1) A student has built a 17-cm-long pinhole camera for a science fair project. S

Question

1) A student has built a 17-cm-long pinhole camera for a science fair project. She wants to photograph her 170-cm-tall friend and have the image on the film be 5.3 cm high. How far should the front of the camera be from her friend?

2) A 1.0-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 50 from the normal. What is the ray's direction of travel in the glass?

3) An underwater diver sees the sun 57 above horizontal. How high is the sun above the horizon to a fisherman in a boat above the diver?

4) A 4.4 m -wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 20 above the horizon. How deep is the pool?

Explanation / Answer

1)   Pinhole camera dimensions are strictly based on similar triangles.
do/di = ho/hi
do/17 cm = 1.7 m/5.3 cm
do = 0.17*1.7/0.053 = 5.453 m

2) sin 50 / 1.5 = sin r =====> r = 30.71

The water doesn't do anything. Any bending as the ray enters the water is balanced by the ray leaving the water. Then there is the effect of the glass. take µ(glass) = 1.5 and thats all you need do.

3) The angles in the equation are not the angles relative to the horizon but are relative to the "normal" (The line that is perpendicular to the surface)
The angle under the water is 90-57= 33
n1 for water is 1.33, n2 for air is 1 Which you seem to understand.
(1.33)(sin (33)) = (1.00)(sin (x2))
Solving for x2= 46.42 degrees. But remember that that is the angle relative to the normal so you have to subtract it from 90 to get the angle relative to the horizon.
and you get 43.58 degrees.

4) Draw a right triangle representing the wall and base of a pool
The base represents the width and known to be 4.4 meters.
The opposite angle from the base is 90-20 = 70 degrees when the shadow just covers the base.

Tan 70 = 4.4 / side height
side height = 4.4 / Tan 70 = 1.6 meters

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