A person with a defective eye-vision is unable to see the objects nearer than 1.5 m. He wants to read books at a distance of 30 cm. Find the nature, focal length and power of the lens he needs in his spectacles.
This person suffers from the defect of hypermetropia.
For him u = -30cm, v = -1.5 m = -150cm
Therefore, focal length of corrective lens to be used by him is
1/f = 1/v- 1/u = 1/-150 - 1/-30 = 4/150 = 37.5cm
The positive sign shows that the lens needed is a convex lens of focal length 37.5 cm.
Hence, power of lens needed
P =1/f = 100/37.5 = 2.67D