Questions on the eye as an optical system

Question 1

State the nature of the image formed on the retina of a normal eye, and name the part of the eye at which most of the refraction happens.

Accepted Answer

The image is real, inverted and diminished (1). Most of the refraction happens at the cornea, the curved front surface, because that is where light meets the largest change in refractive index; the lens makes the smaller adjustment (1).

Question 2

State what is meant by the near point and the far point of an eye, and give the value of each for a normal eye.

Accepted Answer

The near point is the closest object the eye can focus sharply, taken as 0.25 m for a normal adult eye (1). The far point is the furthest object it can focus sharply, which is at infinity for a normal eye (1).

Question 3

State what is meant by accommodation, and name the part of the eye that changes.

Accepted Answer

Accommodation is the eye changing its optical power so that objects at different distances are all focused onto the retina (1). It is the lens that changes: ciliary muscles alter its shape and so its focal length (1).

Question 4

Calculate the power of a converging lens of focal length 0.40 m. A second lens has a power of −4.0 D: calculate its focal length and state what kind of lens it is.

Accepted Answer

P = 1/f = 1/0.40 = +2.5 D (1). For the second lens f = 1/P = 1/(−4.0) = −0.25 m; the negative focal length means it is a diverging lens (1).

Question 5

State the unit of lens power and what it is equivalent to in SI base terms.

Accepted Answer

The dioptre (D), equivalent to one per metre, m⁻¹ (1). The focal length must therefore be in metres before the reciprocal is taken.

Question 6

State what is meant by myopia and by hypermetropia.

Accepted Answer

Myopia (short sight) is an eye that is too powerful for its length, so parallel rays focus in front of the retina and the far point is at a finite distance (1). Hypermetropia (long sight) is an eye with too little power for its length, so rays from a near object would focus behind the retina and the near point is further away than 0.25 m (1).

Question 7

State what is meant by presbyopia, and state the measurement that distinguishes it from hypermetropia.

Accepted Answer

Presbyopia is the loss of accommodation that comes with age as the lens stiffens, so the near point moves out beyond 0.25 m while the power of the relaxed eye is unchanged (1). The far point is the measurement that separates the two: in presbyopia it is still at infinity, so distance vision is normal, whereas a hypermetropic eye is short of power at every distance (1).

Question 8

An eye has its retina 22 mm behind its lens, and its far point is at infinity. Calculate the power of the eye when it is relaxed.

Accepted Answer

Relaxed means the object is at the far point, so u is infinite and 1/u = 0 (1). 1/f = 1/u + 1/v = 0 + 1/0.022 (1), so f = 22 mm and P = 1/0.022 = 45 D (1). Leaving the 22 in millimetres gives a power a thousand times too small.

Question 9

The same eye, with its retina 22 mm behind its lens, now focuses on an object 0.30 m away. Calculate the focal length of the eye in this state.

Accepted Answer

The image is still on the retina, so v = 0.022 m and only f changes (1). 1/f = 1/0.30 + 1/0.022 = 3.33 + 45.45 = 48.79 (1), so f = 20.5 mm (1). 0.30 m is the object distance, so it is u and not v.

Question 10

A short-sighted eye has its far point 0.25 m from the eye. Calculate the power of the spectacle lens needed for distant vision, and state what kind of lens it is.

Accepted Answer

The lens must take an object at infinity and form a virtual image at the far point, so u is infinite and v = −0.25 m (1). 1/f = 0 + 1/(−0.25) (1), so P = −4.0 D, a diverging lens (1). Dropping the minus sign on v gives +4.0 D, which would make the sight worse.

Question 11

A long-sighted eye has its near point 0.75 m from the eye. Calculate the power of the lens that would allow this eye to read at the normal 0.25 m.

Accepted Answer

The lens must take the page at u = 0.25 m and form a virtual image at the eye's own near point, so v = −0.75 m (1). 1/f = 1/0.25 + 1/(−0.75) = 4.00 − 1.33 (1), so P = +2.7 D, a converging lens (1).

Question 12

A person with a normal eye, whose retina is 23 mm behind the lens, looks at a child 1.8 m tall standing 10 m away. Calculate the magnification and the height of the image on the retina.

Accepted Answer

m = v/u = 0.023/10 (1) = 2.3 × 10⁻³ (1). Image height = 1.8 × 2.3 × 10⁻³ = 4.1 mm, about 4 mm (1). The image is inverted.

Question 13

Explain what astigmatism is, and state the three quantities a prescription to correct it must give.

Accepted Answer

The cornea, or sometimes the lens, is not the same shape in every direction across its face, so its power differs between two perpendicular directions and a point object cannot be brought to a point image in both at once (1). The prescription gives the spherical power SPH in dioptres and the additional cylindrical power CYL in dioptres (1), together with the AXIS in degrees from 0 to 180, which states the orientation at which the cylinder is set (1).

Question 14

Explain why rods allow the eye to see in very dim light but resolve much less detail than cones.

Accepted Answer

Many rods are connected to a single nerve fibre, so the light caught by all of them is added together into one signal (1). That summed signal is large enough to register when very little light is arriving, which gives the sensitivity (1). But the brain cannot tell which rod of the group was lit, so detail within the group is lost, whereas each cone in the fovea has effectively its own fibre and so stays a separate signal (1).

Question 15

A spectacle lens for a short-sighted eye sits about a centimetre in front of the eye. Explain why such a lens cannot put an image on the retina, and state where the image it forms has to lie if the eye is to see a distant object clearly.

Accepted Answer

The lens is outside the eye and the image it forms lies outside the eye too, so nothing it does can place an image on the retina (1). What it does instead is present the eye with an object the eye can already focus: it forms a virtual image, on the same side of the lens as the object, and the eye then forms the real image on its own retina (1). For a distant object that virtual image has to lie at the eye's far point, so the image distance is negative and equal in size to the far-point distance (1).

Question 16

An eye has its retina 23 mm behind its lens. Its far point is at infinity and its near point is 0.40 m away. Calculate the power of the eye at each of these two limits, hence the range of power accommodation provides, and the focal length of the eye at its near point. State, with a reason, what these two limits show about this eye.

Accepted Answer

At the far point u is infinite, so P = 1/0.023 = 43.5 D (1). At the near point P = 1/0.40 + 1/0.023 = 2.50 + 43.48 = 46.0 D (1). The range of accommodation is 46.0 − 43.5 = 2.5 D, which is exactly 1/0.40 because the 1/v term never changes (1). At the near point f = 1/46.0 = 21.7 mm (1). The far point is at infinity, so the unaccommodated eye focuses distant objects normally and the eye is not hypermetropic; what the near point at 0.40 m shows is reduced accommodation, 2.5 D against the normal 4.0 D, which is presbyopia (1).

Question 17

A short-sighted eye has its far point 0.20 m away and its near point 0.12 m away. Calculate the power of the lens that corrects the eye for distant vision. With that lens worn, determine how far from the lens the nearest object it can focus now lies, and comment on the result.

Accepted Answer

For distance: u infinite, v = −0.20 m, so P = 1/(−0.20) = −5.0 D and f = −0.20 m (1). With the lens worn, the nearest object is the one whose image the lens forms at the eye's own near point, so v = −0.12 m (1). 1/u = 1/f − 1/v = −5.00 − (−8.33) = 3.33 (1), so u = 0.30 m (1). The corrected near point is 0.30 m, beyond the normal 0.25 m, so the wearer has to hold a book further away than before or take the glasses off to read (1).

Question 18

A faint star is easier to see when you look slightly to one side of it than when you look straight at it, and when seen that way it appears colourless. Explain both observations in terms of the receptors of the retina.

Accepted Answer

Looking straight at the star puts its image on the fovea, which is packed with cones and carries no rods (1). Cones need bright light, so a faint image there produces little response (1). Looking slightly to one side moves the image off the fovea onto the rods, which are far more sensitive because many of them share a nerve fibre and their signals are added together (1). Rods are of a single type only, so they report brightness and carry no information about wavelength (1), and the star therefore appears colourless; colour vision needs the three types of cone, which are not responding at this light level (1).

Question 19

The eye can just resolve two points that subtend an angle of about 3.0 × 10⁻⁴ rad at it. Calculate the smallest separation of two points that can be distinguished at the near point of 0.25 m, and explain what feature of the retina sets this limit.

Accepted Answer

Separation = distance × angle = 0.25 × 3.0 × 10⁻⁴ (1) = 7.5 × 10⁻⁵ m, about 0.08 mm (1). The limit is set by the spacing of the cones in the fovea: two points are seen as separate only if their images fall on two cones with at least one unstimulated cone between them (1). Each fovea cone has its own nerve fibre, so those two signals stay distinct all the way to the brain; on the rods, where fibres are shared, the same two points would merge (1).

Question 20

A long-sighted eye has its near point 1.0 m away and its far point at infinity. It is given a reading lens of power +3.0 D. Determine the nearest and the furthest object distances the eye can focus on while wearing this lens, and comment on the result.

Accepted Answer

f = 1/P = 0.333 m. Nearest: the lens must form its virtual image at the eye's own near point, so v = −1.0 m and 1/u = 1/f − 1/v = 3.0 + 1.0 = 4.0 (1), giving u = 0.25 m (1). Furthest: the eye can focus an object at infinity, so the lens must send its image there, 1/v = 0 and 1/u = 1/f, giving u = f = 0.33 m (1). The whole range of clear vision with these glasses on is 0.25 m to 0.33 m, so they serve for reading and have to be taken off, or looked over, to see anything further away (1).

Question 21

A prescription for one eye reads −3.00 / −1.25 × 180. State what each of the three numbers gives, calculate the total power of the lens in the meridian in which the cylinder acts, and calculate the focal length of the lens in the meridian along which the cylinder adds no power.

Accepted Answer

−3.00 D is the spherical power SPH, −1.25 D is the additional cylindrical power CYL, and 180 is the AXIS in degrees, naming the meridian along which the cylinder adds no power (1). The cylinder acts in the meridian at right angles to that axis, where the two powers add: −3.00 + (−1.25) = −4.25 D (1). Along the axis itself only the spherical power acts, so f = 1/(−3.00) (1) = −0.333 m, and both meridians are diverging, as the negative powers require (1).

Question 22

A short-sighted patient is still wearing glasses of power −1.5 D, made before the myopia worsened. Wearing them, the furthest object seen clearly is 2.0 m away. Determine the far point of the unaided eye, and hence calculate the power the glasses should now have.

Accepted Answer

The glasses form a virtual image of that furthest object at the eye's own far point, so the far point is where that image stands, and being virtual it takes a negative image distance v (1). f = 1/(−1.5) = −0.667 m, and 1/v = 1/f − 1/u = −1.50 − 0.50 = −2.00, so v = −0.50 m: the unaided far point is 0.50 m from the eye (1). A full correction takes an object at infinity to that same far point, so u is infinite and v = −0.50 m, giving P = 1/(−0.50) = −2.0 D (1). It is a diverging lens, and stronger than the pair being worn; those are short of it by 0.5 D, which is exactly why they leave the far point at a finite 2.0 m rather than at infinity (1).