Question

Light passes from benzene (medium1)to water (medium 2) in Figure ₂₀at (a)${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{°}$or (b)${\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{°}$. Find the angle${\mathbf{\theta }}_{\mathbf{2}}$in each case.

Short answer

a) For${\mathrm{\theta }}_1=30°{\mathrm{\theta }}_2=34.33$,

b) For ${\mathrm{\theta }}_1=0°{\mathrm{\theta }}_2=0°$,.

Step-by-step solution

Concept used.

The Snell's law is:

${\mathbf{n}}_{\mathbf{1}}\mathbf{‐}{\mathbf{sin\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{n}}_{\mathbf{2}}\mathbf{‐}{\mathbf{sin\theta }}_{\mathbf{2}}$.

Find the angle θ2 in θ1=30°

a)

Consider the angle${\mathrm{\theta }}_1=30°$ that is made by light when it passes from benzene to water.

The angle${\mathrm{\theta }}_2$is:

$$\begin{gathered} 1.50·\sin \left(30°\right)=1.33·\sin \left({\mathrm{\theta }}_2\right)\sin \left({\mathrm{\theta }}_2\right) \\ =\frac{1.50·\sin \left(30°\right)}{1.33}\sin \left({\mathrm{\theta }}_2\right) \\ =0.56391{\mathrm{\theta }}_2 \\ =34.33° \end{gathered}$$

Find the angle θ2 is θ1=0°

b)

An angle${\mathrm{\theta }}_1=0°$ is made by light when it passes from benzene to water.

The angle ${\mathrm{\theta }}_2$is:

$$\begin{gathered} 1.50·\sin \left(0°\right)=1.33·\sin \left({\mathrm{\theta }}_2\right)\sin \left({\mathrm{\theta }}_2\right) \\ =\frac{1.50·\sin \left(30°\right)}{1.33}\sin \left({\mathrm{\theta }}_2\right) \\ =0{\mathrm{\theta }}_2 \\ =0° \end{gathered}$$