PLEASE ANSWER PARTS A AND B! THANK YOU! PLEASE ANSWER PARTS A AND B! THANK YOU!

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PLEASE ANSWER PARTS A AND B! THANK YOU!

PLEASE ANSWER PARTS A AND B! THANK YOU! (8%) Problem 10: Consider the situation in the figure - a student diver in a pool and an instructor on the edge: outside the water Since the indices of refraction of air and water are different, the light rays coming from the diver and the instructor are refracted at the surface, changing their apparent position with respect to each other. The diver sees the instructor at an apparent angle of theta2 = 25 degree measured from the normal to the interface. Randomized Variables theta2= 25 degree Part (a) Find the height of the instructor?s head above the water in meters, noting that you will first have to calculate the angle of incidence (you can take the indices of refraction to be na = 1 for air and nw = 1.33 for water). Part (b) Find the apparent depth of the diver?s head below water as seen by the instructor in meters, assuming the diver and his image both have the same horizontal distance along the surface of the water.

Explanation / Answer

Here ,

let the angle of refraction in air is r ,

as angle of incidence , i = 25 degree

Using Snell's law

nw * sin(i) = na * sin(r)

1.33 * sin(25) = 1 * sin(r)

r = 34.2 degree

Now . let the height of instructor is h

2/h = tan(34.2)

solving for h

h = 2.94 m

the height of the instructor is 2.94 m

B)

Now, for the apprent depth of the diver head ,

happ = (2*tan(25))/tan(34.2)

happ = 1.372 m

the apprent depth of the divers head is 1.372 m

the apprent depth of head of diver is 1.504 m

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