Question

For a particular liquid–air interface the critical angle (fortotal internal reflection) is $\mathbf{59}\mathbf{.}\mathbf{8}\mathbf{°}$. The index of refraction of air is1.00029.What is the index of refraction of the liquid?

Short answer

The index of refraction of the liquid is$n_1=1.16$

Step-by-step solution

Explain the given information

Given is the critical angle for total internal reflection is $59.8°$ and the refraction index of air is$1.00029$

Explain Snell’s law

Consider that a light beam passing through the two different mediums, the relation between the angle of incidence and the angle of refraction as described by Snell’s law as follows,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$ ……(1)

Find the index of refraction of the liquid

Consider that the light beam travels through high refractive index medium to low refractive index then the angle of incidence is equal to critical angle. So the angle of emergence is $90°$.

Rewrite the Equation (1) as per the above conditions,

$n_1\sin {\theta }_c=n_2\sin \left(90°\right)$

Solving the above equation for $n_1$, as follows,

$n_1=1.16$

Therefore, the index of refraction of the liquid is 1.16.