Question

A man’s near point is 0.5 m and the far point is 3 m. Power of spectacle lenses required for (i ) reading purposes, ( ii) seeing distant objects , respectively are
A) - 2 D and + 3 D
B) + 2 D and - 3 D
C) + 2 D and - 0.33 D
D) - 2 D and + 0.33 D

Answer 1

Hint:The correct spectacle lens for reading purposes should make the man read the letters placed at 25 cm from his eyes ( because 25 cm is the least distance of distinct vision). The correct spectacle lens for seeing distant objects, should make the man see the objects placed at infinity. By identifying object distances and image distances correctly and substituting them in the lens equation power of the spectacle lenses can be obtained.

Complete step by step answer:
i) Power of spectacle lens required for reading purposes
Man’s near point = v = 0.5 m
With the help of the spectacle the man should be able to see the objects placed at 25 cm .
Therefore, u= -25 cm = 0.25 m
Using lens formula:
$
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\dfrac{1}{f} = \dfrac{1}{{( - 0.5)}} - \dfrac{1}{{( - 0.25)}} \\
\dfrac{1}{f} = P = + 2D \\
 $
Therefore, power of spectacle lenses required for reading purposes is 2 D
ii) Power of spectacle lens required for looking at distant objects
Man’s far point = v = -3 m
With the help of the spectacle the man should be able to see the objects placed at infinity.
Therefore, u = infinity
Using lens formula
$
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
  \\
$
As u = infinity and reciprocal of infinity is zero , the lens formula simplifies to:
$
\dfrac{1}{f} = \dfrac{1}{v} \\
P = \dfrac{1}{f} \\
 = \dfrac{1}{{( - 3)}} \\
P = - 0.33D \\
$
Therefore, power of spectacle lenses required for seeing distant objects is -0.33 D
Hence correct choice is option C.

Note:Generally, in these kinds of problems , the main difficulty is to identify the object and image distance. Always the near point or the far point of the defected eye should be considered for image distance. Also, if a person is suffering from myopia or short sightedness, then we can directly say that the far point of the defected eye itself is equal to focal length of the required concave lens.