Question

The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.00 mm , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude.
(Note: The results obtained in the parts A, B and C are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.)

a.Find the radii of curvature of this lens.

b.If an object 11.0 cm tall is placed 29.0 cm from the eye lens, where would the lens focus it?

c.How tall would the image be?

d.Is this image real or virtual?

e.Is it erect or inverted?

Answer 1

Solution : from lens maker formula

1/f = (n-1)( 1/R₁ - 1/R₂)

Here f = 0.008 meter . n = 1.44 . R1 = -R and R2= +R

Replacing the values

1/0.008 = 0.44 * -2/R =

So R = 7.04 mm

Now 1/v -1/u = 1/f

So v = uf / (u+f)

Here u = -290 mm f = 8 mm

So v = 8.27 mm from lens

Size of image = 8 27 x 110 /290 = 3.14 mm

Image is real and inverted.