Question

For a light ray passing from air to water, according to Snell's law:
($n$ represents refractive index of the medium, ${\theta }_{air}$ is the angle of incidence and ${\theta }_{water}$ is the angle of refraction)

• A. $n_{air}\sin \left({\theta }_{air}\right)=n_{water}\sin \left({\theta }_{water}\right)$
• B. $n_{air}\sin \left({\theta }_{water}\right)=n_{water}\sin \left({\theta }_{air}\right)$
• C. $n_{air}\sin \left({\theta }_{air}\right)=n_{water}\sin \left({\theta }_{water}\right)=1$
• D. $n_{air}\sin \left({\theta }_{water}\right)=n_{water}\sin \left({\theta }_{air}\right)=1$

Answer 1

The correct option is A $n_{air}\sin \left({\theta }_{air}\right)=n_{water}\sin \left({\theta }_{water}\right)$

According to Snell's law, for a light ray travelling from one medium to another, the ratio of sines of angle of incidence to the angle of refraction is a constant. This constant is the refractive index of medium 2 with respect to medium one.
$\frac{n_{water}}{n_{air}}=\frac{\sin \left({\theta }_{air}\right)}{\sin ({\theta }_{water}}$
$⟹n_{air}\sin \left({\theta }_{air}\right)=n_{water}\sin \left({\theta }_{water}\right)$