AAMC #10 PS questions 49 and 50

[babx]

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Hey so my AAMC access expired but have paper versions of the test and jot down the correct answers before I lost access. But im not sure if i have the right answers for 49 and 50. I have that it is 49 D and 50 A but I don't undertand either of these answers..so just wondering if they are wrong? If they are right could someone please explain them to me thanks.

If so can someone explain why the answer for #49 is that both angles decrease? If the angle of incidence increases shouldn't the angle of reflection also decrease and the angle of refraction increase?

 

[SN2ed]

Thread moved. These types of questions belong in the Study Q&A forum.

 

[Axon hillock]

Hey so my AAMC access expired but have paper versions of the test and jot down the correct answers before I lost access. But im not sure if i have the right answers for 49 and 50. I have that it is 49 D and 50 A but I don't undertand either of these answers..so just wondering if they are wrong? If they are right could someone please explain them to me thanks.

If so can someone explain why the answer for #49 is that both angles decrease? If the angle of incidence increases shouldn't the angle of reflection also decrease and the angle of refraction increase?

I don't have the test in front of me right now but the angle of incidence is always equal to the angle of reflection. For refraction you just need to work it out using Snell's law, Ni * Sin(Ai) = Nr * Sin(Ar). I'll check out the questions when I get back and put up answers if no one gets to it.

 

[stsa84]

The correct answers are:

49) D, Both Angle 1 and Angle 2 decrease
50) B, 10^0 W/m^2

Reasoning for #49

The question asks what would happen to the reflected angle and the refracted angle if the incident angle were decreased.

The reflected angle is equal to the incident angle, so if the incident angle decreased, the reflected angle would as well.

The refracted angle is always smaller in the slow medium than in the fast medium. The ray of light in this problem is traveling from the fast medium (air) to the slow medium (some mysterious "medium"), so the ray will have a refracted angle of less than the angle of incidence. If the initial angle of incidence is decreased, then the refracted angle will decrease along with it.

Reasoning for #50

120 dB = 10log(I/I0)

12 dB = log ( I / 10^-12)

The log of what number equals 12? The log of 1.0 x 10^12. Therefore, "I" must equal 1, which is equivalent to 10^0.