Chapter 23, Problem 36QAP

To determine

(A)

The speed of light in ice.

Answer to Problem 36QAP

The speed of light in ice is $2.29\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of ice $n=1.309$

Let the speed of light in ice be $v$.

Formula used:

Refractive Index of ice $n=\frac{speedoflightinvacuum}{speedoflightinice}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{1.309}\\ v=2.29\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in ice is $2.29\times {10}^{8}m/s$.

To determine

(B)

The speed of light in acetone.

Answer to Problem 36QAP

The speed of light in acetone is $2.21\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of acetone $n=1.359$

Let the speed of light in acetone be $v$.

Formula used:

Refractive Index of acetone $n=\frac{speedoflightinvacuum}{speedoflightin\text{acetone}}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{1.359}\\ v=2.21\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in acetone is $2.21\times {10}^{8}m/s$.

To determine

(C)

The speed of light in Plexiglas.

Answer to Problem 36QAP

The speed of light in Plexiglas is $1.99\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of Plexiglas $n=1.51$

Let the speed of light in Plexiglas be $v$.

Formula used:

Refractive Index of Plexiglas $n=\frac{speedoflightinvacuum}{speedoflightin\text{Plexiglas}}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{1.51}\\ v=1.99\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in Plexiglas is $1.99\times {10}^{8}m/s$.

To determine

(d)

The speed of light in Sodium Chloride.

Answer to Problem 36QAP

The speed of light in Sodium Chloride is $1.94\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of Sodium Chloride $n=1.544$

Let the speed of light in Sodium Chloride be $v$.

Formula used:

Refractive Index of Sodium Chloride $n=\frac{speedoflightinvacuum}{speedoflightin\text{Sodium Chloride}}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{1.544}\\ v=1.94\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in Sodium Chloride is $1.94\times {10}^{8}m/s$.

To determine

(E)

The speed of light in Sapphire.

Answer to Problem 36QAP

The speed of light in Sapphire is $1.71\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of Sapphire $n=1.76$

Let the speed of light in Sapphire be $v$.

Formula used:

Refractive Index of Sapphire $n=\frac{speedoflightinvacuum}{speedoflightin\text{Sapphire}}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{1.76}\\ v=1.71\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in Sapphire is $1.71\times {10}^{8}m/s$.

To determine

(F)

The speed of light in diamond.

Answer to Problem 36QAP

The speed of light in diamond is $1.24\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of diamond $n=2.419$

Let the speed of light in diamond be $v$.

Formula used:

Refractive Index of diamond $n=\frac{speedoflightinvacuum}{speedoflightin\text{diamond}}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{2.419}\\ v=1.24\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in diamond is $1.24\times {10}^{8}m/s$.

To determine

(G)

The speed of light in water.

Answer to Problem 36QAP

The speed of light in water is $2.25\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of water $n=1.333$

Let the speed of light in water be $v$.

Formula used:

Refractive Index of water $n=\frac{speedoflightinvacuum}{speedoflightin\text{water}}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{1.333}\\ v=2.25\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in water is $2.25\times {10}^{8}m/s$.

To determine

(H)

The speed of light in crow glass.

Answer to Problem 36QAP

The speed of light in crow glass is $1.97\times {10}^{8}m/s$.

Explanation of Solution

Given:

Speed of light in vacuum $c=3.0\times {10}^{8}m/s$

Refractive Index of crow glass $n=1.52$

Let the speed of light in crow glass be $v$.

Formula used:

Refractive Index of crow glass $n=\frac{speedoflightinvacuum}{speedoflightin\text{crow glass}}=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

Using the above formula,

  $\begin{array}{l}v=\frac{c}{n}\\ v=\frac{3.0\times {10}^{8}m/s}{1.52}\\ v=1.97\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the speed of light in crow glass is $1.97\times {10}^{8}m/s$.