For an isosceles prism of angle A and refractive index μ it is...

$i = e$(for minimum deviation)

${r}_{1} + {r}_{2} = A, {r}_{1} = {r}_{2}$

\left(i\right){\delta}_{m} = 2i - A = A$(given)</p><p>$⇒ i = A$</p><p>$⇒ {r}_{1} = \dfrac{A}{2} = \dfrac{i}{2}$</p><p>\left(ii\right)\mu = \dfrac{\sin \left(A\right)}{\sin \left(\dfrac{A}{2}\right)} = 2cos \dfrac{A}{2} ⇒ A = 2\cos^{- 1} \left(\dfrac{\mu}{2}\right)$

\left(iii\right)\mu sin \left({r}_{2}\right) = 1$</p><p>$\sin \left({r}_{2}\right) = \dfrac{1}{\mu}$</p><p>${r}_{1} + {r}_{2} = A$</p><p>${r}_{1} = A - {r}_{2}$</p><p>$= A - \sin^{- 1} \left[\dfrac{1}{\mu}\right]$</p><p>$\sin \left(i\right) = \mu sin \left({r}_{1}\right)$</p><p>$i = \sin^{- 1} \left[\mu sin \left[A - \sin^{- 1} \left[\dfrac{1}{\mu}\right]\right]\right]$</p><p>${i}_{1} = \sin^{- 1} \left[\sqrt{{\mu}^{2} - 1}sin A - cos A\right] = \sin^{- 1} $\left[\mu sin \left(A - {\theta}_{C}\right)\right]$

\left(Here, \mu = 2cos \dfrac{A}{2}\right)$</p><p>\left(iv\right)$ Condition of min. deviation $i = eand{r}_{1} = {r}_{2} = \dfrac{A}{2}$