The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is the range of the power of the eye-lens?

Sol. The child has near point and far point 10 cm and 100 cm respectively. Since, the retina is 2 cm behind the eye-lens, v = 2cm For near point u = – 10 cm = – 0.1 m, v = 2 cm = 0.02 m So, 1/f base near = 1/v – 1/u = 1/0..02 – 1/-0.1 = 50 + 10 = 60D For far point, u = - 100 cm = - 1 m, v = 2 cm = 0.02 m So, 1/f base far = 1/v – 1/u = 1/0.02 – 1/-1 = 50 + 1 = 51D So, the rage of power of the eye-lens is +60D to +51D