A professor reads a greeting card received on his 50th birthday with + 2.5 D glasses keeping the card 25 cm away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter 50 cm away. What power of lens should he now use?

Sol. On the 50^th birthday, he reads the card at a distance 25cm using a glass of +2.5D. Ten years later, his near point must have changed. So after ten years, u = – 50 cm, f = 1/2.5D = 0.4m = 40 cm v = near point now, 1/v – 1/u = 1/f ⇒ 1/v = 1/u + 1/f = 1/-50 + 1/40 = 1/200 So, near point = v = 200cm To read the farewell letter at a distance of 25 cm, U = – 25 cm For lens formula, 1/v – 1/u = 1/f ⇒ 1/f = 1/200 - -/-25 = 1/200 + 1/25 = 9/200 ⇒ f = 200/9 cm = 2/9 m ⇒ Power of the lens = 1/f = 9/2 = 4.5D ∴ He has to use a lens of power +4.5D.