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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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