Thank you for visiting As light shines from air into another medium with an angle of incidence tex i 30 0 circ tex the light bends toward the normal. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
We start with Snell's law, which relates the indices of refraction to the corresponding angles of incidence and refraction:
[tex]$$
n_1 \sin \theta_1 = n_2 \sin \theta_2.
$$[/tex]
Given that the light is coming from air, we have [tex]$n_1 = 1$[/tex], the angle of incidence is [tex]$\theta_1 = 30.0^\circ$[/tex], and the angle of refraction is [tex]$\theta_2 = 22.0^\circ$[/tex]. Substituting these values into Snell's law gives:
[tex]$$
1 \cdot \sin 30.0^\circ = n_2 \sin 22.0^\circ.
$$[/tex]
We know that:
[tex]$$
\sin 30.0^\circ = 0.5.
$$[/tex]
To solve for the index of refraction [tex]$n_2$[/tex], we rearrange the equation:
[tex]$$
n_2 = \frac{\sin 30.0^\circ}{\sin 22.0^\circ} = \frac{0.5}{\sin 22.0^\circ}.
$$[/tex]
Calculating or using a known approximate value for [tex]$\sin 22.0^\circ$[/tex], we have:
[tex]$$
\sin 22.0^\circ \approx 0.3746.
$$[/tex]
Thus,
[tex]$$
n_2 \approx \frac{0.5}{0.3746} \approx 1.33.
$$[/tex]
Therefore, the index of refraction is:
[tex]$$
\boxed{1.33}.
$$[/tex]
[tex]$$
n_1 \sin \theta_1 = n_2 \sin \theta_2.
$$[/tex]
Given that the light is coming from air, we have [tex]$n_1 = 1$[/tex], the angle of incidence is [tex]$\theta_1 = 30.0^\circ$[/tex], and the angle of refraction is [tex]$\theta_2 = 22.0^\circ$[/tex]. Substituting these values into Snell's law gives:
[tex]$$
1 \cdot \sin 30.0^\circ = n_2 \sin 22.0^\circ.
$$[/tex]
We know that:
[tex]$$
\sin 30.0^\circ = 0.5.
$$[/tex]
To solve for the index of refraction [tex]$n_2$[/tex], we rearrange the equation:
[tex]$$
n_2 = \frac{\sin 30.0^\circ}{\sin 22.0^\circ} = \frac{0.5}{\sin 22.0^\circ}.
$$[/tex]
Calculating or using a known approximate value for [tex]$\sin 22.0^\circ$[/tex], we have:
[tex]$$
\sin 22.0^\circ \approx 0.3746.
$$[/tex]
Thus,
[tex]$$
n_2 \approx \frac{0.5}{0.3746} \approx 1.33.
$$[/tex]
Therefore, the index of refraction is:
[tex]$$
\boxed{1.33}.
$$[/tex]
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