Minimum deviation

This section needs expansion with: the derivation of the expression for minimum deviation using Calculus. You can help by adding to it. (June 2020)

In minimum deviation, the refracted ray in the prism is parallel to its base. In other words, the light ray is symmetrical about the axis of symmetry of the prism.^[1]^[2]^[3] Also, the angles of refractions are equal i.e. $r 1 = r 2$ . The angle of incidence and angle of emergence equal each other ( $i = e$ ). This is clearly visible in the graph below.

The formula for minimum deviation can be derived by exploiting the geometry in the prism. The approach involves replacing the variables in the Snell's law in terms of the Deviation and Prism Angles by making the use of the above properties.

From the angle sum of ${\textstyle \triangle OPQ}$ ,

$A+\angle OPQ+\angle OQP=180^{\circ }$ $\implies A=180^{\circ }-(90-r)-(90-r)$ $\implies r={\frac {A}{2}}$

Using the exterior angle theorem in ${\textstyle \triangle PQR}$ ,

$D_{m}=\angle RPQ+\angle RQP$ $\implies D_{m}=i-r+i-r$ $\implies 2r+D_{m}=2i$ $\implies A+D_{m}=2i$ $\implies i={\frac {A+D_{m}}{2}}$

This can also be derived by putting $i = e$ in the prism formula: $i + e = A + δ$

From Snell's law,

$n_{21}={\dfrac {\sin i}{\sin r}}$

$\therefore n_{21}={\dfrac {\sin \left({\dfrac {A+D_{m}}{2}}\right)}{\sin \left({\dfrac {A}{2}}\right)}}$ ^[4]^[3]^[1]^[2]^[5]^{[excessive citations]}

$\therefore D_{m}=2\sin ^{-1}\left(n\sin \left({\frac {A}{2}}\right)\right)-A$

(where $n$ is the refractive index, $A$ is the Angle of Prism and $D m$ is the Minimum Angle of Deviation.)

This is a convenient way used to measure the refractive index of a material(liquid or gas) by directing a light ray through a prism of negligible thickness at minimum deviation filled with the material or in a glass prism dipped in it.^[5]^[3]^[1]

Worked out examples:

More information

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The refractive index of glass is 1.5. The minimum angle of deviation for an equilateral prism along with the corresponding angle of incidence is desired.
Answer: 37°, 49° Solution: Here, $A = 60°$ , $n = 1.5$ Plugging them in the above formula, ${\textstyle {\frac {\sin \left({\frac {60+\delta }{2}}\right)}{\sin \left({\frac {60}{2}}\right)}}=1.5}$ ${\textstyle \implies {\frac {\sin \left(30+{\frac {\delta }{2}}\right)}{\sin(30)}}=1.5}$ ${\textstyle \implies \sin \left(30+{\frac {\delta }{2}}\right)=1.5\times 0.5}$ ${\textstyle \implies 30+{\frac {\delta }{2}}=\sin ^{-1}(0.75)}$ ${\textstyle \implies {\frac {\delta }{2}}=48.6-30}$ ${\textstyle \implies \delta =2\times 18.6}$ ${\textstyle \therefore \delta \approx 37^{\circ }}$ Also, ${\textstyle i={\frac {(A+\delta )}{2}}={\frac {60+2\times 18.6}{2}}\approx 49^{\circ }}$ This is also apparent in the graph below.

The refractive index of glass is 1.5. The minimum angle of deviation for an equilateral prism along with the corresponding angle of incidence is desired.

Answer: 37°, 49°

Solution:

Here, $A = 60°$ , $n = 1.5$

Plugging them in the above formula,

${\textstyle {\frac {\sin \left({\frac {60+\delta }{2}}\right)}{\sin \left({\frac {60}{2}}\right)}}=1.5}$

${\textstyle \implies {\frac {\sin \left(30+{\frac {\delta }{2}}\right)}{\sin(30)}}=1.5}$

${\textstyle \implies \sin \left(30+{\frac {\delta }{2}}\right)=1.5\times 0.5}$

${\textstyle \implies 30+{\frac {\delta }{2}}=\sin ^{-1}(0.75)}$

${\textstyle \implies {\frac {\delta }{2}}=48.6-30}$

${\textstyle \implies \delta =2\times 18.6}$

${\textstyle \therefore \delta \approx 37^{\circ }}$

Also,

${\textstyle i={\frac {(A+\delta )}{2}}={\frac {60+2\times 18.6}{2}}\approx 49^{\circ }}$

This is also apparent in the graph below.

More information Here,

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If the minimum angle of deviation of a prism of refractive index 1.4 equals its refracting angle, the angle of the prism is desired.
Answer: 60° Solution: Here, ${\textstyle \delta =r}$ ${\textstyle \implies \delta ={\frac {A}{2}}}$ Using the above formula, ${\textstyle {\frac {\sin \left({\frac {A+{\frac {A}{2}}}{2}}\right)}{\sin \left({\frac {A}{2}}\right)}}=1.4}$ ${\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\frac {1}{2}}{\frac {1}{\sqrt {2}}}}}$ ${\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\sin 45^{\circ }}{\sin 30^{\circ }}}}$ ${\textstyle \therefore A=60^{\circ }}$

If the minimum angle of deviation of a prism of refractive index 1.4 equals its refracting angle, the angle of the prism is desired.

Answer: 60°

Solution:

Here, ${\textstyle \delta =r}$

${\textstyle \implies \delta ={\frac {A}{2}}}$

Using the above formula,

${\textstyle {\frac {\sin \left({\frac {A+{\frac {A}{2}}}{2}}\right)}{\sin \left({\frac {A}{2}}\right)}}=1.4}$

${\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\frac {1}{2}}{\frac {1}{\sqrt {2}}}}}$

${\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\sin 45^{\circ }}{\sin 30^{\circ }}}}$

${\textstyle \therefore A=60^{\circ }}$

Also, the variation of the angle of deviation with an arbitrary angle of incidence can be encapsulated into a single equation by expressing δ in terms of $i$ in the prism formula using Snell's law:

$\delta =i-A+\sin ^{-1}\left(n\cdot \sin \left(A-\sin ^{-1}\left({\frac {\sin i}{n}}\right)\right)\right)=f(i)$

Finding the minima of this equation will also give the same relation for minimum deviation as above.

Putting $f'(i)=0$ , we get,

${\frac {\cos \left(A-\sin ^{-1}\left({\frac {\sin i}{u}}\right)\right)\cos i}{\sqrt {\left(1-u^{2}\sin ^{2}\left(A-\sin ^{-1}\left({\frac {\sin i}{u}}\right)\right)\right)\left(1-{\frac {\sin ^{2}i}{u^{2}}}\right)}}}=1$ , and by solving this equation we can obtain the value of angle of incidence for a definite value of angle of prism and the value of relative refractive index of the prism for which the minimum angle of deviation will be obtained. The equation and description are given here

For thin prism

In a thin or small angle prism, as the angles become very small, the sine of the angle nearly equals the angle itself and this yields many useful results.

Because $D m$ and $A$ are very small,

${\begin{aligned}n&\approx {\dfrac {\frac {A+D_{m}}{2}}{\frac {A}{2}}}\\n&={\frac {A+D_{m}}{A}}\\D_{m}&=An-A\end{aligned}}$

$\therefore D_{m}=A(n-1)$ ^[1]^[4]

Using a similar approach with the Snell's law and the prism formula for an in general thin-prism ends up in the very same result for the deviation angle.

Because $i$ , $e$ and $r$ are small,

$n\approx {\frac {i}{r_{1}}},n\approx {\frac {e}{r_{2}}}$