Question

The least distance of distinct vision for a far-sighted person is 1 m. The optical power of the lens of his specs, which effectively reduces his LDDV to 25 cm, is:

(a) +4 D
(b) -4 D
(c) +1 D
(d) -1 D

Answer 1

The correct optical power for specs needed to correct the vision of a farsighted person with a LDDV of 1 m to 25 cm is +4 diopters (D). Therefore the correct option is (a) +4 D.

Step-by-step explanation:

The question asks for the optical power required for a farsighted person with a least distance of distinct vision (LDDV) of 1 meter to clearly see objects at 25 cm. The spectacle lens formula to find the power (P) is:

P = 1/f

where f is the focal length in meters. To reduce the LDDV to 25 cm, the lens should have a focal length equal to the new LDDV. Since the LDDV is given in centimeters, it should first be converted into meters (25 cm = 0.25 m).

The required power of the lens is thus:

P = 1/0.25 m = 4 D

The positive sign indicates it is a converging lens, which is suitable for a person with farsightedness (hyperopia). Therefore, +4 D is the correct optical power for the specs.