In this example, we will apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your backyard and look at one of the fish. You see it by sunlight that reflects off the fish and refracts at the water–air interface. If the light from the fish to your eye strikes the water–air interface at an angle of 60.0∘∘ to the interface, what is the angle of refraction of the ray in the air?

What happens to a light wave when it hits a surface that has a higher index of refraction? a)  It reflects with λ/2 phase shift.  b) It reflects with a λ/4 phase shift. c) It's shift is porportional to the angle of reflection.  d) It reflects with it's phase unshifted.

Two light sources of identical strength are placed 10m apart. An object is to be placed at a point P on a line l, parallel to the line joining the light sources and at a distance d meters from it (see the figure). We want to locate P on, so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. a) Find an expression for the intensity l(x) at the point P. b) If d = 5m, use graphs of l(x) and l'(x) to show that the intensity is minimized when x = 5m, that is, when P is at the midpoint of l. c) If d = 10m, show that the intensity (perhaps surprisingly) is not minimized at the midpoint. d) Somewhere between d = 5m and d = 10m there is a transitional value of d at which the point of minimal illumination abruptly changes. Estimate this value of d by graphical methods. Then find the exact value…

A ray of light crosses the boundary between some substance with n = 1.54 and air, going from the substance into air. If the angle of incidence is 29◦ what is the angle of refraction?   Calculate to 1decimal.