Question

Yang can focus on objects 150 cm away with a relaxed eye. With full accommodation, she can focus on objects 20 cm away. After her eyesight is corrected for distance vision, what will her near point be while wearing her glasses?

Answer 1

Answer:

Explanation:

Without wearing glasses , her near point is 20 cm .

for correction of eye

u = infinity ,

v = - 150 cm

f = ?

$\frac{1}{v} -\frac{1}{u} = \frac{1}{f }$

$\frac{1}{-150} -0 = \frac{1}{f }$

f = - 150 cm

He must be wearing glass of focal length of 150 cm .

If near point be x after wearing glass ,

u = x

v = - 20 cm

f = - 150 cm

$\frac{1}{v} -\frac{1}{u} = \frac{1}{f }$

$\frac{1}{-20} -\frac{1}{x} = \frac{1}{-150 }$

$\frac{1}{-20} + \frac{1}{150 }= \frac{1}{x}$

x = 23 cm .

Answer 2

While wearing the glasses, Yang's near point will be 23.08 cm.

Given information:

Yang can focus on objects 150 cm away with a relaxed eye.

With full accommodation, she can focus on objects 20 cm away.

For correction, we have to use a concave lens such that it can make the image of a distant object at 150 cm.

So, the object distance will be infinity, and the image distance will be $v=-150$ cm.

So, the focal length of the lens can be calculated by lens formula as,

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-150}-\dfrac{1}{\infty}=\dfrac{1}{f}\\f=-150\rm\;cm$

Now, after using the lens, the image distance will be $v=-20$ cm. Let u be the near point.

The near point, after correction, can be calculated as,

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-20}-\dfrac{1}{u}=\dfrac{1}{-150}\\\dfrac{1}{u}=\dfrac{1}{150}-\dfrac{1}{20}\\u=23.08\rm\; cm$

Therefore, while wearing the glasses, Yang's near point will be 23.08 cm.

For more details, refer to the ink:

https://brainly.com/question/4419161