Chapter 23, Problem 40P

To determine

The refractive index of the liquid.

Answer to Problem 40P

Solution:

The refractive index of the liquid $n_{liquid}>1.5$ .

Explanation of Solution

Given Info:

Let refractive index of liquid be $n_{liquid}$

Refractive index of air $n_{air}=1.0$

Liquid depth ‘D’ = 8.0 cm

Let the angle of the incident (critical angle) be $\angle i=\angle c$

The distance from just above the spot to the strike point, d = 7.6 cm at a critical angle.

Angle of refraction $\angle r={90}^0$

Formula Used: Apply Snell’s Law

  $\begin{array}{l}{n}_{liquid}\sin \angle i=n_{air}\sin \angle r\\ n_{liquid}\sin \angle c=n_{air}\sin {90}^0\end{array}$

Here, all alphabets are in their usual meanings.

Calculation: Now, we shall calculate the critical angle

  $\begin{array}{l}\tan \angle c=\frac{d}{D}\\ \tan \angle c=\frac{\text{7}\text{.6}\text{cm}}{\text{8}\text{.0}\text{cm}}\\ or,\angle c={43.5}^0\end{array}$

Apply Snell’s Law for liquid-air surface

  $\begin{array}{l}{n}_{liquid}\sin \angle i=n_{air}\sin \angle r\\ n_{liquid}\sin \angle c=n_{air}\sin {90}^0\\ n_{liquid}=\frac{n_{air}\sin {90}^0}{\sin \angle c}\\ n_{liquid}=\frac{1.0\sin {90}^0}{\sin {43.5}^0}\\ n_{liquid}=\frac{1}{0.69}\\ n_{liquid}=1.449\\ n_{liquid}=1.5\end{array}$

For the total internal reflection of light,the light should not exit from the water. Then,the angle of the incident must be greater thanits critical angle. For this, the refractive index must be greater than 1.5, that is, the refractive index should be $n_{liquid}>1.5$ .

Conclusion:

The refractive index of the liquid is $n_{liquid}>1.5$ .