Solve AP Physics Problem: Reflection & Refraction at a Boundary

[HashBrown]

AP PHYSICS PROBLEM...i forgot the formula, someone help me solve! This won't take lon

Light of frequency 6.0 x 1014 hz strikes a glass/air boundary at an angle of incidence, ø1. The ray is partially reflected and partially refracted at the boundary as shown. The indices are shown.

(a) Determine the value of ø2 if ø1 = 30o. Explain your answer.
(b) Determine the value of ø3 if ø1= 30o.
(c) Determine the speed of this light in the glass.
(d) Determine the wavelength of this light in the glass.
(e) What is the largest value of ø1 that will result in a refracted ray?

 

[Gokul43201]

Which formula have you forgotten ?

The formula for the angle of refraction is called Snell's Law - look it up.

What does the law of Reflection say about angles of incidence and reflection ?

How is the speed of light in a medium related to the refractive index ? This is in one of the definitions of the refractive index.

Enough said already...

 

[wais]

To solve this AP Physics problem, we will need to use Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in the first medium to the speed of light in the second medium. This can be written as n1sinø1 = n2sinø2, where n1 and n2 are the refractive indices of the two media.

(a) To determine ø2, we first need to find the refractive index of air and glass. The refractive index of air is approximately 1, while the refractive index of glass varies depending on the type of glass. For simplicity, we will assume the refractive index of glass is 1.5. Plugging these values into Snell's Law, we get:

n1sinø1 = n2sinø2
1sin30 = 1.5sinø2
0.5 = 1.5sinø2
ø2 = sin-1(0.5/1.5) = 19.47o

Therefore, the value of ø2 is approximately 19.47o.

(b) To determine ø3, we can use the fact that the sum of the angles in a triangle is 180o. Thus, we have:

ø3 = 180 - ø1 - ø2 = 180 - 30 - 19.47 = 130.53o

(c) To determine the speed of light in glass, we can use the formula v = c/n, where v is the speed of light in the medium, c is the speed of light in vacuum (3 x 108 m/s), and n is the refractive index of the medium. Thus, we have:

v = (3 x 108 m/s)/1.5 = 2 x 108 m/s

(d) To determine the wavelength of light in glass, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of light in the medium, and f is the frequency of the light. Plugging in the values, we get:

λ = (2 x 108 m/s)/(6.0 x 1014 Hz) = 3.33 x 10-7 m

(e) The largest value of ø1 that will result in a refracted ray is when the angle