Physics - Geometrical Optics Question with Solution | TestHub

The refractive index of a prism is given by the formula:

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Given that $\mu =\cot \left(\frac{A}{2}\right)$, we can write:

$\cot \left(\frac{A}{2}\right)=\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Equating the two expressions for $\mu$:

$\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\sin \left(\frac{A+{\delta }_m}{2}\right)=\cos \left(\frac{A}{2}\right)$

Using the trigonometric identity $\cos (\theta )=\sin (90^{\circ }-\theta )$, we get:

$\sin \left(\frac{A+{\delta }_m}{2}\right)=\sin \left(90^{\circ }-\frac{A}{2}\right)$

Equating the angles:

$\frac{A+{\delta }_m}{2}=90^{\circ }-\frac{A}{2}$

$A+{\delta }_m=180^{\circ }-A$

${\delta }_m=180^{\circ }-2A$

Answer: $(180{)}^{\circ }-2A$