Question

A ray of light passes through a a prism plane; The deviation being equal to angle incidence which is equal to $2\alpha$. It is given that $\alpha$ is the angle of prism and $\mu$ is the refractive inside of the material of prism, then

• A. $cos\alpha =\sqrt{\frac{{\mu }^2-1}{2}}$
• B. $cos\alpha =\sqrt{\frac{{\mu }^2-1}{4}}$
• C. $sin\alpha =\sqrt{{\mu }^2-1}$
• D. $sin\alpha =\sqrt{\frac{{\mu }^2-1}{8}}$

Answer 1

The correct option is D $sin\alpha =\sqrt{\frac{{\mu }^2-1}{8}}$

$S=$ deviation $=2\alpha ($ given $)$

$\therefore 2\alpha =2\alpha +e-d$

$\alpha =e$

and $r_1+{\gamma }_2=$

$\Rightarrow \frac{\sin 2\alpha }{\sin {\gamma }_1}=4,\frac{\sin e}{\sin {\gamma }_2}=A$

$\Rightarrow \sin 2\alpha =4\sin {\gamma }_1\frac{\sin \alpha }{\sin (\alpha -{\gamma }_1)}=u_4$

Now, $\sin \left(\alpha {\gamma }_1\right)=\frac{\sin \alpha }{\mu }$

$\Rightarrow \sin (A-B)=\sin A\cos B-\cos A\sin B$

$\Rightarrow \therefore \left[\sin \alpha \sqrt{1\frac{{\sin }^{2}2\alpha }{{\mu }^2}}\frac{\cos \alpha \sin 2\alpha }{\mu }\right]=\frac{\sin \alpha }{\mu }$

$\{\cos {\gamma }_1=\sqrt{1-{\sin }^2\alpha \alpha }{\mu }^2\}$

$\Rightarrow \sin \alpha \sqrt{{\mu }^2-{\sin }^{2}2\alpha }-\cos \alpha \sin 2\alpha =\sin \alpha$

$\Rightarrow \sqrt{{\mu }^2-{\sin }^{2}2\alpha }-2{\cos }^2\alpha =1$

$\Rightarrow {\mu }^2-{\sin }^{2}2\alpha =1+4{\cos }^4\alpha +4{\cos }^2\alpha$

$\Rightarrow {\mu }^2=1+4{\cos }^2\alpha ({\sin }^2\alpha +{\cos }^2\alpha )$

$\therefore {\cos }^2\alpha =\frac{{\mu }^2-1}{4}$

$\therefore \cos \alpha =\sqrt{\frac{{\mu }^2-1}{4}}$