Red and Violet Light through a Prism (Snell's Law)

AI Thread Summary
A ray of white light entering a 4.0-cm-thick slab of flint glass at an angle of 60° is dispersed into red and violet light, with refractive indices of 1.57 and 1.60, respectively. Using Snell's Law, the angles of refraction for red and violet light are calculated to be 33.5° and 32.8°. The distances traveled by the rays within the glass are determined to be approximately 2.65 cm for red and 2.58 cm for violet, leading to an initial separation of 0.7 mm. However, the correct separation distance is found to be 0.28 mm when accounting for the angle of incidence. The discussion emphasizes the importance of correctly interpreting the angle measurements in relation to the normal of the surface.
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Homework Statement



A ray of white light strikes the surface of a 4.0-cm-thick slab of flint glass (Θ1 = 60°). As the ray enters the glass, it is dispersed into its constituent colors.

Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. Assume nred = 1.57 and nviolet = 1.60.

Θ1 = 60°
n1 = 1 (air)
Θ2violet = ?
Θ2red = ?
nviolet = 1.60
nred =1.57
d (adj) = 4 cm
Δx = ?

Homework Equations



Snell's Law: n1*sin(Θ1) = n2*sin(Θ2)
tan(Θ) = opp/adj

The Attempt at a Solution


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n1*sin(Θ1) = n2*sin(Θ2r)
sin-1((sin 60°)/(1.57)) = Θ2r = 33.5°

n1*sin(Θ1) = n2*sin(Θ2v)
sin-1((sin 60°)/(1.60)) = Θ2v = 32.8°

tan(Θ2r) = opp/adj
dr = tan(33.5°) * 4 cm = 2.65 cm

dv = tan(32.8°) * 4 cm = 2.58 cm

Δx = 2.65 cm - 2.58 cm = 0.7 mm

It says that this answer is not correct, but I am nearly 100% sure this is how it should be done!
 
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Your work looks correct to me. Since no picture is provided, it is important to know how the angle of 60o is measured. I assume it's the angle between the incoming light ray and the normal to the surface of the glass.
 
TSny said:
Your work looks correct to me. Since no picture is provided, it is important to know how the angle of 60o is measured. I assume it's the angle between the incoming light ray and the normal to the surface of the glass.

Yeah, I'm really unsure of why it's marking me wrong. Here's the reference figure: http://i.imgur.com/5r2P1vY.jpg?1
 
Ah. Note that the 60o shown in the figure is not the angle of incidence.
 
TSny said:
Ah. Note that the 60o shown in the figure is not the angle of incidence.

Ah! Thank you so much! The correct answer came out to be 0.28 mm.
 

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