Question

An individual is nearsighted; his near point is 12.0 cm and his far point is 42.0 cm. (a) What lens power is needed to correct his nearsightedness? diopters (b) When the lenses are in use, what is this person's near point? X Enter a number is within 10% of the correct value. This may be due to roundoff error, or intermediate results to at least four-digit accuracy to minimize roundoff error. cm

Answer 1

Given that,

Near point (N.P) = 12.0 cm

Far point (F.P) = 42.0 cm = 0.42 m

And the individual is nearsighted.

A) The power of the lens inorder to correct his nearsighted,

$\small P=\frac{1}{f}$

Where f is the Focal length of the lens. Since the object is in far distance, u = $\small \infty$ and image distance is far point which is. v = -0.42 m

$\small \frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\small \frac{1}{f}=\frac{1}{-0.42}+\frac{1}{\infty}$

Therefore, power,

$\small P= \frac{1}{f}=\frac{1}{-0.42}+0$

$\small P=-2.38D$

Power of the lens, P = -2.38 D

B) Since the person is nearsighted which only affect the far point. There will be no problems with near point. It remains same as the original value.

Therefore the near point of the person after using the lens,

N.P = 12.0 cm

Good luck. Thanks