14 — light Flashcards
Light
Light is a form of energy which consists of waves
-> forms part of electromagnetic spectrum.
Spd of light
C = 3.0 x 10^8m/s
SPD of sound
330m/s
Refraction
Bending of light at the boundary of 2 optical mediums due to different optical densities as it travels from one optical medium into another.
Cause of refraction
Change in speed of light in diff optical mediums
Refractive index
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)
Critical angle
The angle of incidence in an optically denser medium for which the angle of refraction of the less dense medium is 90dg.
Total internal reflection
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
Focal length
Distance betw the optical centre C and the principal focus point F.
Applications of TIR
- Optical fibres
- Periscope
- Binoculars
The laws of reflection
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.
The laws of refraction
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.
Characteristics of image formed by plane mirror
- Virtual
- Image formed by plane mirror and obj r of the same size
- Laterally inverted
- Upright
- Image formed and obj has an equal distance from mirror
Characteristics of image formed when obj is at >2F distance from lens C
Real
Diminished
Inverted
Rays meet at F<x<2F
Applications: Camera
Characteristics of image formed when obj is at 2F distance from lens C
Real
Inverted
Same size
Rays meet at 2F
Applications: photocopier
Characteristics of image formed when obj is at F<x<2F distance from lens C
Real
Inverted
Magnified
Rays meet at >2F
Applications: projector
Characteristics of image formed when obj is at <F1 distance from lens C
Upright
Magnified
Virtual
Image forms behind object, rays form >F1
Applications: magnifying glass
Characteristics of image formed when obj is at F distance from lens C (usually from very far distance)
Inverted
Diminished
Real
Rays converge at F, parallel before lens
Applications: telescope
Define Real image and virtual image
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.
Answ tech for TIR
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.
Answ tech for refraction
When light hits surface x, angle i < angle c, refraction occurs, light emerge from x.
Answ tech for motion of light in medium
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
What happens when a thicker converging lens is used?
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.
Why do images produced by periscopes using mirrors are not as clear as by prism?
Mirror rays give a blurred image due to multiple reflections and refractions at the reflecting surface to form multiple images
Why does light emerge from only a circular area of water?
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).
Answ tech: explain why xx is unable to see y.
The incidence ray of y does not touch the mirrors thus no light rays are reflecting off the mirror into xx’s eyes.
Describe how to use a torch, a semi-circular block of glass and a protractor to determine the critical angle for the glass.
- Position the protractor onto the semi-circular block so that the angle of the light rays can be determined.
- 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.
- Repeat the experiment a few times and calculate the average critical angle.
State how the refractive index of glass shows that light travels faster in a vacuum than in glass.
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.
Images produced by periscopes using mirrors are not as clear as images produced by periscopes using prisms. Explain why [1]
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.
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]
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.
State and explain how the speed of light in x compares with y answering technique
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.
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]
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.
Explain how it can be shown that the image formed is real. [2]
Place a screen at position where light rays intersect.
Only real images can be formed on a screen.
Explain why the light rays near the 2 ends of the lens bend more compared to those near the centre of the lens. [3]
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]
Describe what happens when light is passing thru the surface of the lens. [1]
Refraction occurs. Light rays travel slower in glass than in air thus light rays that pass from air to glass bend towards the normal.
Angle of incidence
Angle between an incident ray and the normal at the point of incidence
Angle of reflection
Angle between a reflected ray and the normal at the point of incidence
Principle of reversibility of light rays
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.
Focal plane
The plane perpendicular to the principal axis on which all parallel rays meet after passing through the lens
Optical centre
The point on the principal axis that is midpoint between the surfaces of the lens
Principal axis
The line passing through the centre of the lens and which s perpendicular to the plane of the lens
Principal focus
The point on the principal axis where all the rays parallel to the principal axis meet after passing though the lens
Normal
A line perpendicular to the reflecting or refracting surface at the point of incidence
Use figure 8.2 to explain why the efficiency of the filament lamp is so low. [1]
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)
Advantages of optical fibre
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
