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Maximum deviation in prism

Maximum deviation in prism :- The angle of deviation of light ray while passing through a glass prism is maximum when the angle of incidence at interface AB is maximum, i.e., when i₁ = 90°. Now as

$\displaystyle \delta ={{i}_{1}}+{{i}_{2}}-A$

$\displaystyle \Rightarrow {{\delta }_{\max }}={{90}^{\circ }}+{{i}_{2}}-A$

Here as i₁ = 90° ⇒ r₁ = C, applying Snell’s Law at interface AC, we get

$\displaystyle \frac{\sin {{r}_{2}}}{\sin {{i}_{2}}}=\frac{1}{\mu }$

$\displaystyle \Rightarrow \sin {{i}_{2}}=\mu \sin {{r}_{2}}$

$\displaystyle \Rightarrow \sin {{i}_{2}}=\mu \sin (A-{{r}_{1}})$

$\displaystyle \Rightarrow \sin {{i}_{2}}=\mu \sin (A-C)$

$\displaystyle \Rightarrow {{i}_{2}}={{\sin }^{-1}}\left[ \mu \sin (A-C) \right]$

Hence maximum deviation (δ_max) is given by

$\displaystyle \Rightarrow {{\delta }_{\max }}={{\sin }^{-1}}\left[ \mu \sin (A-C) \right]+{{90}^{\circ }}-A$

Maximum deviation in prism

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