1. Probably not. If any image is visible it will be diffuse. A surface that feels rough is too irregular to create a coherent reflection of light waves–the lumps and bumps are simply too large.
3. 63 degrees
4. This can be done with a diagram and the premise of alternate interior angles since the opposing sides of the square block are themselves parallel.
5. You have two interfaces to consider. Air –>Glass, then Glass–>Water. The angle of refraction for the first interface becomes the angle of incidence for the second interface.
Theta-refracted-1 = 14.3 degrees = Theta-incident-2
Glass–>Water, theta refracted 2 = 16.4, which makes sense that the final bend is AWAY from normal since water has a slightly smaller index of refraction compared to glass.
6. Your eyes cannot comprehend the bend (refraction) of light cause by the change in medium from water to air, so to your eyes, sunlight reflecting off the fish comes straight out of the water towards you. However, you know that since water is more optically dense than air, the light reflecting off the fish actually bends AWAY from normal as it leaves the water. This fact means that the only possible path the light takes is the one closest to the dock. This also means that you see the fish slightly higher and slightly farther away than it actually is (trace that ray back into the water).
7. Total internal reflection. Theta-critical for water–>glass is 61 degrees
8. Do Snell’s law, solve for n2. n2 = 2.09 (this may not be accurate to real life, I think I made this problem up)
9. Assuming REAL images, image distance is positive, focal length is 12cm, assuming VIRTUAL image (negative i), f=60cm
10. Let’s go with a real image since the question does not specify. M = hi/ho = -i/o. If the image is real the image distance is positive so M = -30/20 = -1.5. Remember the negative means inverted, which makes sense since real images are always upside down.
b. 3.3 m
c. Since the problem in #9 does not specify what type of lens it is, nor whether the image is real or virtual, no, there is no way to tell. However, as indicated in part A, the negative magnification is correct on the assumption of a real image. If we went with the virtual image case then the magnification would turn out postive, which makes sense since virtual images are always upright/erect.
***Disclaimer: I did these quickly on my iphone calc while riding in a car. I hope all the math is correct, but if you just cannot figure out why I’m getting my answers it is possible that my fat fingers hit cos instead of sin once or twice.