A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.

Required:
a. If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
b. What does this tell you about the shape of the nearsighted eye?

1. This distance is greater than for the normal eye.
2. This distance is shorter than for the normal eye.

Answers

Answer 1

Answer:

a) The distance from the cornea vertex to the retina is 2.37 cm

b) This distance is shorter than for the normal eye.

Explanation:

a) Let refractive index of air,

n(air) = x = 1

Let refractive index of lens,

n(lens) = y = 1.4

Object distance, s = 36 cm

Radius of curvature, R = 0.65 cm

The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.

Image distance, s' = ?

(x/s) + (y/s') = (y-x)/R

(1/36) + (1.4/s') = (1.4 - 1)/0.65

1.4/s' = 0.62 - 0.028

1.4/s' = 0.592

s' = 1.4/0.592

s' = 2.37 cm

Distance from the cornea vertex to the retina is 2.37 cm

(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.


Related Questions

A soccer ball is released from rest at the top of a grassy incline. After 2.2 seconds, the ball travels 22 meters. One second later, the ball reaches the bottom of the incline. (Assume that the acceleration was constant.) How long was the incline

Answers

Answer:

x = 46.54m

Explanation:

In order to find the length of the incline you use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]      (1)

vo: initial speed of the soccer ball = 0 m/s

t: time

a: acceleration

You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):

[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex]      (2)

Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:

[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex]       (3)

The length of the incline is 46.54 m

The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax

Answers

Answer:

The distance is  [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

        The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is  [tex]\frac{1}{10}[/tex] of  an arcsec

  This implies that  0.2 grid unit is  [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           [tex]d = \frac{1}{k}[/tex]

substituting values

           [tex]d = \frac{1}{0.02}[/tex]

           [tex]d = 50 \ parsec[/tex]

Note  [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So  [tex]d = 50 * 3.08 *10^{13}[/tex]

     [tex]d = 1.5 *10^{15} \ km[/tex]

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