Chapter 23, Problem 40QAP

To determine

(a)

Refractive Index of second medium $n_2$

Answer to Problem 40QAP

Refractive Index of second medium is $n_2=1.41$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,n_2=\frac{n_1\sin {\theta }_1}{\sin {\theta }_2}\\ or,n_2=\frac{1.0\times \sin 20°}{\sin 14°}\\ or,n_2=\left(\frac{1.0\times 0.342}{0.242}\right)\\ or,n_2=1.41\end{array}$

Hence, Refractive Index of second medium is $n_2=1.41$

Conclusion:

Thus, Refractive Index of second medium is $n_2=1.41$

To determine

(b)

Refractive Index of second medium $n_2$

Answer to Problem 40QAP

Refractive Index of second medium is $n_2=1.46$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,n_2=\frac{n_1\sin {\theta }_1}{\sin {\theta }_2}\\ or,n_2=\frac{1.0\times \sin 30°}{\sin 20°}\\ or,n_2=1.46\end{array}$

Hence, Refractive Index of second medium is $n_2=1.46$

Conclusion:

Thus, refractive index of second medium is $n_2=1.46$

To determine

(c)

Refractive Index of first medium $n_1$

Answer to Problem 40QAP

Refractive Index of first medium is $n_1=0.843$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,n_1=\frac{n_2\sin {\theta }_2}{\sin {\theta }_1}\\ or,n_1=\frac{1.50\times \sin 10°}{\sin 18°}\\ or,n_1=0.843\end{array}$

Hence, Refractive Index of first medium is $n_1=0.843$

Conclusion:

Thus, Refractive Index of first medium is $n_1=0.843$

To determine

(d)

Refractive Index of first medium $n_1$

Answer to Problem 40QAP

Refractive Index of first medium is $n_1=0.843$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,n_1=\frac{n_2\sin {\theta }_2}{\sin {\theta }_1}\\ or,n_1=\frac{1.33\times \sin 25°}{\sin 34.1°}\\ or,n_1=1.001\end{array}$

Hence, Refractive Index of first medium is $n_1=1.001$

Conclusion:

Thus, Refractive Index of first medium is $n_1=1.001$