Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord!

Like

Report

Numerade Educator

Like

Report

Problem 34 Medium Difficulty

For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air: (a) fused quartz, (b) polystyrene, and (c) sodium chloride.

Answer

a. $43.3^{\circ}$
b. $43.3^{\circ}$
c. $43.3^{\circ}$

Discussion

You must be signed in to discuss.
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Farnaz M.

Other Schools

Video Transcript

were asked to find e critical angle for wavelength of light when the substance is surrounded by air. So in sub one is one and brass do this for a fused court. So I called that index of refraction into F Q. And it's 1.45 a brass to do it for B uh, Polly Stream. Well, Police Trina's index of refraction instead p of 1.49 and last raster Do itfor see sodium chloride, which I call ins of S index refraction. And it's 1.544 so we can use the fact that at the critical angle, the incident angle is 90 degrees. So then the sign of the critical angle it will call they to see is going to be equal to the index of refraction of the first substance inside, one divided by the index of refraction, the seven second substance, which is in some F Q for part A. So, um, actually, we're gonna solve for theta to see here, find that this is equal to the inverse sine signed the minus one of the ratio of in some blind toe in some F cube, plugging those values into this expression, we find that this value for the critical angle is 43 0.3 degrees, which we can go ahead and boxing as our solution. For part a part B, we're using the exact same formula. But this time we're just using the instant index of refraction for, um, for the Poly Stream and soapy. So are critical angle for part B is equal to universe sign of the ratio of in someone to in some P plugging those values into this expression, we find that the critical angle for Part B is 42.2 degrees, and then, lastly, we're asked to do the exact same thing once again. But this time we're asked to do this for the medium of sodium chloride. So the critical angle is going to be equal to the inverse sine of in someone to what we called in some s for sodium chloride. We'll try to better s there playing those values. In this expression, we find that this critical angle is equal to 40.4 degrees weekend box. It is our answer to part C in the last part of our questions

University of Kansas
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Farnaz M.

Other Schools