Question

make a diagram to show how hypermetropia is corrected. the near point of a hypermetropic eye is 1 m.what is the power of the lens required to correct this defect .assume that the near point of the normal eye is 25 CM

Answer 1

⭐⭐hey mate!

To find the focal length apply the formula

F = (N.P)(D) / N.P -D

Where ,

N.P = near point

D = least distance

1/F = 1/v -1/u

1/F = 1/-100 - 2/-25

F =100/3

Power = 3/100 × 100 = 3D

#phoenix

Answer 2

#HOPE IT HELPS...^_^