AP Physics 2- 13.3 Refraction- Exam Style questions - FRQs- New Syllabus
Refraction AP Physics 2 MCQ
Unit 13: Geometric and Physical Optics
Weightage : 15–18%
Question
Sugar is then added to the water, resulting in a mixture that has a different index of refraction than water. A student considers two models, Model \( A \) and Model \( B \), for how the sugar mixes with water. The models are shown in Figure \( 2 \).
| Model \( A \) | Model \( B \) | ||
|---|---|---|---|
| \( \theta_i \) | Incident angle of the beam in air | \( \theta_i \) | Incident angle of the beam in air |
| \( \theta_1 \) | Angle the beam makes with the normal in the mixture in Model \( A \) | \( \theta_2 \) | Angle the beam makes with the normal in the top layer in Model \( B \) |
| \( \theta_3 \) | Angle the beam makes with the normal in the middle layer in Model \( B \) | ||
| \( \theta_4 \) | Angle the beam makes with the normal in the bottom layer in Model \( B \) | ||
Most-appropriate topic codes (AP Physics 2):
• Topic \( 13.3 \) — Refraction (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \), Part \( \mathrm{(d)} \))
▶️ Answer/Explanation
(a)
The beam bends toward the normal as it enters the water because \( n_w > n_a \). It then reflects from the horizontal mirror so that the reflected path is symmetric about the vertical line through the point of incidence on the mirror. Finally, when it exits from water to air, it bends away from the normal.
(b)
The observed wavelength decreases as the beam travels from air into the mixture.
In a medium with larger index of refraction, light travels more slowly, while the frequency stays the same at the boundary. Since \( \lambda = \dfrac{v}{f} \), a smaller speed means a smaller wavelength.
(c)(i)
Apply Snell’s law across the boundaries.
From air into the top layer:
\( n_a \sin\theta_i = n_w \sin\theta_2 \)
From the top layer into the middle layer:
\( n_w \sin\theta_2 = n_m \sin\theta_3 \)
From the middle layer into the bottom layer:
\( n_m \sin\theta_3 = n_b \sin\theta_4 \)
Combining these gives
\( n_a \sin\theta_i = n_b \sin\theta_4 \)
so
\( \sin\theta_4 = \dfrac{n_a}{n_b}\sin\theta_i \)
Therefore,
\( \boxed{\theta_4 = \sin^{-1}\!\left(\dfrac{n_a}{n_b}\sin\theta_i\right)} \)
(c)(ii)
The ranking is
\( \boxed{2\ \theta_1 \qquad 1\ \theta_2 \qquad 2\ \theta_3 \qquad 3\ \theta_4} \)
Explanation:
In Model \( B \), the top layer is only water, so \( \theta_2 \) is the angle in the lowest-index liquid region and is therefore the largest.
In Model \( A \), the whole liquid is the mixture with index \( n_m \), and in Model \( B \), the middle layer also has index \( n_m \), so \( \theta_1 = \theta_3 \).
The bottom layer in Model \( B \) has the greatest index of refraction \( n_b \), so the light bends most toward the normal there, making \( \theta_4 \) the smallest.
Since larger index of refraction means smaller angle to the normal for the same incident ray, the order is \( \theta_2 > \theta_1 = \theta_3 > \theta_4 \).
(d)
Both horizontal distances are less than the original one:
\( \boxed{d_A < d_w \qquad \text{and} \qquad d_B < d_w} \)
In both models \( A \) and \( B \), the light travels through regions whose indices of refraction are greater than that of the original pure water case for at least part of the trip. That makes the beam bend more toward the normal than in the original tank.
A smaller angle with the normal means a smaller horizontal component of the path through the tank, so the beam travels a shorter horizontal distance before exiting. Therefore both \( d_A \) and \( d_B \) are less than \( d_w \).