minimum deviation

{{Expand Section|the derivation of the expression for minimum deviation using Calculus|date=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.{{Cite web|url=https://www.a-levelphysicstutor.com/optics-prisms.php|title=Optics-Prism|website=A-Level Physics Tutor}}

Also, the angles of refractions are equal i.e. {{math|1=r₁ = r₂}}. The angle of incidence and angle of emergence equal each other ({{math|1=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.

File:Minimum Deviation.jpg

From the angle sum of $\backslash triangle\; OPQ$,

$A\; +\; \backslash angle\; OPQ\; +\; \backslash angle\; OQP\; =\; 180^\backslash circ$

$\backslash implies\; A\; =\; 180^\backslash circ\; -\; (90\; -\; r)\; -\; (90\; -\; r)$

$\backslash implies\; r\; =\; \backslash frac\{A\}\{2\}$

Using the exterior angle theorem in $\backslash triangle\; PQR$,

$D\_\{m\}\; =\; \backslash angle\; RPQ\; +\; \backslash angle\; RQP$

$\backslash implies\; D\_\{m\}\; =\; i\; -\; r\; +\; i\; -\; r$

$\backslash implies\; 2r\; +\; D\_\{m\}=\; 2i$

$\backslash implies\; A\; +\; D\_\{m\}\; =\; 2i$

$\backslash implies\; i\; =\; \backslash frac\{A\; +\; D\_\{m\}\}\; \{2\}$

This can also be derived by putting {{math|1=i = e}} in the prism formula: {{math|1=i + e = A + δ}}

From Snell's law,

$n\_\{21\}\; =\; \backslash dfrac\{\backslash sin\; i\}\{\backslash sin\; r\}$

$$\backslash therefore\; n\_\{21\}\; =\; \backslash dfrac\{\backslash sin\; \backslash left(\backslash dfrac\{A\; +\; D\_\{m\}\}\{2\}\backslash right)\}\{\backslash sin\; \backslash left(\backslash dfrac\{A\}\{2\}\backslash right)\}$$

{{Cite web|title=Refraction through Prisms|url=http://www.schoolphysics.co.uk/age16-19/Optics/Refraction/text/Prisms_/index.html|website=SchoolPhysics}}{{Cite web|title=Prism|url=http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/prism.html#c2|website=HyperPhysics}}{{Excessive citations inline|date=June 2020}}

$$\backslash therefore\; D\_m\; =\; 2\; \backslash sin^\{-1\}\; \backslash left(n\; \backslash sin\; \backslash left(\backslash frac\{A\}\{2\}\backslash right)\backslash right)\; -\; A$$

(where {{mvar|n}} is the refractive index, {{mvar|A}} is the Angle of Prism and {{mvar|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.{{cite web|url=https://www.mtholyoke.edu/~mpeterso/classes/phys103/geomopti/MinDev.html|title=Minimum Deviation by a Prism|last=Mark A. Peterson|website=mtholyoke|publisher=Mount Holyoke College|url-status=dead|archive-url=https://web.archive.org/web/20190523072850/http://www.mtholyoke.edu/~mpeterso/classes/phys103/geomopti/MinDev.html|archive-date=2019-05-23}}{{Cite book|url=http://ncert.nic.in/NCERTS/l/leph201.pdf|title=Physics Part II Textbook for Class IX|publisher=NCERT|pages=331|language=en|chapter=Chapter Nine, RAY OPTICS AND OPTICAL INSTRUMENTS}}

Worked out examples:

{{collapse top|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, {{math|1=A = 60°}}, {{math|1=n = 1.5}}

Plugging them in the above formula,

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

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

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

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

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

$\backslash implies\; \backslash delta\; =\; 2\; \backslash times\; 18.6$

$\backslash therefore\; \backslash delta\; \backslash approx\; 37^\backslash circ$

Also,

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

This is also apparent in the graph below.

{{collapse bottom}}

{{collapse top|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,

$\backslash delta\; =\; r$

$\backslash implies\; \backslash delta\; =\; \backslash frac\{A\}\{2\}$

Using the above formula,

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

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

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

$\backslash therefore\; A\; =\; 60^\backslash circ$

{{collapse bottom}}

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 {{mvar|i}} in the prism formula using Snell's law:

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

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

Putting $f\text{'}(i)=0$, we get,

$\backslash frac\{\backslash cos\backslash left(A-\backslash sin^\{-1\}\backslash left(\backslash frac\{\backslash sin\; i\}\{u\}\backslash right)\backslash right)\backslash cos\; i\}\{\backslash sqrt\{\backslash left(1-u^\{2\}\backslash sin^\{2\}\backslash left(A-\backslash sin^\{-1\}\backslash left(\backslash frac\{\backslash sin\; i\}\{u\}\backslash right)\backslash right)\backslash right)\backslash left(1-\backslash frac\{\backslash sin^\{2\}i\}\{u^\{2\}\}\backslash right)\}\}=1$, 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 [https://www.desmos.com/calculator/0jkerecdx3 here]

File:Minimum deviation.png

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 {{mvar|D_m}} and {{mvar|A}} are very small,

$$

\begin{align}

n & \approx \dfrac{\frac{A + D_{m}}{2}}{\frac{A}{2}}\\

n & = \frac{A + D_m}{A}\\

D_m & = An - A

\end{align}

$$\backslash therefore\; D\_\{m\}\; =\; A(n\; -\; 1)$$

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 {{mvar|i}}, {{mvar|e}} and {{mvar|r}} are small,

$n\; \backslash approx\; \backslash frac\{i\}\{r\_1\},\; n\; \backslash approx\; \backslash frac\{e\}\{r\_2\}$

From the prism formula,

$$

\begin{align}

\delta & = n r_1 + n r_2 - A \\

& = n(r_1 + r_2) - A \\

& = nA - A \\

& = A(n - 1)

\end{align}

Thus, it can be said that a thin prism is always in minimum deviation.

{{Further|Prism spectrometer}}

{{Expand Section|date=May 2020|virtual simulation, video, detailed explanation, etc}}

Minimum deviation can be found manually or with spectrometer. Either the prism is kept fixed and the incidence angle is adjusted or the prism is rotated keeping the light source fixed.{{Cite web|title=Angle of Minimum Deviation|url=https://www.scribd.com/document/201177761/Angle-of-Minimum-Deviation|website=Scribd|language=en}}{{Cite web|title=Experimental set up for the measurements of angle of minimum deviation using prism spectrometer|url=https://www.researchgate.net/figure/Experimental-set-up-for-the-measurements-of-angle-of-minimum-deviation-using-prism_fig1_326377418|website=ResearchGate|language=en}}{{Cite web |title=Theory of the Prism Spectrometer - Experiment {{!}} UKEssays.com |url=https://www.ukessays.com/essays/physics/theory-prism-spectrometer-experiment-7823.php |access-date=2025-03-11 |website=www.ukessays.com |language=en}}

File:Experiment setup.svg

File:Prisms-disp-diag02.jpg

{{Expand Section||date=May 2020}}

The minimum angle of dispersion for white light is the difference in minimum deviation angle between red and violet rays of a light ray through a prism.

For a thin prism, the deviation of violet light, $\backslash delta\_v$ is $(n\_v-1)A$ and that of red light, $\backslash delta\_r$ is $(n\_r-1)A$. The difference in the deviation between red and violet light, $(\backslash delta\_v-\backslash delta\_r)=(n\_v-n\_r)A$ is called the Angular Dispersion produced by the prism.

{{Further|Rainbow|Sun dog}}

File:Rainbow1.svg

One of the factors that causes a rainbow is the bunching of light rays at the minimum deviation angle that is close to the rainbow angle (42°).{{Cite web|title=Rainbow|url=http://www.schoolphysics.co.uk/age16-19/Optics/Refraction/text/Rainbow_/index.html|website=www.schoolphysics.co.uk}}

thumb

It is also responsible for phenomena like halos and sundogs, produced by the deviation of sunlight in mini prisms of hexagonal ice crystals in the air bending light with a minimum deviation of 22°.{{Cite web|title=Halo 22°|url=http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/halo22.html|website=HyperPhysics}}

{{Portal|physics}}

• Prism
• Refraction
• Geometrical optics

• Minimum Deviation [https://www.khanacademy.org/science/in-in-class-12th-physics-india/in-in-ray-optics-and-optical-instruments/in-in-ray-optics-and-optical-instruments/in-in-refraction-in-prisms/v/minimum-deviation-in-prism?modal=1 Part 1] and [https://www.khanacademy.org/science/in-in-class-12th-physics-india/in-in-ray-optics-and-optical-instruments/in-in-ray-optics-and-optical-instruments/in-in-refraction-in-prisms/v/deriving-prism-formula?modal=1 Part 2] at Khan Academy
• [https://s3.ap-south-1.amazonaws.com/aglasem-cdn/aglasem-doc/703f1f62-4d77-11ea-a344-02f21f5619c4/23.jpg Refraction through a Prism] in NCERT Textbook
• [https://www.mtholyoke.edu/~mpeterso/classes/phys103/geomopti/MinDev.html Minimum Deviation by Prism] {{Webarchive|url=https://web.archive.org/web/20190523072850/http://www.mtholyoke.edu/~mpeterso/classes/phys103/geomopti/MinDev.html |date=2019-05-23 }} by Mark A Peterson, Mount Holyoke College

Category:Geometrical optics

Category:Light