Solving Optics Problems: Finding Angle of Light Refraction and Object Position

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A diver shines light at a 32-degree angle from water to a glass-bottomed boat, seeking to find the angle of refraction as it exits the glass. The correct approach involves using Snell's Law, applying the indices of refraction for water, glass, and air sequentially. The calculations indicate that the angle of light refraction can be determined by setting up the equations for each transition between mediums. Additionally, a concave mirror problem is discussed, where the object distance is calculated based on the magnification of the image being twice the size of the object, leading to an object distance of 28.5 cm. The discussion emphasizes the importance of correctly applying optical formulas to solve these problems.
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A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 32 degrees relative to the normal.
If the indices of refraction of air, water, and glass are 1.0, 1.33, and 1.4 respectively, at what angle does the light leave the glass (relative to its normal)? Answer in units of Degree.

This is how I did it...(1.0)(Sin32)=(1.4)(SinX)
22.24 Degrees = Sin X

Where did I go wrong?


And my other question is:
A concave mirror has a focal length of 19 cm. What is the position of the object if the image is erect and is 2 times larger than the object? Answer in units of cm.

I have no idea what to do for this question.
 
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JassC said:
This is how I did it...(1.0)(Sin32)=(1.4)(SinX)
22.24 Degrees = Sin X

Where did I go wrong?

remember that the ray of light first is refracted from water to glass and next from glass to air.

so we get:

n_{w}\cdot sin(\theta_{w}) = n_{g}\cdot sin(\theta_{g})

then since the refracted ray has an angle \theta equal to the angle after refraction by the glass, the next equation will be:

n_{g}\cdot sin(\theta_{g}) = n_{air}\cdot sin(\theta_{air})

And thus:

n_{w}\cdot sin(\theta_{w}) = n_{air}\cdot sin(\theta_{air})
 
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Alright that helped. Thanks.

Any ideas for the other question :) ?

and there's one more...lol

http://img184.imageshack.us/img184/3126/homework10003vk2.jpg
 
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JassC said:
And my other question is:
A concave mirror has a focal length of 19 cm. What is the position of the object if the image is erect and is 2 times larger than the object? Answer in units of cm.

use the fact that the image is enlarged 2 times. Use the realtion:

M = \frac {i} {o} = \frac {L_{i}} {L_{o}} = 2.0 with M = magnification, i = imagedistance, o = objectdistance and L = length.

Now we know from this relation that the image distance i is twice as large as the object distance o --> i = 2o

Now use the lens equation and substitute i = 20 and solve the equation for o (f is already known):

\frac {1} {19} = \frac {1} {2o} + \frac {1} {o}
 
o = 28.5

Plug that in...i/0 = 2.0

Li = 57?
 
JassC said:
o = 28.5

Plug that in...i/0 = 2.0

Li = 57?

They ask for the position of the object, thus in my opinion the object distance.
 
Hmm...maybe I did the calculations right but 28.5 for object distance is wrong.
 
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