A sphere of optically transparent material is placed in a parallel beam of light (Fig.). The angle of incidence of one of the rays on the surface of the sphere $\varphi ={\tan }^{-1}\frac{4}{3},$ and the angle of its deviation from the original direction after two refractions on the surface of the sphere $\theta =2{\tan }^{-1}\frac{7}{24}$ The refractive index of the material of the sphere is $\frac{X}{3}.$ Find the value of X. 

A beam of light 1A is Incident on the ball at an angle ϕ , passes the ball along the line AB, constituting the angles β with radii AO and BO, so that $\frac{\sin \varphi }{\sin \beta }=n$  

For the ray B2 leaving the ball, $\frac{\sin \beta }{\sin \gamma }=\frac{1}{n}$  
Consider triangle ABC. Obviously, it is isosceles and the angle θ is its outer corner; hence, $\theta =2(\varphi -\beta )=2{\tan }^{-1}\frac{7}{24}$ or $\tan (\varphi -\beta )=\frac{7}{24}$ 
From the expansion formulae of $\tan (\varphi -\beta )$ we get $\tan \beta =\frac{3}{4}$  
Finally, for the refractive index we find $n=\frac{\sin \varphi }{\sin \beta }=\frac{\sqrt{1+\frac{1}{{\tan }^2\beta }}}{\sqrt{1+\frac{1}{{\tan }^2\varphi }}}=\frac{4}{3}$