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The physics of the eye questions
A camera focuses by moving its lens; the eye cannot, because the retina is fixed about 23 mm behind it. Instead the lens changes shape, and when the range of shapes it can take no longer covers near and far vision, a corrective lens supplies the missing power in dioptres.
17 original questions · 51 marks · the physics of the eye notes · Medical physics
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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.
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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).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.
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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).State what is meant by accommodation, and name the part of the eye that changes.
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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).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.
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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).State the unit of lens power and what it is equivalent to in SI base terms.
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The dioptre (D), equivalent to one per metre, m−1 (1). The focal length must therefore be in metres before the reciprocal is taken.State what is meant by myopia and by hypermetropia.
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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).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.
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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.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.
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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). Writing v = 0.30 m is the common error: 0.30 m is where the object is, not where the image is.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.
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.
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.
Explain what astigmatism is, and state the three quantities a prescription to correct it must give.
Explain why rods allow the eye to see in very dim light but resolve much less detail than cones.
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.
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.
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.
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.
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