Solve Refracted Ray Problem: Find Angle of Refraction

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The refractive index of a transparent material can be calculated using the critical angle, with a value of 40.4° yielding an index of refraction of 1.54. When a light ray strikes the material from air at an angle of 37.9°, the angle of reflection is also 37.9°. The angle of refraction for this scenario is calculated to be 23.5°. The discussion also addresses a situation where the light ray exits the material, and participants clarify the application of Snell's law for determining the angle of the refracted ray. The original poster ultimately resolves their confusion with the calculations.
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I figured out the first few, I just can't get the last one!

-The refractive index of a transparent material can be determined by measuring the critical angle when the solid is in air. If θc= 40.4° what is the index of refraction of the material?
1.54

-A light ray strikes this material (from air) at an angle of 37.9° with respect to the normal of the surface. Calculate the angle of the reflected ray (in degrees).
37.9 deg

-Calculate the angle of the refracted ray (in degrees).

23.5 deg

Assume now that the light ray exits the material. It strikes the material-air boundary at an angle of 37.9° with respect to the normal. What is the angle of the refracted ray?
?

Homework Equations


Refractive index x sin (theta)

The Attempt at a Solution


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1.54 x sin (37.9) = .945999What am I doing wrong?
 
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Do you know Snell's law? What you wrote there is incomplete. Wrote the complete law.
 
I figured it out. Thanks
 
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