A ray of blue light is incident on a triangular glass prism as shown - Scottish Highers Physics - Question 11 - 2023

Using angle A determined in part (a)(i), we can now find angle B:

$B = 180 - 60 - A$

Substituting the value of A:

$B = 180 - 60 - 84 = 36$

Thus, angle B is approximately 36 degrees.

The critical angle is defined as the angle of incidence beyond which light cannot pass through a boundary from a denser medium to a less dense medium and instead, is completely reflected back into the denser medium.

To find the critical angle ($\theta_c$), we use Snell's law:

$n_{1} \sin(\theta_c) = n_{2} \sin(90)$

where $n_{1} = 1.53$ (the refractive index of glass for blue light) and $n_{2} = 1.00$ (the refractive index of air). Rearranging gives:

$\sin(\theta_c) = \frac{n_{2}}{n_{1}} = \frac{1.00}{1.53}$

Calculating gives:

$\theta_c = \sin^{-1}(0.653) \approx 40.8$

Thus, the critical angle is approximately 40.8 degrees.

The emerging ray should be drawn at an angle greater than angle B.

Using Snell's law to find the angle of refraction:

$\sin(\theta_r) = n_{1} \sin(\theta_i)$

Where $\theta_i$ is greater than angle B, giving an angle of emergence to be about 68 degrees (approximately). Mark this angle on the diagram, showing the path of the ray as it exits the prism.