Question

In Fig. 35-32a, a beam of light in material 1 is incident on a boundary at an angle of ${\mathbf{30}}^{\mathbf{\text{o}}}$. The extent to which the light is bent due to refraction depends, in part, on the index of refraction ${\mathit{n}}_{\mathbf{2}}$of material 2. Figure 35-32b gives the angle of refraction ${\mathit{\theta }}_{\mathbf{2}}$versus ${\mathit{n}}_{\mathbf{2}}$for a range of possible ${\mathit{n}}_{\mathbf{2}}$values, from ${\mathit{n}}_{\mathbf{a}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{30}$to ${\mathit{n}}_{\mathbf{b}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{90}$. What is the speed of light in material 1?

Short answer

The speed of light in material 1 is $2\times {10}^8\text{m/s}$.

Step-by-step solution

Given information

1. The incident angle of a beam of light in material 1 is,${\theta }_1=30°$ .
2. The angle of refraction of light in material 2 is,${\theta }_2$ .
3. The index of refraction of material 1 is,$n_1$.
4. The index of refraction of material 2 is,$n_2$ .

Diffraction of light

If a ray of light passes through multiple layers of mediums, then it deviates from its direct path, this deviation of light due to mediums is described as ‘diffraction of light’.

The law used to establish a relationship between the angles of incident and refracted waves of light is the Snell’s law.

Speed of light in material 1

According to the Snell's law (the law of refraction),

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

If there is only material 1, then,${\theta }_1={\theta }_2$the expression reduces to,

$n_1=n_2$

From the graph at ${\theta }_2=30°$, the value of$n_2$is,$n_2=1.5$.

So, the value of $n_1$will be,$n_1=1.5$

The formula for the speed with which the light propagates in medium 1 is given by,

$v=\frac{c}{n_1}$

Here, speed of light$c=3\times {10}^8\text{m/s}$.

Putting values,

$$\begin{gathered} v=\frac{3\times {10}^8\text{m/s}}{1.5} \\ v=2\times {10}^8\text{m/s} \end{gathered}$$

Hence, the speed of light in material 1 is$2\times {10}^8\text{m/s}$ .