The critical angle for total internal reflection in Lucite is $41$ degrees. How do you find Lucite’s index of refraction?

Hint:In the above question, we are given the critical angle of Lucite and we have to find the index of refraction. So, use Snell’s law and then apply the condition for total internal reflection.
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\sin {\theta _r}}}{{\sin {\theta _i}}}$ and $\sin {\theta _c} = \dfrac{{{n_2}}}{{{n_1}}}$ where ${\theta _c}$ is the critical angle.

Complete step by step answer:
We are given that critical angle $\left( {{\theta _c}} \right) = 41^\circ $ and we have to find the index of refraction=?
Then solving the question by using the Snell’s law,
${n_1}\sin {\theta _i} = {n_2}\sin {\theta _r}$ where ${n_1}$and ${n_2}$ are the index of refraction of the incident plane and the refraction plane respectively whereas $\sin {\theta _i}$ and $\sin {\theta _r}$ are the angles for incident and the refraction lines made with the normal.

In the question, we are given that the Lucite is having total internal reflection which means that in which ray of light is facing total reflection into another medium. The condition for total internal reflection is ${\theta _r} = 90^\circ $ and ${n_1} > {n_2}$
And we know the index of refraction for air is one. ${n_2} = 1$
Hence, ${n_1}\sin {\theta _c} = \left( 1 \right)\sin \left( {90^\circ } \right)$
${n_1} = \dfrac{1}{{\sin \left( {41^\circ } \right)}} = 1.524$

Hence, the index of refraction for Lucite is $1.524$.

Note:The value for sine is can be calculated with the help of either a scientific calculator or by a logarithm table. But in the final exams, scientific calculators are not allowed, so make sure to learn how to use logarithm tables.