If a diamond has a refractive index of 2.42, what is the speed of light in the diamond? - Leaving Cert Physics - Question e - 2013

To find the speed of light in the diamond (c₂), we rearrange the equation as follows:

$c_2 = \frac{c_1}{n}$

We know that the speed of light in a vacuum (c₁) is approximately $3 \times 10^8 \, \text{m s}^{-1}$. Substituting this and the refractive index (n = 2.42) into the equation gives:

$c_2 = \frac{3 \times 10^8 \, \text{m s}^{-1}}{2.42}$

Calculating this yields:

$c_2 \approx 1.24 \times 10^8 \, \text{m s}^{-1}$

Thus, the speed of light in the diamond is approximately $1.24 \times 10^8 \, \text{m s}^{-1}$.