A person has his nearsightedness corrected by lenses of refractive power −5.00 D. Without his glasses, his near point was 12.5 cm. With his glasses on, what is his near point? Assume the lenses are very close to his eyes.
A) 13.0 cm
B) 20.0 cm
C) 33.3 cm
D) 17.5 cm
E) 7.50 cm

Question

Quizplus · Accepted Answer

The near point of a person with normal vision is 25 cm. The formula for the lens' corrective power is $P = \frac{1}{f}$, where $P$ is the power in diopters and $f$ is the focal length in meters. For a lens with a power of -5.00 D, the focal length is $f = \frac{1}{P} = -0.20$ m or -20 cm. The negative sign indicates that it's a diverging lens, suitable for correcting nearsightedness. The formula for the image distance ($v$) using the lens formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, where $u$ is the object distance (in this case, the near point without glasses, -12.5 cm, taking the sign convention into account). Rearranging for $v$, we get $v = \frac{uf}{u - f}$. Substituting $u = -12.5$ cm and $f = -20$ cm, we find $v = \frac{(-12.5)(-20)}{-12.5 + 20} = \frac{250}{7.5} = 33.3$ cm. This means, with glasses, the person's near point is effectively 33.3 cm from his eyes, allowing him to see objects clearly at a distance where he couldn't without glasses.

A Person Has His Nearsightedness Corrected by Lenses of Refractive