Spread the love

Condition for minimum deviation in a prism

Condition for minimum deviation in a prism :- The angle of deviation of a light ray, while passing through a prism is given by :

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

$\displaystyle \Rightarrow \delta ={{\left( \sqrt{{{i}_{1}}} \right)}^{2}}+{{\left( \sqrt{{{i}_{2}}} \right)}^{2}}-A$

$\displaystyle \Rightarrow \delta ={{\left( \sqrt{{{i}_{1}}} \right)}^{2}}+{{\left( \sqrt{{{i}_{2}}} \right)}^{2}}-2\sqrt{{{i}_{1}}{{i}_{2}}}+2\sqrt{{{i}_{1}}{{i}_{2}}}-A$

$\displaystyle \Rightarrow \delta ={{\left( \sqrt{{{i}_{1}}}-\sqrt{{{i}_{2}}} \right)}^{2}}+2\sqrt{{{i}_{1}}{{i}_{2}}}-A$

Here, $\displaystyle 2\sqrt{{{i}_{1}}{{i}_{2}}}$ cannot be negative and A cannot be zero, hence for deviation(δ) to be minimum, we have,

$\displaystyle {{\left( \sqrt{{{i}_{1}}}-\sqrt{{{i}_{2}}} \right)}^{2}}=0$

$\displaystyle \Rightarrow \left( \sqrt{{{i}_{1}}}-\sqrt{{{i}_{2}}} \right)=0$

$\displaystyle \Rightarrow {{i}_{1}}={{i}_{2}}$

It means in case of minimum deviation, the angle of incidence(i₁) is equal to the angle of emergence(i₂).

Let i₁ = i₂ = i, then applying Snell’s Law at interface AB,

$\displaystyle \frac{\sin i}{\sin {{r}_{1}}}=\mu$ …..(1)

Now applying Snell’s Law at interface AC,

$\displaystyle \frac{\sin {{r}_{2}}}{\sin i}=\frac{1}{\mu }$ …..(2)

Multiplying equations (1) and (2), we get

$\displaystyle \sin {{r}_{1}}=\sin {{r}_{2}}$

$\displaystyle \Rightarrow {{r}_{1}}={{r}_{2}}=r(let)$

Hence in case of minimum deviation, the incident ray PQ passes symmetrically with respect to the refracting faces of the prism and the refracted ray (QR) in the prism is parallel to the base (BC) of the prism as shown in the figure above.

The same condition for minimum deviation in a prism can be derived from the deviation(δ) v/s angle of incidence(i) graph.

From above graph we observe that in case of minimum deviation, i₁ = i₂.

Condition for minimum deviation in a prism

Like this: