Question

A narrow beam of white light enters a prism made of crown glass at a $\mathbf{45}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$ incident angle, as shown in Figure $\mathbf{25}\mathbf{.}\mathbf{57}$. At what angles, ${\mathit{\theta }}_{\mathbf{R}}$and ${\mathit{\theta }}_{\mathbf{V}}$, do the red ($\mathbf{660}\mathbf{}\mathbf{nm}$) and violet ($\mathbf{410}\mathbf{}\mathbf{nm}$) components of the light emerge from the prism?

Short answer

The angle at which the red light emerges from the prism is obtained: $Q_R=53.5^{°}$.

The angle at which the violet light emerges from the prism is obtained: $Q_V=55.2^{\circ }$.

Step-by-step solution

Define Geometric Optics

Geometrical optics, often known as ray optics, is an optics model that describes light propagation using rays. In geometric optics, a ray is an abstraction that can be used to approximate the routes along which light propagates under particular conditions.

Evaluating the angle of refraction

The light is incident at an angle of the value $a$. It undergoes refraction inside the prism. Then assuming the angle of refraction inside the prism be $\beta$and $\gamma$.

So, we obtain:

$$\begin{gathered} ({90}^{°}‐\beta )+({90}^{°}‐\gamma )+{60}^{°}={180}^{°} \\ {180}^{°}‐\beta ‐\gamma +{60}^{°}={180}^{°} \\ \gamma ={60}^{°}‐\beta \end{gathered}$$

For the red light, we have: .$n_2=1.512$

For the violet light, we have: $n_2=1.530$.

Evaluating the angle made by the red light

Using the Snell’s law, we obtain:

$$\begin{gathered} n_1\sin \left(\alpha \right)=n_2\sin \left(\beta \right) \\ \beta ={\sin }^{‐1}\left(\frac{n,\sin \left(\alpha \right)}{n_2}\right) \\ \beta ={\sin }^{‐1}\left(\frac{1\times \sin \left({45}^{°}\right)}{1.512}\right) \\ \beta ={\sin }^{‐1}\left(0.4677\right) \\ \beta =27.883 \end{gathered}$$

With the help of the above equation, we have:

$$\begin{gathered} \gamma ={60}^{°}-\beta \\ \gamma =60-{270883}^{°} \\ \gamma =32.{117}^{°} \end{gathered}$$

Evaluating the angle of the red light making prism

Applying the Snell’s law and we obtain:

$$\begin{gathered} n_2\sin \left(\gamma \right)=n_1\sin \left({\theta }_R\right) \\ {\theta }_R={\sin }^{‐1}\left(n_2\sin \left(\gamma \right)\right) \\ {\theta }_R={\sin }^{‐1}\left(1.512\times \sin \left(32.{117}^{°}\right)\right) \\ {\theta }_R=53.5^{°} \end{gathered}$$

Therefore, the angle at which red light emerges from the prism is:$Q_R=53.5^{\circ }$

Evaluating the angle made by violet light

Using the Snell’s law, we obtain:

$$\begin{gathered} n_1\sin \left(\alpha \right)=n_2\sin \left(\beta \right) \\ \beta ={\sin }^{‐1}\left(\frac{n_1\sin \left(\alpha \right)}{n_2}\right) \\ \beta ={\sin }^{‐1}\left(\frac{1\times \sin \left({45}^{°}\right)}{1.530}\right) \\ \beta =27.527 \end{gathered}$$

With the help of the above equation, we have:

$$\begin{gathered} \gamma ={60}^{\circ }-\beta \\ \gamma ={60}^{\circ }-27.{527}^{°} \\ \gamma =32.473 \end{gathered}$$

Evaluating the angle of the violet light making prism

Applying the Snell’s law and we obtain:

$$\begin{gathered} \left(n_2\right)\sin \left(\gamma \right)=n_1\sin \left({\theta }_v\right) \\ {\theta }_v={\sin }^{‐1}\left((n_2\right)\sin \left(\gamma \right)) \\ {\theta }_v={\sin }^{‐1}\left(1.53\times \sin \left(32.{473}^{\circ }\right)\right) \\ {\theta }_v=55.2^{\circ } \end{gathered}$$

Therefore, the angle at which violet light emerges from the prism is: $Q_V=55.2^{\circ }$.