1.You ride your bike on the pavement & the beach. On the pavement you can ride at 10 miles/hour using a moderate pace. Riding at the same pace on the beach you can go 6.5 miles/hr. Next you ride your bike from the pavement onto the beach. You approach this interface with a 30° angle. What angle will your bike exit onto the beach? (Hint: Use snell’s law)
Solution: In this case one can calculate and equivalent index of refraction of the sand... it is n= 10 mph / 6.5 mph = 1.538462. We'll assume that the pavement is a vacuum and has an index of 1.0
The 30° is intended to be the angle from the normal of the sand / pavement interface, so if one where to ride the bike right at this interface the angle would be 0°. If one were to ride their bike along this interface the angle would be 90°
Using Snell's law n*sinθ=n'*sinθ'. Solve for θ' = arcsin[(n/n')*sinθ] = arcsin[(1.0/1.538462)*sin30] = 18.97°
2.Light goes through a prism with a 18.6° wedge angle and an index of 2.0. How much will the light ray be deviated.
Solution. Utilize the small angle prism equation: D = A(n-1), where D = Deviated angle and A = prism angle D = 18.6°(2 - 1) = 18.6°
Editorial: I should have said the bike entered the interface at a 29.387°.... so both answers would be 18.60°:) |