Question

In the diagram below, a light ray is shown passing from water into air. The indices of refraction are given for water and for air. The angle the light ray makes with the normal in air is shown.
What angle does the light ray make with the normal as it approaches the boundary from the water?
The angles in the drawing are not necessarily drawn with accurate measurements.

• A. ${30.0}^{\circ }$
• B. ${29.7}^{\circ }$
• C. ${33.1}^{\circ }$
• D. ${58.5}^{\circ }$
• E. ${31.5}^{\circ }$

Answer 1

Answer: (E)

${31.5}^{\circ }$

Given : $n_a=1.00$ $n_w=1.33$ ${\theta }_i={44.0}^o$ $sin{44.0}^o=0.695$

Using Snell's law of refraction, $n_a\times sin{\theta }_i=n_w\times sin{\theta }_w$

$\therefore$ $1.00\times 0.695=1.33\times sin{\theta }_w$ $⟹{\theta }_w=si{n}^{-1}\left(0.522\right)={31.5}^o$