Model answers - Topic 5b - refraction and polarisation Flashcards
explain what is meant by refraction
Refraction is the change in direction of a wavefront as it changes speed due to a change in the density of the medium that it is travelling in
define the refractive index
o Refractive index, n = c/v – this is the ratio of the speed of light in a vacuum to the speed of light in the medium
- Explain how to experimentally determine the refractive index of glass
o Measure the angle of refraction,θr for a ray for 6 different values of the angle of incidence, θi
o Plot a graph of sin θr against sin θi.
o sin θr = (1/n2 ) (sin θi ) (n1 = 1 because the first material is air)
o y = m x
o so the gradient is 1/n2 , determine the gradient and n2 = 1/gradient
define critical angle
o The angle of incidence in the denser medium at which the angle of refraction is 90 degrees in the less dense medium: sin c = 1/n
- Describe the condition for total internal reflection to occur
o The light has to be travelling in the denser medium
o Total internal reflection will occur when the angle of incidence is larger than the critical angle
principal focus/focal point define
the point on the principal axis where rays parallel to the principal axis (from a very distant object) will converge
focal length
o Focal length – the distance between the centre of the lens and the principal focus (is negative for diverging lenses and positive for converging lenses)
image distance define
o Image distance, v – the distance between the image and the centre of the lens (is negative for virtual images and positive for real images)
o Object distance
u, distance between object and centre of the lens
magnification
v/u or image height/object heightc
converging lens
lens that causes rays to converge on principal axis - has positive focal length
diverging lens
a lens that causes rays to diverge after refraction - has a negative focal length
power
power = 1/focal length (focal length must be in metres to equal 1 Dioptre)
units = D (dioptres)
what is the power of a diverging lens
negative
when multiple lenses are lined up on the principal axis, how do you find the total power?
the powers add up (and subtract any diverging lenses)
what is a virtual image
an image formed from the apparent divergence of rays from an object, cannot be projected on a screen
what is a real image
an image formed when real rays of light converge to form the image - can be projected on a screen
construct a ray diagram for a diverging lens with the object placed just before the focal length
construct a ray diagram for a converging lens with the object at half the focal length
construct a ray diagram for a converging lens with the object at twice the focal length
- Explain how to estimate the focal length of a converging lens
o Pass parallel rays of light from a distant object through the lens and move a screen behind the lens until the image is focussed. The distance between the lens and the focussed image is an estimate of the focal length.
- Explain how to determine the focal length of a lens accurately
o Place an object (eg. lamp behind cut out shape) on a ruler, 10cm in front of a lens in a lens holder and in front of a screen
o Move the screen until an image is focused on the screen
o Record the positions of the screen, lens and object
o Calculate image distance, v = lens distance – screen distance, and object distance = lens distance – object distance
o Repeat steps 1-4 for five different object distances, u
o Plot a graph of 1/u against 1/v
o 1/u = -1/v + 1/f
o y = m x + c
o the gradient should be -1
o determine the x and y intercepts
o f = 1/ x intercept and f = 1/y intercept
o take an average of the f values
when would the rays from an object converge at the focal point
o Only when the incident rays are parallel to the principle axis
o This happens when the object is at an infinite distance from the lens
- What happens if an object is placed closer to a converging lens than the focal point
o When the object distance is less than the focal length for a converging lens, a virtual image is formed behind the lens
o This is because the rays of light diverge after the lens
o Causing virtual rays to converge on the same side of the lens at the object
o These cannot be projected onto a screen
o The lens is behaving as a magnifying glass
- Draw a ray diagram for a magnifying glass
o Convex lens with object closer than the focal point:
- Describe the situations when a virtual image may form from a real object placed in front of the lens
o All images of real objects made by diverging lenses are virtual
o Images formed by objects placed closer than the focal point for a converging lens
- Explain why a converging lens will not produce an image when the object is placed at the focal point
o If image distance = focal length, u = f, then 1/f = 1/u + 1/v becomes:
o 1/u = 1/u + 1/v
o so 1/v = 0
o so image distance, v would be infinite – there is no image formed
o this is because one ray passes through the focal point on the right hand side of the lens
o and another ray passes through the centre of the lens
o these rays (and all others) are parallel, and therefore do not converge
o nor do they diverge, so a virtual image cannot be made either
Define unpolarised light
o Unpolarized light has oscillations (of the fields) in all directions
- Define polarized light
o Polarized light is EM radiation
o Where the oscillations of the electric field is in one plane
o And that plane contains the direction of propagation
OR
o Polarized light is EM radiation
o Where the oscillations of the electric field is in one direction
o the direction of oscillation is perpendicular to the direction of propagation
- Explain what a polarizing filter does to unpolarized light
o A polarizing filter absorbs all components of the oscillation that are perpendicular to the plane of polarization of the filter
o The filter transmits all components of the oscillation that are parallel to the plane of polarization of the filter
o This reduces the intensity of unpolarized light to half, and produces polarized light, that oscillates in one plane, the plane being the plane of the polarization of the filter (and includes the direction of propagation)
- Explain what a polarizing filter does to polarized light
o A polarizing filter absorbs all components of the oscillation that are perpendicular to the plane of polarization of the filter
o The filter transmits all components of the oscillation that are parallel to the plane of polarization of the filter
o This means that when the plane of polarization of the filter is parallel to the plane of polarization of the polarized light, all of the polarized light is transmitted and the intensity remains the same before and after the filter
o This means that when the plane of polarization of the filter is perpendicular to the plane of polarization of the polarized light, all of the polarized light is absorbed and the intensity becomes zero
Explain why longitudinal waves cannot be polarised
o Longitudinal waves oscillate parallel to the direction of propagation
o So this oscillation direction cannot be removed without also removing the propagation of the wave
o Light from the source is unpolarized – the oscillations occur in all planes / directions
o As the unpolarized light passes through the first filter, the intensity reduces to half
o This is because filter 1 absorbs all oscillations that are perpendicular to the filter’s plane of polarization
o When the second filter is aligned at 0 or 180 degrees to the first filter, all the polarized light (at intensity I0/2) is transmitted
o As the second filter is rotated, the polarized light’s plane of polarization is no longer completely parallel to the filter’s plane of polarization. Perpendicular components of the light’s oscillation are absorbed and parallel components are transmitted.
o When the second filter is aligned at 90 or 270 degrees to the first filter, all the polarized light (at intensity I0/2) is absorbed because the second filter’s plane of polarization is entirely perpendicular to the light’s plane of polarisation, and the intensity is zero
- Explain how to use one filter to demonstrate whether a source of EM radiation is polarized or not
o Hold the filter up to the light source and rotate it
o If the light is polarized, the intensity will cycle from maximum to zero every 90 degrees.
o If the light is unpolarized, the intensity will reduce to half its original value but remain constant as the filter is rotated
82.
o As the filters are placed with their planes of polarization at 90 degrees to one another, all light that has not had its plane of polarization rotated would be absorbed
o So where light is transmitted must indicate that the plane of polarization has been rotated due to stress
o So the brighter areas represent more stress where rotation has happened, and the darker areas represent less stress where rotation has not happened
o This indicates that reflected sunlight is polarized
o meaning its oscillations are only in one plane
o When the polarizing filter’s plane of polarisation is parallel to the plane of polarization of the sunlight, the EM radiation is transmitted, and the intensity is high
o After rotating 90 degrees, the polarizing filter’s plane of polarisation is perpendicular to the plane of polarization of the sunlight, the EM radiation is absorbed, and the intensity is zero
Well done - treat yourself once you finish!
