Question

(a) Given that the angle between the ray in the water and the perpendicular to the water is 25.0ºsize 12{"25" "." 0°} {}, and using information in Figure 25.53, find the height of the instructor’s head above the water, noting that you will first have to calculate the angle of refraction. (b) Find the apparent depth of the diver’s head below water as seen by the instructor. Assume the diver and the diver's image are the same horizontal distance from the normal.

Include a picture and show work please. That way I understand it better.

2.0 m d=2.0 m Figure 25.53 A scuba diver in a pool and his trainer look at each other.

Answer 1

Key points:

In this problem it is asking for two things, first the height of the instructor's head above the water's level, and the apparent" depth of the diver's head below the water's level.

when we say, apparent then it is asking for the depth at which the instructor is observing the diver "not the depth from its actual position" To clarify more we can see that there is two images for the instructor one

which is semi-transparent and one which is solid, same goes for the diver.

The semi-transparent instructor, is what the diver's actually seeing not the instructor's real position, and the semi-transparent diver is what the instructor is seeing not the actual position of the diver.

The reason for both of them are not seeing each other at their real "actual" place, is due to the difference in the refractive index of water and air, which results in bending of the light ray, hence observing each other in different positions.

For part (a):

We need the instructor's head height above from the water level the vertical distance between his eyes and the water surface", in order to calculate his height we use trigonometry.

Where the instructor's height is one side of the triangle, and the distance between the instructor and the point at which the light ray enters the water is one side of the triangle, making a right angle triangle at the instructor,

We need to find the angle which the ray in the air makes with the surface of the water, knowing the angle and the length of the adjacent we can use the following equation to find the height of the instructor's head above the water level.

tan(theta)=opp./adj. (1)

Where, in this case the opposite is the instructor's head height, and the adjacent is given to be 2.0 m, and the angle 8 from the previous problem, was calculate that the angle which the rays makes with the normal is 34.3 degree.

thus the angle the ray makes with the surface of the water is

Binat 90°- 34.3° = 55.7°

Thus, substituting in (1) we can find the instructor's head height.

Calculations:

(a) The height of the instructor's head:

Substituting, the knowns in equation (1) we can find the opposite side of the triangle which corresponds to the height of the instructor's head above the water level, thus we have

height "opp." = 2.0 tan(55.7")

h=2.93 m

Thus, the height of the instructor's head is 2.9 m above the water level.

For part(b):

1

We notice the triangle, whose sides are the horizontal distance between the point at which the rays enters the water, and the vertical distance between the diver's head and the surface of the water.

Making a right angle triangle at the driver's lead.

If we know the angle which the apparent rays "dashed line" makes with the surface of the water, knowing the horizontal distance between the points at which the ray enters the water, and the point perpendicular to the diver's head the adjacent".

We can find the vertical distance between the apparent diver's head and the surface of the water "the opposite".

But, since we don't know the adjacent we can't find the opposite using

equation (1) as we would have two unknowns.

Thus, we must find the adjacent first in order to find the opposite, we ca do so using the triangle formed by the actual diver's position instead, where it is given that the opposite side is 2.0 m the actual depth of the diver". and since both triangles shares the same adjacent, we can use the actual triangle to find the adjacent.

From the triangle of the actual position of the diver, and knowing the angle which the ray makes with the normal in the water to be 25.0°, we can find the angle which the ray in water makes with the surface of the water, which is

Theta diver = 90°- 25.0° = 65°

And, using equation (1) we can find the adjacent which is then can be used to find the apparent depth of the diver's head.

Where, the angle the apparent ray in water "dashed line" makes with the surface of the water is the same angle as that of the angle which the ray in the air makes with the surface of the water (vertical angles),

Thus, knowing the angle and the adjacent we can use equation (1) to find the opposite which would be the apparent depth of the diver's head.

First to find the adjacent side for the triangle which the apparent diver makes, we would use the triangle which the actual position of the diver makes with point at which the ray enters.

Where, it is given that the opposite side is 2.0 m, and knowing the angle which the ray in the water makes with the normal is 25.0°, then the angle which the ray makes with the surface of the water is 65.0°, using equation (1) we can find the adjacent.

Which is the same adjacent sides which the apparent diver makes, now knowing the adjacent and that the angle the apparent rays makes with the surface of the water is 55.7°, then using equation (1) we can calculate the opposite side of the triangle, which is the apparent depth of the diver's head.

The apparent depth of the diver's head: We first find the adjacent side of the triangle, using the triangle which the

actual position of the diver makes, using equation (1) we have

adj.=2/(tan(65°)

adj=0.933 m

Knowing the adjacent we can find the opposite side in the triangle which

the apparent diver makes, which corresponds to the apparent depth of the diver's head, as follows

opp. = 0.93 tan(55.7°)

=1.363 m

Thus, the apparent depth of the diver's head is 1.4 m