Chapter 18, Problem 54A

(a)

To determine

To calculate: The angle ${\theta }_2$ using Snell’s law of refraction.

Answer to Problem 54A

  ${\theta }_2=28.12°$ .

Explanation of Solution

Given:

Refractive index of glass $\left(n_r\right) =\mathrm{1.50.}$

Angle of incidence $\left({\theta }_i\right)=45°$

Refractive index of air $\left(n_i\right)=\mathrm{1.00.}$

  

Formula used:

The Snell’s law provides “the relationship between the angle of incidence and angle of refraction.

Mathematically,

  $n_i\sin {\theta }_i=n_r\sin {\theta }_r$

Calculation:

Rearranging the equation,

  ${\theta }_r={\sin }^{-1}\left(\frac{n_i\sin {\theta }_i}{n_r}\right)$

  ${\theta }_r={\sin }^{-1}\left(\frac{1\times \sin 45°}{1.5}\right)$$$

  ${\theta }_r={\theta }_2=28.12°$

(b)

To determine

To calculate: The angle ${\theta }_1\text{'}$ using elementary geometry.

Answer to Problem 54A

  ${\theta }_1\text{'}=28.12°$

Explanation of Solution

Given:

  ${\theta }_r=28.12°$

  

Formula used:

From the given geometry

  ${\theta }_1{}^1={\theta }_2$

Calculation:

Considering the first case that is by using Snell’s law, the angle of refraction is found. From the geometry the PQ line is horizontal from which the lines with angle ${\theta }_2$ and ${\theta }_1$ are directed and meeting at a point R as shown in the figure.

So, it is very clear that the angle of refraction( ${\theta }_2$ ) is same as that of the angle ( ${\theta }_1{}^1$ ).

Therefore,

  ${\theta }_1{}^1={\theta }_2={28.12}^{\circ }.$

Conclusion:

The angle ${\theta }_1{}^1={28.12}^{\circ }$ from the elemental geometry.

(c)

To determine

To calculate:The ${\theta }_2\text{'}$ .

Answer to Problem 54A

  ${\theta }_2\text{'}=45°$

Explanation of Solution

Given:

Index of glass (material) $\left(n_g\right)=\mathrm{1.50.}$

Angle of incidence $\left({\theta }_i\text{'}\right)=28.12°$

Refractive index of air $\left(n_a\right)=\mathrm{1.00.}$

  

Formula used:

The Snell’s law provides “the relationship between the angle of refractionand angle ofincidence.

Mathematically,

  $n_g\sin {\theta }_1\text{'}=n_a\sin {\theta }_2\text{'}$

Calculation:

  ${\theta }_2\text{'}={\sin }^{-1}\left(\frac{n_g\sin {\theta }_1\text{'}}{n_a}\right)$$$

  ${\theta }_2\text{'}={\sin }^{-1}\left(\frac{1.5\times \sin 28.12°}{1}\right)$

  ${\theta }_2\text{'}=45°$

The angle of ${\theta }_2\text{'}=45°$ .

Conclusion:

The angle of ray leaving is, ${\theta }_2\text{'}=45°$ .