Chapter 23, Problem 114QAP

To determine

The angular separation between the red and blur portion of the spectrum that emerges from the prism

Answer to Problem 114QAP

The angular separation of blue and red color is $1.587°$.

Explanation of Solution

Given info:

Each angle of prism = $A=B=C=60°$ (since, prism is equilateral triangle)
Angle of incidence = $i=60°$

Refractive index of prism for red = $n_{red}=1.51$

Refractive index of prism for blue = $n_{blue}=1.53$

Formula used:

By Snell's law, following equation satisfy for refraction of light,
  $\begin{array}{l}{n}_{1}Sini=n_{2}Sin\text{}r\\ \text{here, }i\text{ and }r\text{ are the angle of incidence and angle of refraction}\\ \text{respectively}\text{.}\end{array}$

Calculation:

   In quadrilateral AQTR,
  $\begin{array}{l}\angle AQT+\angle TRA+\angle RAQ+\angle QTR={360}^o\\ {90}^o+{90}^o+A+T={360}^o\text{since, }\angle RAQ=A;\angle QTR=T\\ T={180}^o-A\mathrm{.......}(1)\\ \text{now, in }\Delta \text{QTR,}\\ \angle QTR+\angle TRQ+\angle RQT={180}^o\\ T+r+r\text{'}={180}^o\\ {180}^o-A+r+r\text{'}={180}^o\\ A=r+r\text{'}\mathrm{........}(2)\end{array}$

At the surface AB,
  $\begin{array}{l}n=\frac{\sin i}{\sin r}\\ r={\sin }^{-1}\left(\frac{\sin i}{n}\right)\mathrm{.........}(3)\end{array}$

Now, at the surface of AC,
  $\begin{array}{l}n\sin r\text{'}=sini\text{'}\\ i\text{'}={\sin }^{-1}\left(n\sin r\text{'}\right)\\ i\text{'}={\sin }^{-1}\left(n\sin (A-r)\right)(\text{from}(2))\\ i\text{'}={\sin }^{-1}\left(n\sin \left(A-{\sin }^{-1}\left(\frac{\sin i}{n}\right)\right)\right)\mathrm{........}(4)(\text{from}(3))\end{array}$

From the geometry, the exterior angle is equal to the sum of opposite interior angle.

So, from the figure,
  $\begin{array}{l}\delta =(i-r)+(i\text{'}-r\text{'})\\ \delta =(i+i\text{'})-(r+r\text{'})\\ \delta =i+{\sin }^{-1}\left(n\sin \left(A-{\sin }^{-1}\left(\frac{\sin i}{n}\right)\right)\right)-A\\ \delta =i-A+{\sin }^{-1}\left(n\sin \left(A-{\sin }^{-1}\left(\frac{\sin i}{n}\right)\right)\right)\\ \text{For}\text{red,}\\ {\delta }_{red}=i-A+{\sin }^{-1}\left(n_{red}\sin \left(A-{\sin }^{-1}\left(\frac{\sin i}{n_{red}}\right)\right)\right)\end{array}$

For blue,
  $\begin{array}{l}{\delta }_{blue}=i-A+{\sin }^{-1}\left(n_{blue}\sin \left(A-{\sin }^{-1}\left(\frac{\sin i}{n_{blue}}\right)\right)\right)\\ \text{Now},\Delta \theta ={\delta }_{blue}-{\delta }_{red}\\ \Delta \theta ={\sin }^{-1}\left(n_{blue}\sin \left(A-{\sin }^{-1}\left(\frac{\sin i}{n_{blue}}\right)\right)\right)-{\sin }^{-1}\left(n_{red}\sin \left(A-{\sin }^{-1}\left(\frac{\sin i}{n_{red}}\right)\right)\right)\\ \Delta \theta ={\sin }^{-1}\left(1.53\sin \left({60}^o-{\sin }^{-1}\left(\frac{\sin {60}^o}{1.53}\right)\right)\right)-{\sin }^{-1}\left(1.51\sin \left({60}^o-{\sin }^{-1}\left(\frac{\sin {60}^o}{1.51}\right)\right)\right)\\ \Delta \theta =41.24748°-39.66045°\\ \Delta \theta =1.587°\end{array}$

Conclusion:

Thus, the angular separation of red and blue color is $1.587°$.