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Lilit [14]
3 years ago
10

A human eye is a convex lens. In normal eyesight, the image of an object in front of our eyes is formed exactly on the retina an

d is inverted. In farsighted and nearsighted eyesight, where is the image formed? Which lens can be use as a corrective measure in each case? You may use online or other resources to help you answer this question. Be sure to record your research sources in the References section near the end of this document.
Physics
1 answer:
Allisa [31]3 years ago
7 0

<u>In farsighted eyesight (Hypermetropia ) </u>

A person  with  hypermetropia  can  see  distant  objects  clearly but cannot see nearby objects distinctly.The light rays from a close by object are focused at a point behind  the retina  

<u> In nearsighted eyesight (Myopia ) </u>

A person with myopia can see nearby objects clearly but cannot see distant objects distinctly.the image of a distant object is formed  in  front  of  the  retina

<u>corrective measure in each case </u>

1. <u>In farsighted eyesight </u>

This defect can be corrected by using a convex lens of appropriate power

2.<u> In nearsighted eyesight </u>

A  concave  lens  of suitable power will bring the image back on to the retina and thus the defect is corrected.


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If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
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For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
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(b) Velocity after a seconds. 
y' = 11 - 3.72t 
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So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
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3 0
3 years ago
The focal length of a lens is inversely proportional to the quantity (n-1), where n is the index of refraction of the lens of th
Ainat [17]

Answer:

46.22 cm

Explanation:

The focal refraction, fr is given by

fr = \frac {c}{(1.572 -1)}  = \frac {c}{0 .572}  

The focal red light is given by

fv = \frac {c}{(1.605 - 1)} = \frac {c}{0.605}

\frac {fv}{fr} = \frac {0.572}{0 .605} = 0.945455

\frac {1}{fr} = \frac{1}{image} + \frac {1}{object} and making fr the subject we obtain

fr = \frac {image * object}{(image + object)} = \frac {24.00 * 55} {(24.0 + 55)} = 16.70886 cm

fv = 0.945455* 16.70886 cm = 15.79747 cm

image = \frac {object * f} {(object - f)} = \frac {15.79747 * 24.0}{(24.0 - 15.79747)} = 46.22222 cm

Therefore, violet image is approximately 46.22 cm

5 0
3 years ago
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