The discussion revolves around determining the critical angle of a specific plastic when it is immersed in water, given its critical angle in air. The original poster presents their calculations and seeks validation of their approach.
The discussion is ongoing, with participants providing hints and prompting the original poster to explore the refractive index of the plastic further. There is no consensus yet, as participants are still evaluating the original poster's approach and calculations.
The original poster's calculations are based on the assumption of known refractive indices for air and water, but the refractive index of the plastic is not provided, which may affect the outcome.
fiddler crab said:No. I think it should be:
<< complete solution deleted by berkeman >>
No.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?