Light

Question 1

Laws of reflection (2)

Answer 1

1) The angle of incidence is always equal to the angle of reflection
2) The incident ray, the reflected ray and the normal all lie on the same plane

Question 2

Describe an experiment to verify the law of reflection (7)

Answer 2

1) A plain white paper is placed on the wooden board
2) A mirror is placed in the middle of a paper and the normal and incident ray is drawn
3) 2 pins are placed on the incident ray
4) The images of the two pins are viewed in such a way that they are aligned
5) 2 more pins are placed so that they are aligned with the first two pins
6) The reflected ray is drawn by joining the third and fourth pin
7) The angle of incidence and reflection are measured and compared. The experiment is a success if they are approximately equal

Question 3

Characteristics of an image formed in the mirror (5)

Answer 3

1) Image has the same size and shape of object
2) The image is upright
3) The distance of the image behind the mirror is the same as the distance of he object in front of mirror
4) The image is virtual
5) The image is laterally inverted

Question 4

Describe an experiment to locate the position of an image (3)

Answer 4

1) Two incident rays are drawn
2) By measuring the incident angles, the reflected rays are drawn
3) By projecting the reflected rays inside the mirror, the point of intersection gives the location of the image

Question 5

Precautions when doing the experiment to verify law of reflection

Answer 5

1) Pins must be straight
2) Thin lines must be drawn
3) Avoid parallax error when measuring angles

Question 6

Define refraction of light

Answer 6

When a ray of light travels from one medium to another it changes speed and direction

Question 7

What occurs when a ray moves from a denser to less dense medium?

Answer 7

The speed and angle of refraction will increase and the ray moves away from the normal

Question 8

What occurs when a ray moves from a less dense to a denser medium?

Answer 8

The speed and angle of refraction will decrease and the ray moves towards the normal

Question 9

Laws of refraction (2)

Answer 9

1) Sin i ÷ sin r = constant

2) The incident ray, the normal line and the refracted ray are found on the same plane

Question 10

What is refractive index?

Answer 10

It is a measure of the capacity of an object to deviate light towards the normal

Question 11

Formula for refractive index

Answer 11

R.I = sin i ÷ sin r (where angle i is equal to angle in air)

Question 12

What is a critical angle?

Answer 12

It is the incident ray in the denser medium for which the refracted angle is equal to 90º

Question 13

When does total internal refraction occur?

Answer 13

If the incident angle is greater than the critical angle, the ray will be reflected back into glass

Question 14

Application of total internal refraction

Answer 14

In optical fibres for communication

Question 15

What are the advantages of optic fibres over copper wires

Answer 15

1) More signals may be transmitted simultaneously
2) Signals are clearer because there is no interference
3) Less energy is lost
4) It is cheaper

Question 16

Uses of lens (3)

Answer 16

1) Camera
2) Magnifying glass
3) Our eyes

Question 17

What are the rules of locating the position of an image by construction?

Answer 17

1) Any ray passing through the optical centre will be undeviated
2) All rays parallel to the principle axis will pass through the focussing point

Question 18

Differences between a real and virtual image

Answer 18

Real images are formed on different side of the object, are inverted and can be obtained on a screen whereas virtual images are formed on the same side of the object, are upright and cannot be obtained on a screen

Question 19

What is the focal length?

Answer 19

The length between the optical centre and the focusing point

Question 20

Describe an experiment to estimate the focal length of a converging lens (3) labelled diagram

Answer 20

1) The white paper is moved slowly away from the lens until a sharp image is formed
2) The perpendicular distance between the cardboard and lens is measured using a meter ruler. This represents the focal length
3) The experiment is repeated and the average value is calculated

Question 21

How are rays from a very far object?

Answer 21

Parallel

Question 22

What is short sight defect and the correction?

Answer 22

Eyes cannot see far objects because rays meet before retina. A diverging lens must be used

Question 23

What is long sight defect and its correction?

Answer 23

Eyes cannot see close objects because rays meet after retina. A converging lens must be used

Question 24

2 Formulas for magnification

Answer 24

1 - height of image ÷ height of object

2- distance between image and optical centre ÷ distance between object and optical centre

Question 25

Experiment to determine the magnification (and focal length) of a convex lens

Answer 25

1) The white screen is placed 1m from the lamp box
2) The lens is placed very near to the lamp box and then we adjust it until a sharp image is formed on the screen
3) The distance between image and optical centre and distance between object and optical centre are noted
4) The magnification is calculated
(5- focal length) Focal length may be calculated using uxv÷u+v