91Ó°ÊÓ

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

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 ${\mathrm{θ}}_{\mathrm{B}}$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:

${\mathrm{n}}_1{\mathrm{²õ¾\pm ²Ôθ}}_1={\mathrm{n}}_2{\mathrm{²õ¾\pm ²Ôθ}}_2$

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

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

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

$n_3\sin {\mathrm{θ}}_3=n_2\sin {\mathrm{θ}}_2$

Hence,

$n_1\sin {\mathrm{θ}}_1=n_3\sin {\mathrm{θ}}_3$

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

$\begin{array}{l}16{\mathrm{²õ¾\pm ²Ôθ}}_1=1.3\sin {90}^{∘}\\ {\mathrm{²õ¾\pm ²Ôθ}}_1=\frac{1.3}{1.6}\\ {\mathrm{θ}}_1=\underset{0}{{\sin }^{-1}}0.812\\ {\mathrm{θ}}_1=54.3^{{∘}^{}}\end{array}$

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

If${\mathrm{θ}}_{\mathrm{A}}$is decreased, thentheangle of refraction in material 2, i.e.,data-custom-editor="chemistry" ${\mathrm{θ}}_2$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

$n_1\sin {\mathrm{θ}}_B=n_2\sin {\mathrm{θ}}_2$

But,

$$\begin{gathered} {\mathrm{θ}}_2={90}^{∘}-{\mathrm{θ}}_{\mathrm{c}} \\ {\mathrm{²õ¾\pm ²Ôθ}}_2=\sin \left({90}^{∘}-{\mathrm{θ}}_{\mathrm{c}}\right) \\ {\mathrm{²õ¾\pm ²Ôθ}}_2={\mathrm{³¦´Ç²õθ}}_{\mathrm{c}} \\ {\mathrm{³¦´Ç²õθ}}_{\mathrm{c}}=\sqrt{1-{\sin }^2{\mathrm{θ}}_{\mathrm{c}}} \end{gathered}$$

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

$n_2\sin {\mathrm{θ}}_c=n_3\sin {\mathrm{θ}}_3$

We have ${\mathrm{θ}}_3={90}^{∘}$

$\sin {\mathrm{θ}}_c=\frac{n_3}{n_2}$

We have

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

We have

$n_{1}sin{\mathrm{θ}}_B=n_{2}sin{\mathrm{θ}}_2$

But,

$\sin {\mathrm{θ}}_2=\cos {\mathrm{θ}}_c$

And

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

$$\begin{gathered} {\mathrm{n}}_1{\mathrm{²õ¾\pm ²Ôθ}}_{\mathrm{B}}={\mathrm{n}}_2\left(\sqrt{1-{\left(\frac{{\mathrm{n}}_3}{{\mathrm{n}}_2}\right)}^2}\right) \\ {\mathrm{²õ¾\pm ²Ôθ}}_{\mathrm{B}}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}×\left(\sqrt{1-{\left(\frac{{\mathrm{n}}_3}{{\mathrm{n}}_2}\right)}^2}\right) \\ {\mathrm{²õ¾\pm ²Ôθ}}_{\mathrm{B}}=\frac{1.8}{1.6}×\left(\sqrt{1-{\left(\frac{1.3}{1.8}\right)}^2}\right) \\ {\mathrm{²õ¾\pm ²Ôθ}}_{\mathrm{B}}=1.125×0.69 \\ {\mathrm{θ}}_{\mathrm{B}}={\sin }^{-1}0.78 \\ {\mathrm{θ}}_B=51.1^{∘} \end{gathered}$$

Hence, the value of incident angle ${\mathrm{θ}}_B$ will be $51.1^{∘}$

If${\mathrm{θ}}_B$is decreased, thentheangle of refraction in material 2, i.e.,${\mathrm{θ}}_2$ 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.