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Ronch [10]
2 years ago
15

Approximating the eye as a single thin lens 2. 75 cmcm from the retina, find the eye's near-point distance if the smallest focal

length the eye can produce is 2. 20 cm.
Physics
1 answer:
RSB [31]2 years ago
5 0

The near-point distance of the eye is 11 cm.

<h3>What is the eye's near-point distance?</h3>

The near-point distance of the eye is the closest possible distance an object can be from the eye in order for its image to be formed on the retina. It can also be termed the closest distance of accommodation.

The near-point distance of the eye in the given scenario can be calculated using the lens formula given below:

1/f = 1/v + 1/u

where;

f = focal length

v = image distance

u = object distance

From the data provided;

f = 2.20 cm

v = 2.75 cm

u = ?

Solving for u:

1/u = 1/f - 1/v

1/u = 1/2.20 - 1/2.75

1/u = 0.91

u = 11 cm

In conclusion, the lens formula is used to determine the eye's near-point distance.

Learn more about eye's near-point distance at: brainly.com/question/16391605

#SPJ1

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snow_tiger [21]

The astronaut final velocity is 0.06 m/s to the right

Explanation:

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At the beginnig, the total momentum of the two is zero, since they are at rest:

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where

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The total momentum is conserved, so

p_i = p_f\\0=mv+MV

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3 years ago
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vovangra [49]
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aliya0001 [1]
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 
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