PHYS 1008 Optical Instruments

Objectives: by the end of this you will be able to Predict the behaviour of compound lens systems Understand the eye Understand telescopes Understand microscopes Explain the limitations of optical instruments

Compound systems

• Draw a diagram (!) with object and a guess at the light-path
• Find out where the first lens produces an image
• Use this image as the object for the second lens (if the object distance is negative, it means that it is just a virtual object)
• Find out where the final image is.

e.g. Two convex lenses, f = 10 cm for both are separated by 8 cm. If there is an object 10 cm high at 40 cm, where is the final image what kind is it how high is it?

The Eye

• Works differently from any other optical instrument (such as camera, telescope..) in that focussing is performed by deforming the lens by the eye muscles. Means that eye can be focussed (ideally) from a far point of ∞ down to a near point of 25 cm.

• Really 2-lens system (cornea does most of focussing)
• Iris cuts down light
• Retina detects light
• Eye is filled with vitreous humour

Common eye-problems:

• Short-sight/Near-sightedness/myopia: caused by too strong a lens, corrected by concave lens.

• Long-sight/far-sightedness/hyperopia: caused by too weak a lens, corrected by convex lens

• Presbyopia: inability to focus, corrected by bi-focals
• Astigmatism: eye is not perfectly spherical, corrected by cylindrical lens.

• a near-sighted person can only see objects clearly at a distance of 30 cm:
• what corrective lens would be needed so that he could see objects at a large distance?
• If his near point was at 10 cm, where would it be wearing glasses?

• Note that opticians usually quote the power of a lens in diopters:
• power in diopters = 1/focal length in metres
• can add power in diopters for a compound lens: e.g. 2 lenses of 10 and 30 cm focal length: what is power of each and f for the combination?

Simple magnifying glass

• The object is inside the focal length. e.g. the object is 2 mm high, f = 5 cm, s = 4.8 cm

• Where is the image?
• What is M?
• This magnification is not really meaningful, since the image is much further away: instead

  We really use a magnifying glass because we cannot bring an object nearer to the eye than x ~ 25 cm (the near point). Hence what we get is angular magnification M = θ = x θ' f What is it in this case?

Limits to Optical Instruments

• Since light is a wave, it is possible for it to take different paths to the same point. This leads to interference, if one path is ½λ longer than the other

• Means that a point will be spread out into a disk, and hence the image of two close objects will overlap

θ = 1.22 λ
         D 

• Stars closer than this cannot be resolved into pairs. Similarly microscopes cannot resolve objects which are much smaller than λ (see below)
• e.g you want to be able to see a dime at 10 km: what is the angular size?
• How large a telescope would you need (assume green light?)
• At what distance could you see headlights as separate (assume they are 2 m apart, and your pupil has a diameter of .5 cm.

• ---

  Also there are problems due to the lenses themselves

Chromatic aberration:

• Glass disperses light into constituent wavelengths, so get different foci for different wavelengths

• Overcome by exploiting different dispersive powers of different glasses (usually "crown" and "flint" glass)
• In fact, objective is usually sandwich of several different glasses

• Thin lens formula assumes that deflections are small (i.e. sin(θ) = θ)
• In practice, light from outer region of lens is more strongly focussed than central part

• Effect is blurred image

• One solution is to eliminate light from outside of lens by stopping down

• Unfortunately this reduces the area, so you get less light. Only real way out is to go to non-spherical lenses/mirrors. e.g. for a telescope, spherical mirror gives blurred image

• Can be corrected by going to a parabolic mirror (but this only works for objects at infinity)

Telescopes

• Simplest is two lens refractor. Objective is large lens at "front", brings light to a primary focus. Eyepiece is used to magnify image. Magnification: if f₀ is focal length of objective f₁ is focal length of objective M = f₀ f₁

• e.g. typically objective might have f₀ = 150 cm, eyepiece .75 cm
• Light-Collecting power ∝ area of objective
• i.e. want to maximize R

• Newtonian reflectors Replace objective lens by mirror: Advantages: No chromatic aberration Support for whole of mirror Only one surface matters

• Disadvantage: Spherical Aberration remains

• "Best" standard telescope is Schmidt- Cassegrain Corrector plate is very weak aspherical lens to exactly (well, almost) compensate for spherical abberation

• Gives compact design, with folded light-path, eyepiece at end, wide field of view. This is (essentially) what is used in the Hubble space telescope

A spectacular picture from the Hubble: large masses can bend light,

• so a very large cluster of galaxies (about 1013 MSun) can act as a (lousy!) lens. This cluster is producing multiple images of a much more distant galaxy.

Compound Microscope.

• Objective produces magnified real image of object, which is then magnified by the ocular (eyepiece). Usually have final image at ∞.

• Overall angular magnification M = m₁M₂ =- s₁'x f₁f₂

• Where is the image produced by the objective?
• What is m₁?
• Where should ocular be placed so the final image is at ∞ ?
• What should the length of the barrel be?

Note an important practical problem: resolving power for a microscope is \color{red}{ RP = \frac{{0.6\lambda }}{{n\sin \left( \alpha \right)}}} This says that we can get below λ by using a wide lens (making α large) and by increasing the refractive index n: e.g. putting the object in oil.: e.g. in the last example, what is the resolving power if the objective has a diameter of 3 mm, λ = 500nm (green light), n = 1.6 (typical oil) and α = 45°? Roughly: cannot resolve anything with size < λ

• And this finishes classical physics. Haven't we understood a lot!
• But now there are a few problems