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Problem 59

A beam of light traveling horizontally is made of…

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Problem 60

A rainbow is produced by the reflection of sunlight by spherical drops of water in the air. $\textbf{Figure P33.60}$ shows a ray that refracts into a drop at point $A$, is reflected from the back surface of the drop at point $B$, and refracts back into the air at point $C$. The angles of incidence and refraction, $\theta_a$ and $\theta_b$ , are shown at points $A$ and $C$, and the angles of incidence and reflection, ua and ur , are shown at point $B$. (a) Show that $\theta^B_a$ = $\theta^A_b$ , $\theta^C_a$ = $\theta^A_b $, and $\theta^C_b$ = $\theta^A_a$ . (b) Show that the angle in radians between the ray before it enters the drop at $A$ and after it exits at $C$ (the total angular deflection of the ray) is $\Delta= 2\theta^A_a - 4\theta^A_b + \pi$. ($Hint$: Find the angular deflections that occur at $A$, $B$, and $C$, and add them to get $\Delta$.) (c) Use Snell's law to write $\Delta$ in terms of $A$ a and n, the refractive index of the water in the drop. (d) A rainbow will form when the angular deflection $\Delta$ is $stationary$ in the incident angle $\theta^A_a$ -that is, when $d\Delta/d\theta^A_a$ = 0. If this condition is satisfied, all the rays with incident angles close to $\theta^A_a$ will be sent back in the same direction, producing a bright zone in the sky. Let $\theta_1$ be the value of $\theta^A_a)$ for which this occurs. Show that $cos^2\theta_1$ = $1 \over 3$ ($n^2$ - 10). ($Hint$: You may find the derivative formula $d($arcsin$ u(x)22/dx$ = ${(1 - u^2)^{-1/2}}(du/dx)$ helpful.) (e) The index of refraction in water is 1.342 for violet light and 1.330 for red light. Use the results of parts (c) and (d) to find $\theta_1$ and $\Delta$ really never } for violet and red light. Do your results agree with the angles shown in Fig. 33.19d? When you view the rainbow, which color, red or violet, is higher above the horizon?

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Problem 58

Given small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light refracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass ($n$ = 1.52) and place a drop of the liquid on the top surface of the block. You shine a laser beam with wavelength 638 nm in vacuum at one side of the block and measure the largest angle of incidence $\theta_a$ for which there is total internal reflection at the interface between the glass and the liquid ($\textbf{Fig. P33.58}$). Your results are given in the table:

What is the refractive index of each liquid at this wavelength?

Answer

$$1.40$$


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Video Transcript

to solve this question, we have to relate. Teacher A With the refraction index, it's off liquids A, B and C. We begin by remembering that a total internal reflection is the same as an angle off refraction off 90 degrees. Therefore, in these interface between liquid and gas, the refraction angle is 90 degrees, so we can apply Snell's law there by using Snell's law. There we get that the refraction index off the liquid times this sign off the angle of refraction. Izzy Kohstuh refraction index off glass times the incident angle which, according to my drawing and Steve the seat, notice that our is 90 degrees on the sign off. 90 is one. So these is equals to one and we get n equals two and glass times this sign off the angle off incidence. No, As you can see from the figure, we can do this on the normals, meet at a 90 degree angle. The reform you can see that t to C and T Toby are such that they at 90 degrees. Then this sign off details see is equals to the co sign off the meat. So the index off a fraction off the liquid is the index off infraction off glass times they curl. Sign off, detail me. And now we have to relate. Victor, be with Tito A. In order to do that, we have toe applies now's law in these interface. By doing that, we get the following the index of refraction off air times. The angle off incidents teeter a is equals to the index off refraction off less times the sign off the angle of refraction in these cases, Toby. Then notice that an air is in question. Well, so we can solve this equation for the sign off. To me down the sign off the baby is equal to the sign off detail. Eight. Divided by the refraction index off glass. Using this equation, we can calculate it'll be for each off these three liquids and then using the value off it'll be, we can calculate device off the refraction index. Before doing that. Let me organize my board for liquid A. The sign off it'll be is equals to the sign off 52 degrees divided by 1.50 truth, and these results in an angle theta be off approximately 31 point to true six degrees. Then the refractive index off liquid A is given by 1.50 truth times. They have signed off 31.2 to 6. These results seen approximately one 0.3. Now for a liquid B, we have the sign off. The Derby equals to the sign off 44.3 degrees, divided by 1.50. Choose these results in an angle. Tita, be off approximately 27 points. 35 38 degrees. Then they refractive index off. With me, it's 1.53 times They cursed Sign off 27.35 38 which results in approximately 1.35 now for liquid. See, we have sign off. It'll be being equal to the sign off 36.3, divided by 1.50 troops. The results in Tita be off approximately 22 0.922 degrees show. Their refractive index off liquid see is 1.52 times the co sign off 22.922. The results in an attractive index off approximately one 0.4 leases down. Start to this question

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