The cornea behaves as a thin lens of focal length approximately 1.80 cm , althou

Question

The cornea behaves as a thin lens of focal length approximately 1.80 cm , although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.00 mm . (Note: The results obtained here are not strictly accurate, because, on one side, the cornea has a fluid with a refractive index different from that of air.)

If this focal length is in air, what is the radius of curvature of the back side of the cornea?

The closest distance at which a typical person can focus on an object (called the near point) is about 25.0 cm , although this varies considerably with age. Where would the cornea focus the image of an 15.0 mm -tall object at the near point?

What is the height of the image in part B?

Explanation / Answer

We know that from the lens makers formula

1/f =(n-1)[1/R1-1/R2]

given that f =0.0180m, n =1.38   and R1 =0.005m and now we have to solve for R2

1/0.0180 =(1.38-1)[1/0.005-1/R2]

1/R2 =200-146.2 =53.8m

Now R2 =0.0186m =1.86cm =18.6mm

So, the front is curves outward with a radius of 5mm and the back is curved with a radius of 18.6mm

b)

We know that

1/u+1/v =1/f

u =25cm

f =1.80cm

1/25+1/v =1/1.80

v =1.94cm

c)

The magnification is given by

m =v/u =-hi/ho

1.94/25 =hi/15

then the height of the image is given by hi =1.164mm

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