3.18. Defects caused by reflections We have assumed that the paraxial approximation applies in that all rays make small angles with respect to the optical axis. We also assume that the rays strike close to th e centre of the mirror. In real systems the rays hit the whole mirror. We find that rays that hit near the outside of mirror are focused at a different point when compared with those that are close to the centre. This causes a blurring of the image, and is known as spherical aberration. Note that to show the effect the reflection of the outer rays has been greatly exaggerated. The normals for the outer rays come from the same point as those for the inner rays. This problem is eliminated by use of a parabolic mirror.

3.19. Refraction and refractive index The wave speed of light in a vacuum is 3 x10 8 m/s. When light propagates through a medium the interaction between the atoms in the material and the light waves causes a reduction in the speed of the wave. This change in speed is quantified by the refractive index n

3.20. Dispersion and Refraction The refractive index n( ) is wavelength dependent which means that different colours travel with speeds through a material. This is known as dispersion.

3.20. Dispersion and refraction When light propagates through a material the speed of light is reduced owing to the response of the medium to the electric and magnetic fields. Although the wave speed changes the energy of the wave does not change. As E = hf the frequency remains constant. This means that the wavelength must reduce. The effective wavelength in the material is given by  /n. There is a phase shift imposed by the material. If the material of length l and refractive index n the phase shift is 2πnl/

3.21. Refraction at an interface Consider a perfectly flat interface between two transparent materials. The wave speed in material 1 is v 1 and the refractive index is n 1. The wave speed in material 2 v 2 and the refractive index is n 2. A plane wave propagates from material 1 to material 2. The wave is incident at an angle i to the interface. The transmitted wave makes a angle r to the interface. All the rays and normals lie in the same plane - the plane of incidence

This sytem can be understood by use of Huygens’ Principle. Consider the incident wave front AB. Point A is in contact with the interface. In the time taken for point B to move to D the point A moves to C to form the refracted wave front CD. Time, T, to for A to move to C is T = AC/v 2 Time, T, to for B to move to D is T = BD/v 1 So AC/v 2 = BD/v 1 But AC = ADsin r And BD = ADsin i

So sin r/v 2 = sin i/v 1 Hence but n 2 =c/v 2 and n 1 = c/v 1 Hence This is Snell’s law. When light goes from a less dense to a more dense medium n 2 >n 1 the rays are refracted towards the normal i>r. When light goes from a more dense to a less dense medium n 1 >n 2 the rays are refracted away from the normal r>i.

3.22. Deviation caused by refraction Consider a parallel plate of glass of thickness t and refractive index n in air. This plate causes a deviation of the beam from its original path. The angular deviation  = i - r The spatial deviation d can be found from the figure d = ce Triangle ace gives –ce = acsin(i-r) Triangle abc gives –ac = ab/cosr But ab = t So ac =t/cosr Hence d = ce = tsin(i-r)/cosr

3.23. A method of measuring the refractive index A plate of glass has a thickness t and refractive index n. A mark is placed under the plate and the mark is viewed from above the plate. The rays that come from the mark are refracted at the interface between the plate and air. The rays are refracted away from the normal. Thus the mark appears to be inside the block. The refractive index n is given by

3.24. Total Internal Reflection What happens when light passes from a more dense medium to a less dense medium? Here n 1 > n 2 From Snell’s law we have As n 1 > n 2 Then r > i Hence the ray is refracted away from the normal. When r = 90˚ the refracted ray moves along the interface. The angle of incidence at which this happens is the critical angle  c.  c is given by

3.24. Total Internal Reflection Here n 1 > n 2 i <  c, r < 90˚ i =  c, r = 90˚ i >  c, tir When r = 90˚ the refracted ray moves along the interface. The angle of incidence at which this happens is the critical angle  c. For angles of incidence greater than the critical angle light cannot exit at the boundary and so is reflected. This is total internal reflection.

3.25. How an optical fibre works An optical fibre relies on total internal reflections to guide light down the fibre. The fibre consists of a core of refractive index n co and a cladding layer of refractive index n cl (n co > n cl ). There is a specific angle  at the input facet which if exceeded means that light is no longer guided down the fibre. This is given by the following

3.26. Refraction and prisms Consider a prism of apex angle A and refractive index n. What happens when monochromatic light is incident on the prism? The beam is deviated from its original path by the effects of refraction at both surfaces. The total angular deviation d t is given by d t = d 1 + d 2 With d1 and d2 the angular deviation at both surfaces So d t = (i 1 - r 1 ) + (i 2 - r 2 )

3.27. Minimum deviation There is an angle of incidence i whereby the beam path through the prism is symmetric. At this angle of incidence the deviation is a minimum As the beam path is symmetric d min = 2i -2 r Now from the figure we find that 2r = A So d min = 2i -A Hence 2i = A+d min Thus the refractive index n is given by We assume that the prism is in air

3.28. Prisms and Polychromatic Light So far we have assumed that monochromatic light has been incident on the prism. When polychromatic light is shone on to the prism we find that the light is dispersed. The dispersion occurs because the refractive index is wavelength dependent. The refractive index for blue light is greater than that of red light. Hence the blue light is refracted more at the prism interfaces and so experiences a greater deviation when compared to the red light. The dispersive power  of a prism is given by