## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) We can see the marks from the bottom of the tank coming up. (b) The highest mark we can see is $60~cm$
(a) The light rays from the marks near the top of the tank meet the water surface at a large angle from the normal. Since these angles are larger than the critical angle, these light rays do not emerge from the water. The light rays from the marks near the bottom of the tank meet the water surface at a smaller angle from the normal. Since these angles are smaller than the critical angle, these light rays emerge from the water. Therefore, we can see the marks from the bottom of the tank coming up. (b) We can find the critical angle: $n_1~sin~\theta_1 = n_2~sin~90^{\circ}$ $sin~\theta_1 = \frac{n_2}{n_1}$ $\theta_1 = sin^{-1}~(\frac{n_2}{n_1})$ $\theta_1 = sin^{-1}~(\frac{1.00}{1.33})$ $\theta_1 = 48.75^{\circ}$ We can find the highest level that we can see in the tank: $\frac{65~cm}{y} = tan~48.75^{\circ}$ $y = \frac{65~cm}{tan~48.75^{\circ}}$ $y = 57~cm$ Since there are marks every 10 cm, the highest mark we can see is $60~cm$