Chapter 24, Problem 16P

To determine

The angle at which each beam leaves the prism.

Answer to Problem 16P

  ${\theta }_{d1}=68.4°$ and ${\theta }_{d2}=65.2°$ .

Explanation of Solution

Given info:

Angle of incidence, $i=45°$

Apex angle of prism, $\varphi =60°$

First wavelength, ${\lambda }_1=450nm$

Second wavelength, ${\lambda }_2=650nm$

The refractive index of the silicate flint glass for ${\lambda }_1$ is $n_1=1.64$

The refractive index of the silicate flint glass for ${\lambda }_2$ is $n_2=1.62$

Formula used:

The angles of refraction in the glass: $n_{1}sin{\theta }_1=n_{2}sin{\theta }_2$

Where,

  $n_1$ is Refractive index of first medium

  $n_2$ is Refractive index of second medium

  ${\theta }_1$ is angle of incidence

  ${\theta }_2$ is angle at which light transmitted to another medium.

Calculation:

Calculation for the refraction at the first surface: It is having that $n_{air}sin{\theta }_a=nsin{\theta }_b$

Now, substitute the give values:

  $\left(1.00\right)sin\left(45°\right)=\left(1.642\right)sin{\theta }_{b1}$

  ${\theta }_{b1}=25.51°$

  $\left(1.00\right)sin\left(45°\right)=\left(1.619\right)sin{\theta }_{b2}$

  ${\theta }_{b2}=25.90°$

Now, calculate the angle of incidence at the second surface from $\left(90°-{\theta }_b\right)+\left(90°-{\theta }_c\right)+A=180°$ . This gives

  ${\theta }_{c1}=A-{\theta }_{b1}=60.00°-25.51°=34.49°$

  ${\theta }_{c2}=A-{\theta }_{b2}=60.00°-25.90°=34.10°$

Now, calculation for the refraction at the second surface. For this: $nsin{\theta }_c=n_{air}sin{\theta }_d$ . This gives,

  $\left(1.642\right)sin\left(34.49°\right)=\left(1.00\right)sin{\theta }_{d1}$

  ${\theta }_{d1}=68.4°$ from the normal

  $\left(1.619\right)sin\left(34.10°\right)=\left(1.00\right)sin{\theta }_{d2}$

  ${\theta }_{d2}=65.2°$ from the normal

Conclusion: The required angles are ${\theta }_{d1}=68.4°$ and ${\theta }_{d2}=65.2°$ .