The critical angle of a specific plastic in air is established at 37.3 degrees. When this plastic is immersed in water, the calculation for the new critical angle involves the refractive indices of air (n1 = 1.00) and water (n2 = 1.33). The formula used is n1sin(θ) = n2sin(θ), leading to a calculated critical angle of approximately 27.1 degrees. However, the discussion emphasizes the importance of understanding the relationship between the refractive index and the critical angle rather than providing complete solutions.
PREREQUISITESStudents in physics or engineering, educators teaching optics, and anyone interested in the practical applications of light behavior in different media.
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?