AS2 2.3 Lenses

Question 1

What are the two types of lenses?

Answer 1

1. Converging ( convex )
2. Diverging ( concave )

Question 2

What is the optical centre?

Answer 2

It is the centre of a lens, denoted as O or C

Question 3

What is the principal axis?

Answer 3

It is a line, passing through the optical centre at right angles to the centre of the lens

Question 4

What is the focal point?

Answer 4

It is the point of which rays travelling parallel to the principal axis coverage and meet after passing through a convex lens

Question 5

Lenses Acronym

Answer 5

Need Nearsightedness
Medical Myopia
Doctors. Diverging
For. Farsightedness
Healthy. Hypermetropia
Children. Converging

Question 6

Focal Length

Answer 6

Distance from O to F ie from the centre of the lens to the principal focus/focal point

Question 7

Focal length in diverging lens

Answer 7

A diverging lens will have a negative focal length, this is because the principal focus/focal point is a virtual one

Question 8

Real Image

Answer 8

• an image formed by the actual intersection of ray
• can be formed on a screen
• is on the opposite side of the lens from the object

Question 9

Virtual Image

Answer 9

• an image formed by the apparent intersection of rays
• can not be formed on a screen
• is on the same side of the lens as the object

Question 10

Image in converging(convex) lens

Answer 10

They can produce real or virtual images

Question 11

Images in diverging(concave) lens

Answer 11

Can produce only virtual images

Question 12

Inverted Images

Answer 12

• if an image is above the axis it will be upright
• if an image is formed below the axis it will be inverted
• real images are always inverted, and virtual images are always upright

Question 13

Images formed by converging lens with the object beyond 2f

Answer 13

Image position - between F and 2F
Real Image
Diminished
Inverted

Question 14

Images formed by converging lens with the object at 2F

Answer 14

Image position - at 2F
Real
Same size
Inverted

Question 15

Images formed by converging lens with the object between F and 2F

Answer 15

Image position - beyond 2F
Real
Enlarged
Inverted

Question 16

Images formed by converging lens with the object at F

Answer 16

Image position - at infinity

Question 17

Images formed by converging lens with the object inside F

Answer 17

Image position - varies
Virtual
Enlarged
Upright

Question 18

Images formed by diverging lens with the object anywhere

Answer 18

Image position - inside F
Virtual
Diminished
Upright

Question 19

Focal length equation

Answer 19

1/u + 1/v = 1/f
u = object distance
v = image distance
f = focal length

Question 20

Lens equation rules

Answer 20

• the distance to a virtual image or virtual focal point is always negative
• the distance to a real image or real focal point is always positive
• the focal length of a converging lens is positive
• the focal length of a diverging lens is negative

Question 21

Magnification equation

Answer 21

m = v / u or m = hi / ho
m = magnification
v or hi = image height or distance
u or ho = object height or distance

Question 22

Magnified images

Answer 22

Magnified images are always virtual

Question 23

Far point and near point of a normal eye

Answer 23

Far point is infinity
Near point is 0.25m

Question 24

Short sightedness

Answer 24

Myopia
Their far point is much closer than infinity making objects far away blurry
Their near point may be slightly closer than 0.25m
There are two reasons to cause myopia;
- the lens of the eye is too powerful
- the eye is too long
This causes rays to meet before the retina/back of the eye

Question 25

Long sightedness

Answer 25

Hypermetropia Their far point is at infinity Their near point is much further away than 0.25m This means that things close up are blurry There are two reasons that cause this; - the lens is too weak - the eye is too short This causes rays to meet “behind” the retina

Question 26

Correcting myopia ( shortsightedness )

Answer 26

- f will always be negative to the persons far point To calculate new range of vision u must first find f and U will be infinity and V will be the far point Then calculate the NP by using the original near point as V and must be negative and the F from the first equation and then using this to find V ( new range of vision will have far point at infinity and near point further from before)

Question 27

Correcting Hypermetropia ( longsightedness )

Answer 27

You must first calculate the value of F by using the normal eyes near point ( 0.25m ) as U and the persons near point as V and it must be negative and F should be positive Then the new far point will be the same as the focal length

Question 28

Sign convention summary

Answer 28

Focal length for converging lens; **f is +** Focal length for diverging lens; **f is -** Virtual image; V is - Real Image; V is +

Question 29

Power of a lens equation

Answer 29

P = 1 / f P = dioptre, D f = focal length (must be m)