Answer:
Approximately [tex]22.8^{\circ}[/tex].
Explanation:
Let
- [tex]\theta_i[/tex] be the angle of incidence, and
- [tex]\theta_r[/tex] be the angle of refraction.
By Snell's Law, [tex]n_i \, \sin \theta_i = n_r \, \sin \theta_r[/tex],
where
- [tex]n_i[/tex] is the refractive index of the medium where the light comes from, and
- [tex]n_r[/tex] is the refractive index of the medium that the light enters.
In this case,
- The light initially travels in the air. The refractive index of the air is approximately [tex]1.00[/tex] (about the same as that of vacuum.) Hence, [tex]n_i \approx 1.00[/tex].
- The light enters into glass, which (according to the question) has a refractive index of [tex]1.50[/tex]. That is: [tex]n_r = 1.50[/tex].
Also, the question states that the angle of refraction is [tex]15^\circ[/tex]. By Snell's Law,
[tex]1.00 \, \sin \theta_i = 1.50\, \sin \left(15^\circ\right)[/tex].
Solve for [tex]\theta_i[/tex], the angle of incidence.
[tex]\sin \theta_i \approx 1.50 \, \sin\left(15^\circ\right) \approx 0.388229[/tex].
[tex]\implies \theta_i \approx 22.8^\circ[/tex].
Hence, the angle of incidence is approximately [tex]22.8^{\circ}[/tex].
