Q106P In Fig. 33-78, where n1=1.70, n2... [FREE SOLUTION]

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Chapter 33: Q106P (page 1009)

In Fig. 33-78, where $n_{1} = 1 .70$ , $n_{2} = 1 .50$ ,and $n_{3} = 1 .30$ ,light refracts from material 1 into material 2. If it is incident at point A at the critical angle for the interface between materials 2 and 3, what are (a) the angle of refraction at pointBand (b) the initialangle $胃$ ?If, instead, light is incident atBat the critical angle for the interface between materials 2 and 3, what are (c) the angle of refraction at pointAand (d) the initial angle $胃$ ? If, instead of all that, light is incident at point Aat Brewster鈥檚 angle for the interface between materials 2 and 3, what are (e) the angle of refraction at point B and (f) the initialangle $胃$ ?

Short Answer

Expert verified

a) The angle of refraction at point $B$ is $35.1 掳 .$

b) The initial angle $胃$ is $49.9 掳 .$

c) The angle of refraction at point $A$ is. $35.1 掳$

d) The initial angle $胃$ is $26.1 掳 .$

e) The angle of refraction at point is $60.7 掳 .$

f) The initial angle $胃$ is $35.3 掳 .$

Step by step solution

Listing the given quantities

Refractive index, $n_{1} = 1.7$

Refractive index, $n_{2} = 1.5$

Refractive index, $n_{3} = 1.3$

Understanding the concepts of angle of the refraction

We can use Snell鈥檚 law to find the required angles of refraction and the initial angles.

Formula:

$n_{1} 鈥塻颈苍胃_{1} = n_{2} 鈥塻颈苍鈥� 胃_{2}$

(a) Calculations of theangle of refraction at point B

The angle of incidence $胃_{B, 1}$ at $B$ is the component of the critical angle at $A .$

From the figure, its sine angle is equal to the cosine of the critical angle. $胃_{c}$

$sin 鈥� 胃_{B, 1} = 肠辞蝉鈥� 胃_{c}$

Using Snell鈥檚 law,

$n_{2} sin 胃_{c} = n_{3} 蝉颈苍鈥� 90 掳$

$s i n s i n 胃_{c} = \frac{n_{3}}{n_{2}}$

Therefore,

$\begin{matrix} sin 鈥� 胃_{B, 1} = 肠辞蝉鈥� 胃_{c} \\ = \sqrt{1 鈭� 胃_{c}} \\ = \sqrt{1 鈭� {(\frac{n_{3}}{n_{2}})}^{2}} \end{matrix}$

Since

$n_{3} sin 鈥� 胃_{B, 2} = n_{2} 鈥塻颈苍鈥� 胃_{B, 1}$

Thus, the angle of refraction $胃_{B, 2}$ at $B$ becomes

$\begin{matrix} 胃_{B, 2} = \sin^{鈭� 1} (\frac{n_{2}}{n_{3}} \sqrt{1 鈭� {(\frac{n_{3}}{n_{2}})}^{2}}) \\ = \sin^{鈭� 1} \sqrt{{(\frac{n_{2}}{n_{3}})}^{2} 鈭� 1} \\ = \sin^{鈭� 1} (\sqrt{{(\frac{1.50}{1.30})}^{2} 鈭� 1} \\ = 35.1 掳 \end{matrix}$

Hence, the angle of refraction at point $B$ is $35.1 掳 .$

(b) Calculations of theinitial angle

Using Snell鈥檚 law,

$\begin{array}{l} n_{1} 鈥塻颈苍鈥� 胃 = n_{2} 蝉颈苍鈥� 胃_{c} \\ n_{1} 鈥塻颈苍鈥� 胃 = n_{2} (\frac{n_{3}}{n_{2}}) \end{array}$

$\begin{matrix} 胃 = \sin^{鈭� 1} (\frac{n_{3}}{n_{1}}) \\ = \sin^{鈭� 1} (\frac{1.3}{1.7}) \\ = 49.9 掳 \end{matrix}$

Hence, the initial angle $胃$ is $49.9 掳 .$

(c) Calculations of theangle of refraction at point A

The angle of incidence $胃_{A, 1}$ at $A$ is the complement of the critical angle at. $B$

Its sine is given by

$\begin{matrix} 蝉颈苍鈥� 胃_{A, 1} = cos 鈥� 胃_{c} \\ = \sqrt{1 - {(\frac{n_{3}}{n_{2}})}^{2}} \end{matrix}$

$n_{3} 蝉颈苍鈥� 胃_{A, 2} = n_{2} 蝉颈苍鈥� 胃_{A, 1}$

Therefore,

$\begin{matrix} 胃_{A, 2} = \sin^{鈭� 1} ((\frac{n_{2}}{n_{3}}) \sqrt{1 鈭� {(\frac{n_{3}}{n_{2}})}^{2}}) \\ = \sin^{鈭� 1} (\sqrt{{(\frac{n_{2}}{n_{3}})}^{2} 鈭� 1}) \\ = \sin^{鈭� 1} \sqrt{{(\frac{1.50}{1.30})}^{2} 鈭� 1} \\ = 35.1 掳 \end{matrix}$

Hence, the angle of refraction at point $A$ is. $35.1 掳$

(d) Calculations of theinitial angle

Using Snell鈥檚 law, we get

$\begin{matrix} n_{1} 鈥塻颈苍鈥� 胃 = n_{2} 鈥塻颈苍鈥� 胃_{A, 1} \\ = n_{2} \sqrt{1 鈭� {(\frac{n_{3}}{n_{2}})}^{2}} \\ = \sqrt{n_{2}^{2} 鈭� n_{3}^{2}} \end{matrix}$

$\begin{matrix} 胃 = \sin^{鈭� 1} \frac{\sqrt{n_{2}^{2} 鈭� n_{3}^{2}}}{n_{1}} \\ = \sin^{鈭� 1} \frac{\sqrt{{(1.5)}^{2} 鈭� {(1.3)}^{2}}}{1.7} \\ = 26.1 掳 \end{matrix}$

Hence, the initial angle $胃$ is $26.1 掳 .$

(e) Calculations of theangle of refraction at point B

The angle of incidence $胃_{B, 1}$ at $B$ is the complement of Brewster angle at $A .$

Its sine is given by

$蝉颈苍鈥� 胃_{B, 1} = \frac{n_{2}}{\sqrt{n_{2}^{2} + n_{3}^{2}}}$

So, the angle of refraction $胃_{B, 2}$ at $B$ will be

$n_{3} 鈥塻颈苍鈥� 胃_{B, 2} = n_{2} 鈥塻颈苍鈥� 胃_{B, 1}$

$\begin{matrix} sin 鈥� 胃_{B, 2} = \frac{n_{2}}{n_{3}} 蝉颈苍鈥� 胃_{B, 1} \\ = \frac{n_{2}^{2}}{n_{3} \sqrt{n_{2}^{2} + n_{3}^{2}}} \end{matrix}$

$\begin{matrix} 胃_{B, 2} = \sin^{鈭� 1} (\frac{{(1.5)}^{2}}{(1.3) \sqrt{{1.3}^{2} + {1.5}^{2}}}) \\ = 60.7 掳 \end{matrix}$

Hence, the angle of refraction at point $B$ is $60.7 掳 .$

(f) Calculations of theinitial angle

Using Snell鈥檚 law,

$n_{1} 鈥塻颈苍鈥� 胃 = n_{2} 鈥塻颈苍鈥� 胃_{B r e w s t e r}$

$蝉颈苍鈥� 胃_{B r e w s t e r} = \frac{n_{3}}{\sqrt{n_{2}^{2} + n_{3}^{2}}}$

$n_{1} 鈥塻颈苍鈥� 胃 = n_{2} \frac{n_{3}}{\sqrt{n_{2}^{2} + n_{3}^{2}}}$

$\begin{matrix} 胃 = \sin^{鈭� 1} (\frac{n_{2} n_{3}}{n_{1} \sqrt{n_{2}^{2} + n_{3}^{2}}}) \\ = \sin^{鈭� 1} (\frac{1.5 脳 1.3}{1.7 \sqrt{{1.5}^{2} + {1.3}^{2}}}) \\ = 35.3 掳 \end{matrix}$