Sometimes a person cannot clearly see objects close up or far away. to correct this type of vision, bifocals are often used. the top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. a person can clearly see objects only if they are located between 45 cm and 161 cm away from her eyes. bifocal lenses are used to correct her vision. what power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?

P= -0.62 D

Explanation:

We know that lens equation

1/f=1/u+1/v

Given that person can see object only when object is located between 45 and 161 cm so we can say that far point of that person is at 161 cm.

So we need to take that objects ray are coming from and infinity and will focus at 161 cm away.

So now by putting the values

1/f=1/∞+1/-161    (negative because image will be virtual ,by sign convention)

So f=-161 cm

We know that inverse of focus(in meter) length is called power of lens.

So power of lens P=1/f

  P=-1/1.61

P= -0.62 D

well clustered together.

points close on a scatter diagram