Question: In Fig. 33-58, light from rayArefracts from material $\mathbf{1}\left(n_1=1.60\right)$into a thin layer of material $\mathbf{2}\mathbf{(}{\mathbf{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{)}$ , crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3 .(a) What is the value of incident angle ${\mathbf{\theta }}_{\mathbf{A}}$ ? (b) If ${\mathbf{\theta }}_{\mathbf{A}}$ is decreased, does part of the light refract into material 3 ? Light from ray B refracts from material 1into the thin layer, crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3. (c) What is the value of incident angle${\mathbf{\theta }}_{\mathbf{B}}$ ? (d) If${\mathbf{\theta }}_{\mathbf{B}}$ is decreased, does part of the light refract into material 3 ?

$$\begin{gathered} {\mathrm{n}}_1=1.60 \\ {\mathrm{n}}_2=1.80 \\ {\mathrm{n}}_3=1.30 \end{gathered}$$

Using Snell’s law, we can find the angle ${\text{θ}}_A$as it will transmit through material 1 to material 2. Using Snell’s law for material 2 and 3, we can equate Snell’s law for material 1 and 3. From this, we can conclude whether a part of light refracts into material 3.

Using Snell’s law, we can find the angle ${\mathrm{\theta }}_{\mathrm{B}}$as it will transmit through material 1 to material 2. Using Snell’s law for material 2 and 3, we can equate Snell’s law for material 1 and 3. From this, we can conclude whether a part of light refracts into material 3.

Formula:

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2$

For material 1 and 2, using Snell’s law, we can write that

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

For material 2 and 3, using Snell’s law, we can write that

$n_3\sin {\theta }_3=n_2\sin {\theta }_2$

Hence,

$n_1\sin {\theta }_1=n_3\sin {\theta }_3$

Angle of refraction between material 2 and 3, i.e.,${\theta }_3={90}^{\circ }$

$\begin{array}{l}16{\mathrm{sin\theta }}_1=1.3\sin {90}^{\circ }\\ {\mathrm{sin\theta }}_1=\frac{1.3}{1.6}\\ {\mathrm{\theta }}_1=\underset{0}{{\sin }^{-1}}0.812\\ {\mathrm{\theta }}_1=54.3^{{\circ }^{}}\end{array}$

Hence,the value of incident angle${\mathrm{\theta }}_{\mathrm{A}}$is$54.3^{\circ }$.

If${\mathrm{\theta }}_{\mathrm{A}}$is decreased, thentheangle of refraction in material 2, i.e.,data-custom-editor="chemistry" ${\mathrm{\theta }}_2$will also decrease and it will become less thanthecritical angle.

Therefore, there will be some transmission of light into material 3.

We can draw a ray diagram for the given problem as follows

From the diagram, we can say that

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

But,

$$\begin{gathered} {\mathrm{\theta }}_2={90}^{\circ }-{\mathrm{\theta }}_{\mathrm{c}} \\ {\mathrm{sin\theta }}_2=\sin \left({90}^{\circ }-{\mathrm{\theta }}_{\mathrm{c}}\right) \\ {\mathrm{sin\theta }}_2={\mathrm{cos\theta }}_{\mathrm{c}} \\ {\mathrm{cos\theta }}_{\mathrm{c}}=\sqrt{1-{\sin }^2{\mathrm{\theta }}_{\mathrm{c}}} \end{gathered}$$

Applying Snell’s law for material 2 and 3, we get

$n_2\sin {\theta }_c=n_3\sin {\theta }_3$

We have ${\theta }_3={90}^{\circ }$

$\sin {\theta }_c=\frac{n_3}{n_2}$

We have

$\begin{array}{l}\cos {\theta }_c=\sqrt{1-{\sin }^2{\theta }_c}\\ \cos {\theta }_c=\sqrt{1-{\left(\frac{n_3}{n_2}\right)}^2}\\ \end{array}$

We have

$n_{1}sin{\theta }_B=n_{2}sin{\theta }_2$

But,

$\sin {\theta }_2=\cos {\theta }_c$

And

$\cos {\theta }_c=\sqrt{1-{\left(\frac{n_3}{n_2}\right)}^2}$

$$\begin{gathered} {\mathrm{n}}_1{\mathrm{sin\theta }}_{\mathrm{B}}={\mathrm{n}}_2\left(\sqrt{1-{\left(\frac{{\mathrm{n}}_3}{{\mathrm{n}}_2}\right)}^2}\right) \\ {\mathrm{sin\theta }}_{\mathrm{B}}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\times \left(\sqrt{1-{\left(\frac{{\mathrm{n}}_3}{{\mathrm{n}}_2}\right)}^2}\right) \\ {\mathrm{sin\theta }}_{\mathrm{B}}=\frac{1.8}{1.6}\times \left(\sqrt{1-{\left(\frac{1.3}{1.8}\right)}^2}\right) \\ {\mathrm{sin\theta }}_{\mathrm{B}}=1.125\times 0.69 \\ {\mathrm{\theta }}_{\mathrm{B}}={\sin }^{-1}0.78 \\ {\theta }_B=51.1^{\circ } \end{gathered}$$

Hence, the value of incident angle ${\theta }_B$ will be $51.1^{\circ }$

If${\theta }_B$is decreased, thentheangle of refraction in material 2, i.e.,${\theta }_2$ will also decrease. This will cause the angle of incidence at material 2 and 3 to be greater than the critical angle, which concludes that, there will not be any transmission of light into material 3.