The refractive index of a liquid is 1.35, what is the critical angle of the liquid? - Leaving Cert Physics - Question e - 2007

To find the critical angle ( \theta_c ) of a liquid given its refractive index ( n_g ), we can use Snell's law, expressed as:

$n_g = \frac{1}{\sin(\theta_c)}$

Here, we need to rearrange this formula to solve for the critical angle:

$\sin(\theta_c) = \frac{1}{n_g}$

Substituting the given refractive index of the liquid (1.35):

$\sin(\theta_c) = \frac{1}{1.35}$

Now, calculating the sine value:

$\sin(\theta_c) \approx 0.7407$

To find the angle, take the inverse sine:

$\theta_c = \arcsin(0.7407) \approx 47.8^{\circ}$