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 8.00mm , 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 14.0cm tall is placed 27.0cm 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?

a) R = 7.04 mm , b)  q = 0.848 cm , c) h’= -1.635 cm , d)  image is rea amd e) inverted image

Explanation:

a) In optics we have equations that relate the radius of curvature and focal length

         1 / f = (n-1) (1/R1 -1/R2)

They indicate that the two radii of curvature have the same magnitude, notice that the radius of the left is positive, but the radius of the right is negative

         1 / f = (n-1) 2/R

         R = 2 (n-1) f

         R = 2 (1.44 -1) 8

         R = 7.04 mm

b) Distance to the object (p) 14.0cm, let's calculate the distance to the image (q)

         1/p + 1/q = 1/f

         1/q = 1/f  - 1/p

         1/q = 1/0.8 - 1/14.0

         1/q = 1.25  -0.0714  = 1,179

          q = 0.848 cm

The image is 0.848 cm behind the lens

c) The height of the image can be found through magnification

          M = h’/ h = - q / p

          h’= -q / p h

          h’= - 0.848/14.0  27.0

          h’= -1.635 cm

d) That image is real because the light rays pass through it, you can also see it because the distance to the image is positive

e) The found value of the image height is negative, so the inverted image

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.