Question

Constants A student's far point is at 17.0 cm, and she needs glasses to view her computer screen comfortably at a distance of 53.0 cm

Answer 1

SOL)

Given the far point of student is s^’= -17cm

1cm=10-2m

S^’=-17/100= -0.17m

S^’=-0.17m

She needs glasses to view her computer screen comfortably at a distance of S=53cm

S=53cm/100

S=0.53m

The object distance, image distance and focal length are related by the formula is

1/f=1/s+1/s^’

1/f=s^’+ s /s. s^’

f=s .s^’/s^’+ s

  =(0.53m) (-0.17m)/(-0.17m)+(0.53m)

= -(0.0901m)/0.7m

f= - (0.12871)m

  The power of her glasses is

     1/f=1/ (-0.12871m)

     1/f= -7.7694diopters

When she wears diverging lenses the power of lenses is negative