Solutions: Medical Option - the EYE
(a) Rays of light from a point source, 25 cm from a normal eye, enter the eye and are focused forming an image on the retina. State where most refraction occurs and explain why.
The greatest refraction occurs at the boundary between the air and cornea. This is because at that interface there is the
greatest change in refractive index of the materials.
(b) Explain two changes which occur when the eye responds to a wide change in intensity of illumination from very dim light to bright light.
In bright light the cones are able to be used, whereas in dim light there is only a high enough light intensity for the rods to be activated. There is a greater density of cones on the central area of the retina, the rods are more widely spaced and only more concentrated than cones on the peripery of vision, therefore images seen in bright light have higher definition.
In bright light the iris reacts to make the pupil smaller This limits the intensity of light falling on retina so that the cones are not 'bleached out' because they recieve above their maxium intensity of light.
(3 marks maximum)
(c) Explain the term accommodation when applied to the eye.
Accomodation is the ability of the eye to change the lens shape/power. This allows the eye to form a clearly focused image for a range of object distances by adjusting the power of the total optical system to suit the purpose.
(d) A person suffering from long sight is just able to see clearly an object placed 25 cm from the eye when a correcting lens of power +2.4 D is used. Calculate the least distance from the eye to an object that the eye could just see clearly without the use of the correcting lens.
u = 0.25 m
P = 2.4 D
P = 1/f = 1/u + 1/v
2.4 = 1/0.25 + 1/v
1/v = 2.4 - 4 = -1.6
v = -0.625
The least distance from the eye to an object that the eye could just see clearly without the use of the correcting lens is 63 cm
(Total 9 marks)