Question

The index of refraction of light in passing from a vacuum into dense flint glass is 1.66. If the angle of incidence is 50°, determine the angle of refraction.

Short answer

The angle of refraction is $27.{45}^{\mathrm{o}}$.

Step-by-step solution

Step 1. Given Information  

The index of refraction of light in passing from a vacuum into dense flint glass is 1.66. If the angle of incidence is 50°.

We have to determine the angle of refraction.

Step 2. The Snell's law tell us that when you have two medium with refraction index n1 and n2 the equation is n1sinθ1=n2sinθ2.

With ${\mathrm{\theta }}_1\mathrm{and}{\theta }_2$the angle in the first medium and in second medium respectively.

For us the initial angle ${\mathrm{\theta }}_1={50}^{\mathrm{o}}\mathrm{and}{n}_2=1.66$.

In order to find ${\mathrm{\theta }}_2$we need to know $n_1$.

Assume$n_1=1$for air.

Step 3. Then our equation is

$1·\sin \left({50}^o\right)=1.66·\sin \left({\mathrm{\theta }}_2\right)$

Divide by 1.66 on both side

$$\begin{gathered} \frac{\sin \left({50}^o\right)}{1.66}=\frac{1.66}{1.66}·\sin \left({\mathrm{\theta }}_2\right) \\ \frac{\sin \left({50}^o\right)}{1.66}=\sin \left({\mathrm{\theta }}_2\right) \\ \sin \left({\mathrm{\theta }}_2\right)=\frac{\sin \left({50}^o\right)}{1.66} \\ {\mathrm{\theta }}_2={\sin }^{-1}\left(\frac{\sin \left({50}^o\right)}{1.66}\right) \end{gathered}$$

Step 4. After solving the equation is θ2=sin-1sin50o1.66

$$\begin{gathered} {\mathrm{\theta }}_2\approx {\sin }^{-1}\left(\frac{0.766}{1.66}\right) \\ {\mathrm{\theta }}_2\approx {\sin }^{-1}\left(0.461\right) \\ {\mathrm{\theta }}_2\approx 27.{45}^{\mathrm{o}} \end{gathered}$$

The angle of refraction is $27.{45}^{\mathrm{o}}$.