What's the Critical Angle of Plastic in Water?

AI Thread Summary
The critical angle of a plastic in air is 37.3 degrees, and the discussion revolves around finding its critical angle when immersed in water. The initial calculation used the formula n1sin(theta) = n2sin(theta), yielding an incorrect result of 27.1 degrees. Participants emphasize the importance of understanding the relationship between refractive indices and critical angles. To solve the problem correctly, one must first determine the refractive index of the plastic. The conversation highlights the need for guidance rather than complete solutions in homework discussions.
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Homework Statement



The critical angle of a certain plastic in air is 37.3 degrees. Wats the critical angle of the same plastic if its immersed in water?

Homework Equations



n1sin( theta)= n2sin(theta)

The Attempt at a Solution


Nair= 1.00
1.00sin37.3/1.33 = sin(theta)
sin(theta)= .455
inverse sin (theta) = 27.1 degrees


Is that right?
 
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No. I think it should be:
<< complete solution deleted by berkeman >>
 
Last edited by a moderator:
fiddler crab said:
No. I think it should be:
<< complete solution deleted by berkeman >>

Please do not post complete solutions to homework/coursework questions. Instead, provide hints and tutorial advice. The original poster (OP) must do the buld of the work. The Rules link at the top of the page explains how we treat homework here on the PF.
 
xswtxoj said:

Homework Statement



The critical angle of a certain plastic in air is 37.3 degrees. Wats the critical angle of the same plastic if its immersed in water?

Homework Equations



n1sin( theta)= n2sin(theta)

The Attempt at a Solution


Nair= 1.00
1.00sin37.3/1.33 = sin(theta)
sin(theta)= .455
inverse sin (theta) = 27.1 degrees


Is that right?
No.
What is the relation between the refractive index and the critical angle?
Find the refractive index of the plastic.
 

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