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AP Physics 2: Light Reflection and Refraction

Created by @nathanedwards
 at November 3rd 2023, 12:12:20 pm.
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#physics
#refraction
#total-internal-reflection

AP Physics 2 Exam Question

A beam of light is incident on a glass block with an angle of incidence of 30°. The glass block has an index of refraction of 1.5. The beam of light enters the glass block and undergoes both reflection and refraction.

a) Calculate the angle of refraction for the beam of light as it enters the glass block.

b) After entering the glass block, the beam of light strikes the air-glass interface at an angle of 60°. Determine the angle of reflection for the beam of light.

c) Calculate the critical angle for total internal reflection to occur at the air-glass interface.

d) If the beam of light exceeds the critical angle, what will be the fate of the light ray?

Answer and Explanation

a) To calculate the angle of refraction, we can use Snell's law, which states:

n1n2=sin⁡(angle of incidence)sin⁡(angle of refraction)\frac{{n_1}}{{n_2}} = \frac{{\sin(\text{{angle of incidence}})}}{{\sin(\text{{angle of refraction}})}}

Here, n1n_1 and n2n_2 are the indices of refraction of the initial medium and the new medium respectively.

Given that the angle of incidence θ1=30∘\theta_1 = 30^\circ and the index of refraction of the glass block n2=1.5n_2 = 1.5, we can substitute these values into Snell's law to find the angle of refraction θ2\theta_2:

11.5=sin⁡(30∘)sin⁡(θ2)\frac{{1}}{{1.5}} = \frac{{\sin(30^\circ)}}{{\sin(\theta_2)}}

Simplifying, we have:

sin⁡(θ2)=1.51⋅sin⁡(30∘)=1.5⋅0.5=0.75\sin(\theta_2) = \frac{{1.5}}{{1}} \cdot \sin(30^\circ) = 1.5 \cdot 0.5 = 0.75

To find the value of θ2\theta_2, we take the inverse sine of 0.75:

θ2=sin⁡−1(0.75)≈48.59∘\theta_2 = \sin^{-1}(0.75) \approx 48.59^\circ

Therefore, the angle of refraction for the beam of light as it enters the glass block is approximately 48.59∘48.59^\circ.

b) To determine the angle of reflection for the beam of light, we need to use the law of reflection, which states that the angle of incidence is equal to the angle of reflection.

Given that the angle of incidence at the air-glass interface is 60∘60^\circ, the angle of reflection will also be 60∘60^\circ.

Therefore, the angle of reflection for the beam of light is 60∘60^\circ.

c) The critical angle θc\theta_c is the angle of incidence that produces an angle of refraction of 90∘90^\circ (i.e. the refracted ray is parallel to the boundary surface). To calculate the critical angle, we can use Snell's law again:

n1n2=sin⁡(θc)sin⁡(90∘)\frac{{n_1}}{{n_2}} = \frac{{\sin(\theta_c)}}{{\sin(90^\circ)}}

Since sin⁡(90∘)=1\sin(90^\circ) = 1, the equation simplifies to:

n1n2=sin⁡(θc)\frac{{n_1}}{{n_2}} = \sin(\theta_c)

Substituting the given values, where n1=1n_1 = 1 (index of refraction for air) and n2=1.5n_2 = 1.5 (index of refraction for glass):

11.5=sin⁡(θc)\frac{{1}}{{1.5}} = \sin(\theta_c)

Taking the inverse sine of both sides:

θc=sin⁡−1(11.5)≈41.81∘\theta_c = \sin^{-1}\left(\frac{{1}}{{1.5}}\right) \approx 41.81^\circ

Therefore, the critical angle for total internal reflection to occur at the air-glass interface is approximately 41.81∘41.81^\circ.

d) If the beam of light exceeds the critical angle, total internal reflection occurs. This means that the light ray will not escape from the glass and will instead be reflected back into the glass. The light wave will bounce off the boundary surface, creating an effect similar to a mirror reflection, but within the medium itself.

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