due to focal length becoming shorter when converging lens are thicker. Hence, light rays converge more.

Question 24

Why do images produced by periscopes using mirrors are not as clear as by prism?

Answer 24

Mirror rays give a blurred image due to multiple reflections and refractions at the reflecting surface to form multiple images

Question 25

Why does light emerge from only a circular area of water?

Answer 25

Light from the lamp can only emerge from the water from an angle of incidence of 0dg to 49dg since critical angle for light from water to air = 49dg. (Calculation) The result is a shape formed by all the incident rays. When viewed from above, light emerges from a circular area with centre O and radius OY which can be calculated via tan(opp/adj).

Question 26

Answ tech: explain why xx is unable to see y.

Answer 26

The incidence ray of y does not touch the mirrors thus no light rays are reflecting off the mirror into xx’s eyes.

Question 27

Describe how to use a torch, a semi-circular block of glass and a protractor to determine the critical angle for the glass.

Answer 27

1. Position the protractor onto the semi-circular block so that the angle of the light rays can be determined. 2. Shine light ray at different angles radially into the block on the curved surface at different incident angles until one of the incident angles produces a refracted ray which is 90dg to the normal. The incident angle will be the critical angle of the glass block. 3. Repeat the experiment a few times and calculate the average critical angle.

Question 28

State how the refractive index of glass shows that light travels faster in a vacuum than in glass.

Answer 28

The refractive index of glass is greater than 1. Thus, from n = c/v, we obtain c = nv, light travels n times as fast in a vacuum as compared to in glass.

Question 29

Images produced by periscopes using mirrors are not as clear as images produced by periscopes using prisms. Explain why [1]

Answer 29

There are 2 images produced from mirrors. Weak reflected rays from the smooth surface and incident rays which are refracted causes the formation of 2 images that may overlap with each other. Prisms only produce 1 image which is from the refracted incident ray thus produces clearer image.

Question 30

The refractive index of glass for red light is 1.52 and for blue light is 1.56. Suggest and explain what effect this difference in refractive index has on focal length. [2]

Answer 30

A larger refractive index suggests a smaller angle of refraction, resulting in a decrease in focal length. Thus, the focal length of the lens for blue light is less than that of red light.

Question 31

State and explain how the speed of light in x compares with y answering technique

Answer 31

Speed of light in x is faster than speed of light in y since the refractive index of x is less than that of y.

Question 32

State and explain whether the critical angle for light travelling from glass into the cladding is equal to, larger than or smaller than 42dg. [2]

Answer 32

Critical angle for light travelling from glass into cladding should be larger than 42dg. Since critical angle c is given by n = 1/sin c, since refractive index of cladding is less than that of glass, if refractive index is a smaller value, calculated critical angle will be larger.

Question 33

Explain how it can be shown that the image formed is real. [2]

Answer 33

Place a screen at position where light rays intersect. Only real images can be formed on a screen.

Question 34

Explain why the light rays near the 2 ends of the lens bend more compared to those near the centre of the lens. [3]

Answer 34

By Snell's law, ratio of sine of angle of incidence to that of refraction is constant for a substance. [1] Thus, if angle of incidence increases, deviation of light passing thru lens increases. [1] Light rays are incident on the surface of the convex lens at different angles and near the ends of the lens, angles of incidence of light rays are larger than those near centre of lens. [1]

Question 35

Describe what happens when light is passing thru the surface of the lens. [1]

Answer 35

Refraction occurs. Light rays travel slower in glass than in air thus light rays that pass from air to glass bend towards the normal.

Question 36

Angle of incidence

Answer 36

Angle between an incident ray and the normal at the point of incidence

Question 37

Angle of reflection

Answer 37

Angle between a reflected ray and the normal at the point of incidence

Question 38

Principle of reversibility of light rays

Answer 38

States that regardless of how many times a light ray has been reflected or refracted, it will follow the same path when its direction is reversed.

Question 39

Focal plane

Answer 39

The plane perpendicular to the principal axis on which all parallel rays meet after passing through the lens

Question 40

Optical centre

Answer 40

The point on the principal axis that is midpoint between the surfaces of the lens

Question 41

Principal axis

Answer 41

The line passing through the centre of the lens and which s perpendicular to the plane of the lens

Question 42

Principal focus

Answer 42

The point on the principal axis where all the rays parallel to the principal axis meet after passing though the lens

Question 43

Normal

Answer 43

A line perpendicular to the reflecting or refracting surface at the point of incidence

Question 44

Use figure 8.2 to explain why the efficiency of the filament lamp is so low. [1]

Answer 44

From the figure, the area of the graph from 400 nm to 700 nm is smaller as compared to the area of the entire graph from 200 nm to 1800 nm. As such, the efficiency of the lamp is low as the only useful output power comes from the 400 nm to 700 nm range of the electromagnetic spectrum in the form of visible light. (Note: question says ‘Use Fig. 8.2 to explain…’, thus area of the graph must be used and must know what the area represents. In addition, useful power output for a lamp is to produce light)

Question 45

Advantages of optical fibre

Answer 45

1.greater bandwidth 2 less signal degradation than copper wires 3. Light weight, thin and more durable than copper wires 4. Higher data security than copper wires 5. Lower cost than copper wires