Chapter 23, Problem 42QAP

To determine

(a)

Critical angle $i_c$

Answer to Problem 42QAP

Critical angle is $i_c=41.8°$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Let the critical angle be $i_c$

Refractive index of plastic $n_1=1.50$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

  $\begin{array}{l}or,\sin i_c=\frac{n_2\sin r}{n_1}\\ or,i_c={\sin }^{-1}\left(\frac{n_2\sin r}{n_1}\right)\\ or,i_c={\sin }^{-1}\left(\frac{1.0\times \sin 90°}{1.50}\right)\\ or,i_c=41.8°\end{array}$

Hence, the critical angle is $i_c=41.8°$

Conclusion:

Thus, the critical angle is $i_c=41.8°$

To determine

(b)

Critical angle $i_c$

Answer to Problem 42QAP

Critical angle is $i_c=48.8°$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Let the critical angle be $i_c$

Refractive index of water $n_1=1.33$

Refractive index of air $n_2=1.00$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

  $\begin{array}{l}or,\sin i_c=\frac{n_2\sin r}{n_1}\\ or,i_c={\sin }^{-1}\left(\frac{n_2\sin r}{n_1}\right)\\ or,i_c={\sin }^{-1}\left(\frac{1.0\times \sin 90°}{1.33}\right)\\ or,i_c=48.8°\end{array}$

Hence, the critical angle is $i_c=48.8°$

Conclusion:

Thus, the critical angle is $i_c=48.8°$

To determine

(c)

Critical angle $i_c$

Answer to Problem 42QAP

Critical angle is $i_c=58.5°$

Explanation of Solution

Given:

Angle of refraction $r=90°$

Let the critical angle be $i_c$

Refractive index of glass $n_1=1.56$

Refractive index of water $n_2=1.33$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

  $\begin{array}{l}or,\sin i_c=\frac{n_2\sin r}{n_1}\\ or,i_c={\sin }^{-1}\left(\frac{n_2\sin r}{n_1}\right)\\ or,i_c={\sin }^{-1}\left(\frac{1.33\times \sin 90°}{1.56}\right)\\ or,i_c=58.5°\end{array}$

Hence, the critical angle is $i_c=58.5°$

Conclusion:

Thus, the critical angle is $i_c=58.5°$

To determine

(d)

Critical angle $i_c$

Answer to Problem 42QAP

Critical angle is $i_c=40.2°$

Explanation of Solution

Given:

NOTE: in the given question, light goes from air to glass which is impossible at critical angle. It should be from glass to air.

Angle of refraction $r=90°$

Let the critical angle be $i_c$

Refractive index of air $n_2=1.00$

Refractive index of glass $n_1=1.55$

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin i_c=n_2\sin r$

  $\begin{array}{l}or,\sin i_c=\frac{n_2\sin r}{n_1}\\ or,i_c={\sin }^{-1}\left(\frac{n_2\sin r}{n_1}\right)\\ or,i_c={\sin }^{-1}\left(\frac{1.00\times \sin 90°}{1.55}\right)\\ or,i_c=40.2°\end{array}$

Hence, the critical angle is $i_c=40.2°$

Conclusion:

Thus, the critical angle is $i_c=40.2°$