Chapter 23, Problem 49QAP

To determine

(a)

Calculate the range of the index of refraction of the material for visible light.

Answer to Problem 49QAP

Refractive index for visible light ranges from $\text{1}\text{.50 to 1}\text{.57}$

Explanation of Solution

Given:

For a certain optical medium the speed of light varies from a low value of $1.9\times {10}^{8}m/s$ for violet light to a high value of $\text{ 2}\text{.00}\times {10}^{8}m/s$ for red light.

Concept Used:

The relationship between refractive index and the speed of light is given as follows:

  $n=\frac{c}{v}$

Where

  $\begin{array}{l}n:\text{Refractive index}\\ c:\text{Speed of light in vaccuum = }3.00\times {10}^{8}m/s\\ v:\text{Speed of light in medium}\end{array}$

Calculation:

As per the given problem,

Speed of violet light, $v_{\text{violet}}=1.9\times {10}^{8}m/s$

Speed of red light, $v_{\text{red}}=\text{ 2}\text{.00}\times {10}^{8}m/s$

Refractive index corresponding to violet light $n_{\text{violet}}$is calculated as follows:

  $\begin{array}{l}{n}_{\text{violet}}=\frac{c}{v_{\text{violet}}}\\ n_{\text{violet}}=\frac{\text{3}\text{.00}\times {10}^8}{1.9\times {10}^8}=1.57\end{array}$

Refractive index corresponding to red light $n_{red}$is calculated as follows:

  $\begin{array}{l}{n}_{\text{red}}=\frac{c}{v_{\text{red}}}\\ n_{\text{red}}=\frac{\text{3}\text{.00}\times {10}^8}{2.00\times {10}^8}=1.50\end{array}$

Conclusion:

Refractive index for visible light ranges from $\text{1}\text{.50 to 1}\text{.57}$

To determine

(b)

A white light is incident on the medium from air, making an angle of 30.0° with the normal. Compare the angles of refraction for violet light and red light.

Answer to Problem 49QAP

Angle of refraction for violet light= $18.54°$

Angle of refraction for red light= $19.45°$

Explanation of Solution

Given:

For a certain optical medium the speed of light varies from a low value of $1.9\times {10}^{8}m/s$ for violet light to a high value of $\text{ 2}\text{.00}\times {10}^{8}m/s$ for red light.

Formula used:

Snell's law or the law of refraction is expressed as

  $n_1\sin {\theta }_i=n_2\sin {\theta }_r$

Where

  $\begin{array}{l}{n}_1:\text{Refractive index of medium1}\\ n_2:\text{Refractive index of medium2}\\ {\theta }_i:\text{Angle of incidence}\\ {\theta }_r:\text{Angle of refraction}\end{array}$

  $\sin {\theta }_r=\frac{n_1\sin {\theta }_i}{n_2}$

Calculation:

As per the given problem, medium 1 is air and medium 2 is the optical medium.

  $n_1=1$

Angle of incidence, ${\theta }_i=30°$

Refractive index of violet light, $n_{\text{violet}}=1.57$

Substituting the values of $n_{\text{violet}}$, $n_1$ and ${\theta }_i$ in $\sin {\theta }_r=\frac{n_1\sin {\theta }_i}{n_2}$, we get

  $\begin{array}{l}\sin {\theta }_{r(violet)}=\frac{n_1\sin {\theta }_i}{n_{violet}}\\ \sin {\theta }_{r(violet)}=\frac{1\times \sin 30}{1.57}\\ \sin {\theta }_{r(violet)}=0.318\\ \Rightarrow {\theta }_{r(violet)}={\sin }^{-1}(0.318)\\ {\theta }_{r(violet)}=18.54°\end{array}$

Refractive index of red light, $n_{red}=1.50$

Substituting the values of $n_{red}$, $n_1$ and ${\theta }_i$ in $\sin {\theta }_r=\frac{n_1\sin {\theta }_i}{n_2}$, we get

  $\begin{array}{l}\sin {\theta }_{r(red)}=\frac{n_1\sin {\theta }_i}{n_{red}}\\ \sin {\theta }_{r(red)}=\frac{1\times \sin 30}{1.50}\\ \sin {\theta }_{r(red)}=0.333\\ \Rightarrow {\theta }_{r(red)}={\sin }^{-1}(0.333)\\ {\theta }_{r(red)}=19.45°\end{array}$

Conclusion:

Angle of refraction for violet light= $18.54°$

Angle of refraction for red light= $19.45°$

To determine

(c)

Repeat the previous part when the incident angle is 60.0°.

Answer to Problem 49QAP

Explanation of Solution

Given:

For a certain optical medium the speed of light varies from a low value of $1.9\times {10}^{8}m/s$ for violet light to a high value of $\text{ 2}\text{.00}\times {10}^{8}m/s$ for red light.

Formulaused:

Snell's law or the law of refraction is expressed as

  $n_1\sin {\theta }_i=n_2\sin {\theta }_r$

Where

  $\begin{array}{l}{n}_1:\text{Refractive index of medium1}\\ n_2:\text{Refractive index of medium2}\\ {\theta }_i:\text{Angle of incidence}\\ {\theta }_r:\text{Angle of refraction}\end{array}$

  $\sin {\theta }_r=\frac{n_1\sin {\theta }_i}{n_2}$

Calculation:

As per the given problem, medium 1 is air and medium 2 is the optical medium.

  $n_1=1$

Angle of incidence, ${\theta }_i=60°$

Refractive index of violet light, $n_{\text{violet}}=1.57$

Substituting the values of $n_{\text{violet}}$, $n_1$ and ${\theta }_i$ in $\sin {\theta }_r=\frac{n_1\sin {\theta }_i}{n_2}$, we get

  $\begin{array}{l}\sin {\theta }_{r(violet)}=\frac{n_1\sin {\theta }_i}{n_{violet}}\\ \sin {\theta }_{r(violet)}=\frac{1\times \sin 60}{1.57}\\ \sin {\theta }_{r(violet)}=0.551\\ \Rightarrow {\theta }_{r(violet)}={\sin }^{-1}(0.551)\\ {\theta }_{r(violet)}=33.43°\end{array}$

Refractive index of red light, $n_{red}=1.50$

Substituting the values of $n_{red}$, $n_1$ and ${\theta }_i$ in $\sin {\theta }_r=\frac{n_1\sin {\theta }_i}{n_2}$, we get

  $\begin{array}{l}\sin {\theta }_{r(red)}=\frac{n_1\sin {\theta }_i}{n_{red}}\\ \sin {\theta }_{r(red)}=\frac{1\times \sin 60}{1.50}\\ \sin {\theta }_{r(red)}=0.333\\ \Rightarrow {\theta }_{r(red)}={\sin }^{-1}(0.577)\\ {\theta }_{r(red)}=35.23°\end{array}$

Conclusion:

Angle of refraction for violet light= $33.43°$

Angle of refraction for red light= $35.23°$