Question

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 ?

Short answer

1. The value of incident angle ${\mathrm{\theta }}_{\mathrm{A}}$ is $54.3^{\circ }$.
2. If${\mathrm{\theta }}_{\mathrm{A}}$is decreased, there will be transmission of light into material 3.
3. The value of incident angle${\mathrm{\theta }}_{\mathrm{B}}$will be$51.1^{\circ }$
4. If ${\mathrm{\theta }}_{\mathrm{B}}$ is decreased, there will be no transmission of light into material 3.

Step-by-step solution

Given

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

Understanding the concept

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:

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

(a) Calculate the value of incident angle θA

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$54.3^{\circ }$.

(b) Find out if a part of the light refracts into material 3, If  θA is decreased

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.

(c) Calculate the value of incident angle  θB

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 $51.1^{\circ }$

(d) Find out if a part of the light refracts into material 3, If θB  is decreased

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.