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Difficulty: HardRefraction of Light, Total Internal Reflection, and Prisms

A light ray strikes the first face of a glass prism of refracting angle 30∘30^\circ at normal incidence (i=0∘i = 0^\circ). If the refractive index of the glass is 1.501.50, calculate the angle of emergence, in degrees, as the ray leaves the second face into air. (Take arcsin⁡(0.75)=48.6∘\arcsin(0.75) = 48.6^\circ)

Answer: 48.6 degrees

Answer

The angle of emergence of the light ray as it exits the prism is 48.6∘48.6^\circ.
Because the ray is incident normally at the first surface, it continues undeviated into the glass (r1=0∘r_1 = 0^\circ). By prism geometry, the angle of incidence at the second face is equal to the apex angle of the prism (r2=A=30∘r_2 = A = 30^\circ). Applying Snell's Law at the glass-air boundary gives 1.50sin⁡(30∘)=1.00sin⁡(e)1.50 \sin(30^\circ) = 1.00 \sin(e), leading to sin⁡(e)=0.75\sin(e) = 0.75, which evaluates to an emergent angle of 48.6∘48.6^\circ.

Step-by-Step Solution

1
Determine the angle of refraction at the first surface
r1=0∘r_1 = 0^\circ
Light entering a surface normally (i1=0∘i_1 = 0^\circ) passes straight through without bending.
2
Find the angle of incidence at the second surface inside the prism using prism geometry
r2=30∘r_2 = 30^\circ
For any triangular prism, the refracting angle A=r1+r2A = r_1 + r_2. Since r1=0∘r_1 = 0^\circ, r2=A=30∘r_2 = A = 30^\circ.
3
Apply Snell's Law at the second interface (glass to air)
sin⁡(e)=0.75\sin(e) = 0.75
nglasssin⁡(r2)=nairsin⁡(e)  ⟹  1.50×sin⁡(30∘)=1.00×sin⁡(e)n_{\text{glass}} \sin(r_2) = n_{\text{air}} \sin(e) \implies 1.50 \times \sin(30^\circ) = 1.00 \times \sin(e).
4
Calculate the emergent angle ee
e=48.6∘e = 48.6^\circ
Taking the inverse sine of 0.750.75 yields e=arcsin⁡(0.75)=48.6∘e = \arcsin(0.75) = 48.6^\circ.

Key Concept

Prism Geometry and Snell's Law Refraction
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Topics

Refraction of Light, Total Internal Reflection, and Prisms

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