a-submerged-scuba-diver-looks-un-toward-the-calm-surface-of-a-freshwater-lake-and-notes-that-the-sun

Question

A submerged scuba diver looks un toward the calm surface of a freshwater lake and notes that the Sun appears to be $35^{\circ}$ from the
vertical. The diver's friend is standing on the shore of the lake. At
what angle above the horizon does the friend see the sun?

William Affel · Accepted Answer

all right, so we have a scuba diver for this problem. It&#x27;s underneath the water underneath the surface of the water, and he sees the sun an angle of 35 degrees below the vertical that we&#x27;ve placed above his head. And we&#x27;re wondering if his buddies on the shore I would angle above the horizontal or both. Horizon, does his buddy see the sun? So we have to realize first is that the scuba diver isn&#x27;t actually seeing a ray of light that&#x27;s coming directly from the sun. So he&#x27;s not directly seeing the sun. He&#x27;s seeing a ray of light that comes from the sun and then refracts through the water. So first we have to figure out what angles are going on in this refraction in order to figure out what angle but the rise in the sonnets. So we&#x27;re gonna draw our normal line at the point of a fraction. And by having these two vertical lines, we can clearly see that this angle Fada is congruent to this angle, which is gonna be our angle of a fraction as the late great comes from the sun through the water. So you know this angle here is 35 degrees, and now we can set up Snell&#x27;s law. We know that here the index of refraction of air it was about one in the index of refraction of water is 1.33 but you can find in your textbook let&#x27;s set up Snell&#x27;s law. We can say that. And one or the end of air times a sign a fatal one. Just the angle that we&#x27;re looking for. It&#x27;s gonna be the angle of incidence here. It&#x27;s gonna be equal to end, too. The index of refraction of water How was the sine of the angle of refraction? 35. Now, if you plug in our indexes over fraction and solve for theta one, we&#x27;ll find that state a one is gonna be equal to the universe. Sign of 1.33 or index of refraction of water and to time is the sign of 35 or angle over fraction all the vital about one, and this should give you an angle of 49 0.8 degrees. But that&#x27;s not the angle that our buddy in the shore is seeing. That&#x27;s the angle from the vertical being with everybody on the shore is seeing is the angle from the horizontal, this angle phi. But these two angles of his compliments with one another. So you say that 90 minus 49.8 gives you an answer about 40 degrees. That&#x27;s what our body sees over here. He sees an angle of 40 degrees above the horizon.

Aatish Gupta · Answer

So for this problem, we have a diver, and he&#x27;s underwater. And we have some light that is coming in at some angle. Theta one and is being observed had an angle of Okay, So, um yes. So it&#x27;s being observed to be coming in at an angle off 45 degrees below the horizontal. So we want to know s So that means that this is also 45 degrees. We want to know what is the actual data? What is the actual value of data One. So we know that. And one sign so you don&#x27;t want is equal to end to sine theta two. You know that the first medium is air, so this is just equal to one. And the end to of water is 1.333 So our theta one is going to be equal to the inverse sign of, um, 1.333 times sign of 45 degrees. So this will give us an angle off. Okay, 70 0.48 degrees. Oh, um, so that means that, uh, the angle Sorry. This is not data one. Uh, this This is state a one. Oh. So we want to know what is the actual elevation of the sun above the horizon. So that&#x27;s gonna be 90 minus state one. So we&#x27;re gonna call the state of too. So 90 minus data one is equal to eight. It too, which will give us an angle off. We&#x27;ll call the state of three. To make things less confusing is 19.5 degrees. That is our final answer.

Zulfiqar Ali · Answer

according to Snell&#x27;s law. And one sign Peter one is equal to and do sign pleaded to here. And one is that effective index Full material with incident light and into is the index of refraction for radiantly refracted light on Peter one is the angle of incidence and Peter to east angle over a fraction. Well, the angle is measured from the normal, A line drawn perpendicular to the surface at the point where the incident Reese strikes the surface, so in one is equal to one. Uh, this is for you, hear, And and to is equal to 1.333 And this is for water right now. Using. And one sign plead Awan is equal to into sign pleaded to by substituting value One motive. My guy signed Peter one, um is equal to 1.333 multiplied by sine of 45 degrees 45 degrees. So here pleaded to is equal to sign of 45 degrees. Right, And this is equal to zero point 94 to 57 Axel plead. A one is equal to 70. Find four degrees. This is from the working gold from the article now uh, from the horizontal. The angle ease. So from strong what it&#x27;s on tour. Well, we have, like two degree minus 70 point for eight degrees, which is equal to 19.51 degrees.

Samantha Baker · Answer

Okay, So in this scenario, you have a lake and you have a diver who sees a ray in the transmitted ray angle is given to us as 42 degrees. We know the angle of reflection. I&#x27;m sorry. Index of refraction and air would be 1.0 And the index of refraction of the transmitted ray and with water is 1.33 And we want to know the angle, um, of the ray from the sun as it hits the lake. So the angle of incidents is what we&#x27;re trying to sell foreign. We&#x27;re going to use Snell&#x27;s law to solve this. So the angle of incidents times sign. I&#x27;m sorry. Let&#x27;s back up here. Got ahead of myself. Okay? The index ever fraction for the Air Times Signed data for the angle of incidence is equal to the index of refraction for the transmitted wave. Times sign data of the transmitted wave. We can plug in what we know. Sign plea that I 1.33 sign 42. And now we need to solve for angle of incidents which equals the inverse sign of 1.33 sign 42 divided by 1.0 and that tells us the sun, um, hits the lake at an angle of incidence of 62.9 degrees.

Adnan Gill · Answer

So we have an interface between water and air, and a ring off light meets the normal at some angle. Taylor one, which is unknown. But once it goes into the air and water, the two is equal to 28 degrees. So the refractive index off air What 10 1 is equal to one and the refractive, and that&#x27;s a water and two is equal to one point beneath everything. Snell&#x27;s losses and one sign off fatal one is equal to end to sign off there, too. So one times side off Failla one is equal to 1.333 times. Sign off 28 degrees. Sophia one is equal to signing worse off 1.333 Sign off 28 degrees that comes out to be equal. Do 38.7 degrees.

Suzanne W. · Answer

An underwater scooper scuba diver sees the sun at an apparent angle of 45° from the vertical. What is the actual direction of the sun? So we&#x27;ll start by using snails law in one sign of fate of one equals into kind of faded too. So we&#x27;re looking for the angle of incidence. So into sign a faded too. Over in one. So one is Sorry into is water. So 1.33 And one is, there&#x27;s just one sign of three. Data to data to is 45°.. So once this is the inverse sine of all of this. So the inverse sine. This gives us 70.5°.. But this is not the angle that is the direction of the Sun. This is the angle of reflection, very refraction. This angle We want the angle that the sun actually is, so it&#x27;s 90- that. Do we have 19.5° above the horizontal because the sun is coming in that way.

Problem

A pond with a total depth (ice + water) of 3.25 $…

Problem 50 Medium Difficulty

A submerged scuba diver looks un toward the calm surface of a freshwater lake and notes that the Sun appears to be $35^{\circ}$ from the
vertical. The diver's friend is standing on the shore of the lake. At
what angle above the horizon does the friend see the sun?

Answer

40 $\mathrm{deg}$

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