Q2. A hypermetropic person has a near point of 100 cm. The person wants to read at 25 cm. The minimum power of the correcting lens close to the eye is:
Options

Question

Q2. A hypermetropic person has a near point of 100 cm. The person wants to read at 25 cm. The minimum power of the correcting lens close to the eye is:

G1 Physics · Accepted Answer

GIVEN: Old near point = 100 cm; desired reading distance = 25 cm; lens close to eye. Step 1: The convex spectacle lens must produce a virtual image 100 cm in front of it when the book is at 25 cm. Step 2: Using Cartesian signs: u = −0.25 m and v = −1.00 m. Step 3: Lens power P = 1/f = 1/v − 1/u (distances in metres). Step 4: P = (−1/1.00) − (−1/0.25) = −1 + 4 = +3 dioptres. Step 5: The positive sign indicates a converging lens, as required for hypermetropia. FINAL ANSWER: +3 D. Option C.