Consider the human eye as a thin lens (the eye lens) and a screen (the retina),
ID: 1785602 • Letter: C
Question
Consider the human eye as a thin lens (the eye lens) and a screen (the retina), which is situated at a distance of 17 mm behind the lens.
a) Calculate the dioptric power of thin contact lenses that allow advancing a person’s near point from 150 cm to 25 cm. Is this person near-sighted or far-sighted? And is the contact lens converging or diverging?
b) Calculate the dioptric power of contact lenses that allow a person who sees objects at a distance larger than 1.5 m blurred to see object at infinite distance sharp. Is this person near-sighted or far-sighted? And is the contact lens converging or diverging?
c) What type of eyeglasses (converging or diverging) should a normal sighted person wear under water to see objects sharp? The refractive index of water is 1.33, the refractive index of the eye lens is 1.38.
Explanation / Answer
given human eye is a lens and retina is a screen
distance of screen from the lens, v = 17mm = 1.7 cm
a. initial near point = 150 cm
final near point = 25 cm
hence object distance = u = -25 cm
image distance, v = -150 cm
so, focal length of the lens be f
then from thinlens formula
1/v - 1/u = 1/f
-1/150 + 1/25 = 1/f
f = 30 cm
power of lens = 1/f in m = 3.333 D
the lens is converging lens with +ve focal length
this man is far sighted as near by objects appear blurry top him
b. object distance u = infinity
image distacne, v = -150 cm
hence
1/v - 1/u = 1/f
f = -150 cm = -1.5 m
hjence power = -0.6667 D
the lens is diverging and trhe person is near sighted
c. to see objects under water sharp a normal person will normall not need lens as eyelens adjusts its power to see objects clearly