A far-sighted student has a near point of 309 cm. Calculate the focal length of the glasses needed so the near point will be normal (25 cm). Neglect the space between the eyes and the eye-glasses. Answer in units of cm.

27.2 cm

Explanation:

Note: For the student to see an object clearly at 25 cm, from the eye the image must be formed 309 cm on the same side of the lens, at his near point.

From mirror Formula,

1/f = 1/v+1/u

Where f = focal length of the glasses, v = image distance, u = object distance.

f = vu/(v+u) Equation 1

Given: u =  25 cm, v = -309 cm ( virtual image)

Substitute into equation 1

f = [25×(-309)]/[25+(-309)]

f = -7725/-284

f = 27.2 cm.

Hence the focal length of the glass = 27.2 cm.

object 2 because all the other ones are wrong