Q7E - University Physics with Modern Physics

Question

A parallel beam of light in air makes an angle of 47.5º with the surface of a glass plate having a refractive index of 1.66. (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface of the glass?

Step-by-Step Solution

Answer

1. The angle between the reflected part of the beam and the surface of the glass is $47.5°$
2. The angle between the refracted part of the beam and the surface of the glass is ${\theta }_b=24.0°$

Step 1: (a) Determination of the angle between the reflected part of the beam and the surface of the glass.

The law of reflection states,

${\theta }_i={\theta }_r$

Now, refer to the image below,

               

From, the figure and the law of reflection,

$$\begin{gathered} {\theta }_a={\theta }_r \\ =42.5° \end{gathered}$$  

Thus, the reflected angle with respect to the glass is,

  $$\begin{gathered} 90.0°-{\theta }_r \\ =47.5° \end{gathered}$$

Step 2: (b) Determination of the angle between the refracted part of the beam and the surface of the glass.

According to the Snell’s Law, the reflection, refraction and the normal lie on the same plane and 

$n_{a}sin{\theta }_a=n_{b}sin{\theta }_b$

  

Therefore,

$$\begin{gathered} sin{\theta }_b=\frac{n_a\sin {\theta }_a}{n_b} \\ =\frac{\left(1.00\right)\left(\sin 42.5°\right)}{1.66} \\ =0.4070 \end{gathered}$$

  

Thus, the refraction angle with respect to glass is ${\theta }_b=24.0°$