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Consider a light ray traveling between air and a diamond cut in the shape shown in Figure P22.42. (a) Find the critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and the outside air. (b) Consider the light ray incident normally on the top surface of the diamond as shown in Figure P22.42. Show that the light traveling toward point $P$ in the diamond is totally reflected. (c) If the diamond is immersed in water, find the critical angle at the diamond–water interface. (d) When the diamond is immersed in water, does the light ray entering the top surface in Figure P22.42 undergo total internal reflection at $P$ ? Explain. (e) If the light ray entering the diamond remains vertical as shown in Figure P22.42, which way should the diamond in the water be rotated about an axis perpendicular to the page through $O$ so that light will exit the diamond at $P$ ? (f ) At what angle of rotation in part (e) will light first exit the diamond at point $P$ ?

a. $24.4^{\circ}$

b. $\text { Because the incident angle }(35.0 \text { degrees) is greater than the critical angle }(24.4 \text { degrees), therefore the light will totally reflected}$

c. $33.4^{\circ}$

d. $\text { Because the incident angle }(35.0 \text { degrees) is greater than the critical angle }(33.4 \text { degrees), therefore the light will totally reflected}$

e. $\begin{array}{l}{\text { The angle of incident must be less than the critical angle }(33.4 \text { degrees). Therefore we should rotate the diamond }} \\ {\text { clockwise to decrease the incident angle. }}\end{array}$

f. no answer available

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four Part of our question, it says, to find the critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and the outside air. Okay, so I have the incident or the index of refraction for Aeryn Sunday as 1.0, the index of refraction for diamond in sub d as 2.419 and the index of refraction for water, which would use later as 1.333 So the critical angle is when the, uh, the index every fraction is at least equal to 90 degrees. So using Snell's law, we find that the critical angle then is defined as they to see equal to the inverse sine of the medium that it's traveling, um, into inside a divided by in sub d. This is inverse I in here, so sign of the minus one. We find then that the angle is equal to 24 0.4 degrees, which can be boxed in as their solution. For part a part, beef says, consider the light ray incident, normally on the top surface of the diamond, has shown a figure peed 22.42 and show that the light traveling between point P in the diamond is totally reflected. Okay, so the angle of incident in the, uh in the images 35 degrees and the critical angle they to see we just found his 24.4 degrees here. The critical angle is less than the angle of incident. And when the angle of incidence is greater than the critical angle, the light undergoes total internal reflection. Therefore, the light undergo sternal total internal reflection. So for part B, we can say that 35 degrees is greater than the critical angle day to see, and then we'll type this out. This total internal reflection. That's our solution for part B. Part C says if the diamond is immersed in water, find the critical angle the diamond water interface. So again, we're going to use the exact same formula that we use for part C except now are index of refraction zehr gonna be diamond in water. So this is inverse sine of the ratio of the second medium that it travels into, which is the water divided by the index of refraction of the medium. It's leaving, which is the diamond and sub d Let's see here. Yeah. Okay, So calculating this, we find that this is equal to 33 0.4 degrees inequity boxed in as their solution for parts seat, part D Ask did ask us to do something similar that we did The part be, um, in which it wants us to figure out our wants us to figure out if the ray is totally internally reflected. But again, since the angle of incidence is 35 degrees and that's a greater than the critical angle we just found. It is totally internally reflected, so we can say 35 degrees is greater than the critical angle. Thus totally internal reflection. It's a total making boxes in his air solution for party. Okay, party says, uh, if the rate entering the diamond remains vertical as shown in p 24.42 which would, uh which way should the diamond and the water be rotated about an access perpendicular to the page through. Oh, so the light will exit the diamond at p. Okay, so the light ray exits from the diamond when the angle of incident is less than the critical angle in order to decrease the angle of incident. The diamond is rotated in a clockwise direction and the diamond is rotated in a clockwise direction such that the angle of incidence is less than the critical angle. And that's what we're gonna go ahead and type out as our answer. So the diamond is rotated clockwise so that the angle of incidence is less No is less then, huh? Critical angle, which we've been calling data see, So we can just go ahead and say Okay to see that's your answer for Part E. And then lastly, go ahead box that in to indicate that first and then lastly, Part F wants us to figure out how much we need to rotate that eso the angle that we want to rotate It will call it data. Actually, let's call the angle we want to rotate. Fine bye is equal to 35 degrees, which is the angle of incident minus the critical angle. They had a Sikh, and this is the critical angle that we used in part See here. So this is the 33.4 degrees. So we find this is equal to one 0.6 year degrees. This would be clockwise. Oh, let's go back to that page. This would be in a clockwise direction, so it's type out clockwise connected Boxing is their solution for part F in the final answer to the question.