Snell's Law: \[ n_1 \sin\theta_i = n_2 \sin\theta_r \] where:
Light travels at different speeds in different media. In vacuum, light speed is \(c = 3 \times 10^8 \text{ m/s}\), the maximum possible speed in physics.
The refractive index \(n\) measures a medium's optical density and is defined as:
\[
n = \frac{c}{v}
\]
where \(v\) is light speed in the medium. Higher \(n\) indicates greater optical density.
What is the refractive index of a medium where light speed is \(1.5 \times 10^8 \text{ m/s}\)?
\[ n = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2 \]
Find light speed in water with refractive index \(n = 1.33\).
\[ 1.33 = \frac{3 \times 10^8}{v} \quad \Rightarrow \quad v = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8 \text{ m/s} \]
Refraction occurs at interfaces between media with different refractive indices, causing light bending.
Snell's Law:
\[
n_1 \sin\theta_i = n_2 \sin\theta_r
\]
where:
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A light ray in glass (\(n_1 = 1.52\)) strikes a glass-water interface (\(n_2 = 1.32\)) at \(25^\circ\). Find the refraction angle.
\[ 1.52 \sin 25^\circ = 1.32 \sin\theta_r \quad \Rightarrow \quad \theta_r = \arcsin\left(\frac{1.52}{1.32} \sin 25^\circ\right) \approx 29.1^\circ \]
What incidence angle in air (\(n_1 = 1\)) yields a \(30^\circ\) refraction angle in water (\(n_2 = 1.32\))?
\[ 1 \cdot \sin\theta_i = 1.32 \sin 30^\circ \quad \Rightarrow \quad \theta_i = \arcsin(1.32 \times 0.5) \approx 41.3^\circ \]
A light ray enters a glass block of width \(w\) at angle \(\theta_i\) and emerges at angle \(\theta_i'\).
a) Show \(\theta_i' = \theta_i\) and that incident/emerging rays are parallel.
b) Find lateral displacement \(d\) in terms of \(\theta_i\), \(n_1\), \(n_2\), and \(w\).
c) Calculate \(d\) for \(n_1=1\), \(n_2=1.55\), \(w=3\text{ cm}\), \(\theta_i=32^\circ\).
a) Apply Snell's Law at both interfaces:
\[
\begin{aligned}
\text{Point A: } & n_1 \sin\theta_i = n_2 \sin\theta_j \\
\text{Point B: } & n_2 \sin\theta_k = n_1 \sin\theta_i'
\end{aligned}
\]
Since normals are parallel, \(\theta_j = \theta_k\). Thus:
\[
n_1 \sin\theta_i = n_2 \sin\theta_j = n_2 \sin\theta_k = n_1 \sin\theta_i' \quad \Rightarrow \quad \theta_i' = \theta_i
\]
Hence, rays are parallel.
b) From geometry:
\[
\begin{aligned}
AB &= \frac{w}{\cos\theta_j}, \quad \alpha = \theta_i - \theta_j \\
\sin\alpha &= \frac{d}{AB} = \frac{d\cos\theta_j}{w} \\
\Rightarrow d &= \frac{w \sin(\theta_i - \theta_j)}{\cos\theta_j}
\end{aligned}
\]
where \(\theta_j = \arcsin\left(\frac{n_1}{n_2} \sin\theta_i\right)\).
c) Substitute values:
\[
\begin{aligned}
\theta_j &= \arcsin\left(\frac{1}{1.55} \sin 32^\circ\right) \approx 20.0^\circ \\
d &= \frac{3 \sin(32^\circ - 20^\circ)}{\cos 20^\circ} \approx 0.7 \text{ cm}
\end{aligned}
\]