The diagram shows a ray of light as it leaves a rectangular block of glass - Leaving Cert Physics - Question b - 2012

To have the ray of light emerge parallel to the side of the glass block, the angle of incidence when entering the block must match the critical angle, $\theta_c$. The critical angle is given by:

$\theta_c = \sin^{-1}\left(\frac{n_1}{n_2}\right)$

Substituting the values:

$\theta_c = \sin^{-1}\left(\frac{1.0}{1.5}\right) \approx 41.81°$.

Thus, for the light to emerge parallel, we must achieve:

$θ\ = 41.81°$.

The speed of light in a medium is calculated using the formula:

$c_g = \frac{c}{n}$

Where:

• $c \approx 3 \times 10^8 \text{ m s}^{-1}$ (speed of light in vacuum),
• $n = 1.5$ (refractive index of glass).

Substituting the values:

$c_g = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m s}^{-1}$.