Please show in detail how to solve. Thank you! HW07 3 webassign.ne /web/Student/

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Please show in detail how to solve. Thank you!

HW07 3 webassign.ne /web/Student/Assignment-Responses/submit?dep 9408009#Q2 c Google shown from above in the figure below is from side A of the container and enters the water at one corner of a rectangular box filled with water. A laser beam starts 8 cm position X. You can ignore the thin walls of the container Side B Water (top view) Side A (a) If x 17 cm, does the laser beam refract back into the air through side B r reflect from side B back into the water? C. refract back into air O reflect from side B back in the water Determine the angle of refraction or reflection. 47 (b) Repeat part a for x 11 cm. O refract back into air reflect from side B back in the water Determine the angle of refraction or reflection. (c) Find the minimum value of x for which the laser beam passes through side B and emerges into the air. Cm Submit Answer Save Progress Practice Another Version

Explanation / Answer

Part A)

We need the angle of incidence

tan(angle) = 8/17

angle = 25.2o

Angle of incidence - 90 - 25.2 = 64.8o

Then by Snells law

n1sin(i) = n2sin(r)

1(sin 64.8) = 1.33(sin r)

r = 42.87o

That means it strike the other side at 90 - 42.87 = 47.130

So, by Snell's Law again

1.33(sin 47.13) = 1(sin r)

r = 77.1o

Part B)

tan(angle) = 8/11

angle = 36o degrees

Then the angle of incidence is 90 - 36 = 54o

By Snell's Law 1(sin 54) = 1.33(sin r)

r = 37.45o

That means it will strike and reflect off the back wall at 90 - 37.45 = 52.5o

Part C)

The critical angle is from

sin(i) = 1/1.33

i = 48.75o

The refracted angle is 90 - 48.75 = 41.25o

Then from Snells Law

1(sin i) = 1.33(sin 41.25)

i = 61.27o

The angle from the beam would need to be 90 - 61.27 = 28.7o

tan(28.7) = 8/x

x = 14.6 cm

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