Chapter 23, Problem 43QAP

To determine

(A)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=1.34$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=48.5°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 48.5°}\right)\\ or,n_1=1.34\end{array}$

Hence, the refractive index of medium is $n_1=1.34$

Conclusion:

Thus, the refractive index of medium is $n_1=1.34$

To determine

(B)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=1.37$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=47.0°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 47.0°}\right)\\ or,n_1=1.37\end{array}$

Hence, the refractive index of medium is $n_1=1.37$

Conclusion:

Thus, the refractive index of medium is $n_1=1.37$

To determine

(C)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=1.48$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=42.6°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 42.6°}\right)\\ or,n_1=1.48\end{array}$

Hence, the refractive index of medium is $n_1=1.48$

Conclusion:

Thus, the refractive index of medium is $n_1=1.48$

To determine

(d)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=1.74$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=35.0°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 35.0°}\right)\\ or,n_1=1.74\end{array}$

Hence, the refractive index of medium is $n_1=1.74$

Conclusion:

Thus, the refractive index of medium is $n_1=1.74$

To determine

(E)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=1.21$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=55.7°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 55.7°}\right)\\ or,n_1=1.21\end{array}$

Hence, the refractive index of medium is $n_1=1.21$

Conclusion:

Thus, the refractive index of medium is $n_1=1.21$

To determine

(F)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=1.61$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=38.5°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 38.5°}\right)\\ or,n_1=1.61\end{array}$

Hence, the refractive index of medium is $n_1=1.61$

Conclusion:

Thus, the refractive index of medium is $n_1=1.61$

To determine

(G)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=2.65$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=22.2°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 22.2°}\right)\\ or,n_1=2.65\end{array}$

Hence, the refractive index of medium is $n_1=2.65$

Conclusion:

Thus, the refractive index of medium is $n_1=2.65$

To determine

(H)

Refractive index of medium $n_1$

Answer to Problem 43QAP

Refractive index of medium is $n_1=1.04$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Critical angle be ${\theta }_c=75.0°$

Let refractive index of medium be $n_1$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_c=n_2\sin r$

  $\begin{array}{l}or,n_1=\frac{n_2\sin r}{\sin i_c}\\ or,n_1=\left(\frac{1.0\times \sin 90°}{\sin 75.0°}\right)\\ or,n_1=1.04\end{array}$

Hence, the refractive index of medium is $n_1=1.04$

Conclusion:

Thus, the refractive index of medium is $n_1=1.04$