lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens

The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?

29.96cm

Explanation:

Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.

Now using the lens equation as follows;

  -------------(i)

Where;

f = focal length of the lens

v = image distance as seen by the lens

u = object distance from the lens

From the question;

v = -151cm        [-ve since the image formed is virtual]

u = 25cm

Rewrite equation (i) to have;

Substitute the values of v and u into the equation;

f = 29.96cm

The focal length should be 29.96cm

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it