Q48P - University Physics with Modern Physics

Question

A -45°-45°-90° prism is immersed in water. A ray of light is incident normally on one of its shorter faces. What is the minimum index of refraction that the prism must have if this ray is to be totally reflected within the glass at the long face of the prism?

Step-by-Step Solution

Answer

The refractive index of glass at the long face of the prism is 1.88.

Step 1: Prism refractive index and Snell’s law.

We -45°-45°-90° as shown in the figure below. The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by the speed of light in the material. The prism is dropped into the water and as the ray is totally reflected, therefore, the incident angle is 90°.

Now, we can use Snell’s law in the next form

 

                                                                    $n_{glass}\sin {\theta }_{glass}=n_{water}\sin {\theta }_{water}$

 

Where ${\theta }_{glass}$ represents the critical angle which is 45° and . $n_{water}=1.33$ . Now, solve Snell’s law for $n_{glass}$ and plug the values for $n_{water},{\theta }_{glass},{\theta }_{water}$ .

                                                                          $$\begin{gathered} n_{glass}=\frac{n_{water}\sin {\theta }_{water}}{{\theta }_{glass}} \\ =\frac{\left(1.33\right)\left(\sin 90°\right)}{\sin 45°} \\ =1.88 \end{gathered}$$ 

Step 2: Figure and Conclusion.

                                                 

Hence the refractive index of glass is 1.88.