What will be the refractive index of the material of the prism if the angle of minimum deviation for prism of angle is?

Answer

Hint : In this solution, we will use the relation of the refractive index of the prism with the angle of prism and angle of minimum deviation. By taking the assumption that the angle of minimum deviation of a prism is equal to its refracting angle, we will obtain a range of possible refractive index for the prism.Formula used: In this solution, we will use the following formula: $\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}} $ where $ \mu $ is the refractive index, $ A $ is the angle of the prism, and $ {\delta _m} $ is the angle of minimum deviation.

Complete step by step answer

So, $ \sqrt 2 < \mu < 2 $ which corresponds to option A.

Note

Option 2 : lies between 2 and √2

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CONCEPT:

• Prism: It is a piece of glass or other transparent material cut with precise angles and plane faces, useful for analyzing and reflecting light.
  • An ordinary triangular prism can separate white light into its constituent colors, called a spectrum.
  • The angle of deviation (δ) decreases with an increase in the angle of incidence (i) up to a particular angle. 
  • This angle of incidence where the angle of deviation in a prism is minimum is called the Minimum Deviation Position of the prism and that very deviation angle is known as the Minimum Angle of Deviation

where r1 and r2 are the refractive angles.

The formula for the angle of minimum deviation:

δmin = i + e - A

where, i = angle of incidence, e = angle of emergence, A = prism angle.

EXPLANATION:

From the above discussion,

for minimum deviation δmin = A then

2A = i + e

in case of δ min i = e

\(2A = 2i,{r_1} = {r_2} = \frac{A}{2}\)

i = A = 90° 

From snell's law

sin i = n sin r1

sin A = n sin A/2

By trigonometry, sinθ = 2 sin θ /2 cos θ /2

2 sin A/2 cos A/2 = n sin A/2

2 cos A/2 = n 

when A = 90° = imin

nmin = √ 2

i = A = 0, nmax = 2

The correct option is b.

• Refractive index(μ ): It is a dimensionless number that describes how fast light travels through the material.
  • It is defined as the ratio of c and v where c is the speed of light in vacuum and v is the phase velocity of light in the medium. 
• Snell's law: It states that the ratio of the sine of the angles of incidence and transmission is equal to the ratio of the refractive index of the materials at the interface.

Formula:

μ = c/v

where c = speed of light, v = phase velocity of light.

sin i = sin r

where, i =  angle of incidence, r = angle of refraction

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Condition when the angle of deviation is minimal in a prism

In a prism, the angle of deviation (δ) decreases with increase in the angle of incidence (i) up to a particular angle. This angle of incidence where the angle of deviation in a prism is minimum is called the minimum deviation position of the prism and that very deviation angle is known as the minimum angle of deviation (denoted by δmin, Dλ, or Dm).

The angle of minimum deviation is related with the refractive index as:

n 21 = sin ⁡ ( A + D m 2 ) sin ⁡ ( A 2 ) {\displaystyle n_{21}={\dfrac {\sin \left({\dfrac {A+D_{m}}{2}}\right)}{\sin \left({\dfrac {A}{2}}\right)}}}

This is useful to calculate the refractive index of a material. Rainbow and halo occur at minimum deviation. Also, a thin prism is always set at minimum deviation.

Formula

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. r1 = r2. And, 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 △ O P Q {\textstyle \triangle OPQ}

A + ∠ O P Q + ∠ O Q P = 180 ∘ {\displaystyle A+\angle OPQ+\angle OQP=180^{\circ }}

⟹ A = 180 ∘ − ( 90 − r ) − ( 90 − r ) {\displaystyle \implies A=180^{\circ }-(90-r)-(90-r)}

⟹ r = A 2 {\displaystyle \implies r={\frac {A}{2}}}

Using the exterior angle theorem in △ P Q R {\textstyle \triangle PQR}

D m = ∠ R P Q + ∠ R Q P {\displaystyle D_{m}=\angle RPQ+\angle RQP}

⟹ D m = i − r + i − r {\displaystyle \implies D_{m}=i-r+i-r}

⟹ 2 r + D m = 2 i {\displaystyle \implies 2r+D_{m}=2i}

⟹ A + D m = 2 i {\displaystyle \implies A+D_{m}=2i}

⟹ i = A + D m 2 {\displaystyle \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 = sin ⁡ i sin ⁡ r {\displaystyle n_{21}={\dfrac {\sin i}{\sin r}}}

∴ n 21 = sin ⁡ ( A + D m 2 ) sin ⁡ ( A 2 ) {\displaystyle \therefore n_{21}={\dfrac {\sin \left({\dfrac {A+D_{m}}{2}}\right)}{\sin \left({\dfrac {A}{2}}\right)}}}

∴ D m = 2 sin − 1 ⁡ ( n sin ⁡ ( A 2 ) ) − A {\displaystyle \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 Dm 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][6]

Worked out examples:

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, sin ⁡ ( 60 + δ 2 ) sin ⁡ ( 60 2 ) = 1.5 {\textstyle {\frac {\sin \left({\frac {60+\delta }{2}}\right)}{\sin \left({\frac {60}{2}}\right)}}=1.5} ⟹ sin ⁡ ( 30 + δ 2 ) sin ⁡ ( 30 ) = 1.5 {\textstyle \implies {\frac {\sin \left(30+{\frac {\delta }{2}}\right)}{\sin(30)}}=1.5} ⟹ sin ⁡ ( 30 + δ 2 ) = 1.5 × 0.5 {\textstyle \implies \sin \left(30+{\frac {\delta }{2}}\right)=1.5\times 0.5} ⟹ 30 + δ 2 = sin − 1 ⁡ ( 0.75 ) {\textstyle \implies 30+{\frac {\delta }{2}}=\sin ^{-1}(0.75)} ⟹ δ 2 = 48.6 − 30 {\textstyle \implies {\frac {\delta }{2}}=48.6-30} ⟹ δ = 2 × 18.6 {\textstyle \implies \delta =2\times 18.6} ∴ δ ≈ 37 ∘ {\textstyle \therefore \delta \approx 37^{\circ }} Also, i = ( A + δ ) 2 = 60 + 2 × 18.6 2 ≈ 49 ∘ {\textstyle i={\frac {(A+\delta )}{2}}={\frac {60+2\times 18.6}{2}}\approx 49^{\circ }} This is also apparent in the graph below.

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, δ = r {\textstyle \delta =r} ⟹ δ = A 2 {\textstyle \implies \delta ={\frac {A}{2}}} Using the above formula, sin ⁡ ( A + A 2 2 ) sin ⁡ ( A 2 ) = 1.4 {\textstyle {\frac {\sin \left({\frac {A+{\frac {A}{2}}}{2}}\right)}{\sin \left({\frac {A}{2}}\right)}}=1.4} ⟹ sin ⁡ ( 3 A 4 ) sin ⁡ ( A 2 ) = 1 2 1 2 {\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\frac {1}{2}}{\frac {1}{\sqrt {2}}}}} ⟹ sin ⁡ ( 3 A 4 ) sin ⁡ ( A 2 ) = sin ⁡ 45 ∘ sin ⁡ 30 ∘ {\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\sin 45^{\circ }}{\sin 30^{\circ }}}} ∴ A = 60 ∘ {\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 e in terms of i in the prism formula using Snell's law:

δ = i − A + sin − 1 ⁡ ( n ⋅ sin ⁡ ( A − sin − 1 ⁡ ( sin ⁡ i n ) ) ) {\displaystyle \delta =i-A+\sin ^{-1}\left(n\cdot \sin \left(A-\sin ^{-1}\left({\frac {\sin i}{n}}\right)\right)\right)}

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

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 Dm and A are very small,

n ≈ A + D m 2 A 2 n = A + D m A D m = A n − A {\displaystyle {\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}}}

∴ D m = A ( n − 1 ) {\displaystyle \therefore D_{m}=A(n-1)}

Interestingly, 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 ≈ i r 1 , n ≈ e r 2 {\displaystyle n\approx {\frac {i}{r_{1}}},n\approx {\frac {e}{r_{2}}}}

From the prism formula, δ = n r 1 + n r 2 − A = n ( r 1 + r 2 ) − A = n A − A = A ( n − 1 ) {\displaystyle {\begin{aligned}\delta &=nr_{1}+nr_{2}-A\\&=n(r_{1}+r_{2})-A\\&=nA-A\\&=A(n-1)\end{aligned}}}

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

Experimental determination

Further information: Prism spectrometer

This section needs expansion with: virtual simulation, video, detailed explanation, etc. You can help by adding to it. (May 2020)

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.[7][8][9][10][11]

Minimum angle of dispersion

This section needs expansion. You can help by adding to it. (May 2020)

The minimum angle of dispersion[12] for white light is the difference in minimum deviation angle between red and violet rays of a light ray through a prism.[2]

For a thin prism, the deviation of violet light, δ v {\displaystyle \delta _{v}}

( n v − 1 ) A {\displaystyle (n_{v}-1)A}

δ r {\displaystyle \delta _{r}}

( n r − 1 ) A {\displaystyle (n_{r}-1)A}

( δ v − δ r ) = ( n v − n r ) A {\displaystyle (\delta _{v}-\delta _{r})=(n_{v}-n_{r})A}

Applications

Further information: Rainbow and Sun dog

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°).[3][13]

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°.[3][14]

See also

• 
  physics portal

• Prism
• Refraction
• Geometrical optics

References

1. ^ a b c d "Chapter Nine, RAY OPTICS AND OPTICAL INSTRUMENTS". Physics Part II Textbook for Class IX (PDF). NCERT. p. 331.
2. ^ a b c "Optics-Prism". A-Level Physics Tutor.
3. ^ a b c d e Mark A. Peterson. "Minimum Deviation by a Prism". mtholyoke. Mount Holyoke College. Archived from the original on 2019-05-23.
4. ^ a b "Refraction through Prisms". SchoolPhysics.
5. ^ a b "Prism". HyperPhysics.
6. ^ "Determination of the refractive index of the material of the prism". BrainKart.
7. ^ "Angle of Minimum Deviation". Scribd.
8. ^ "Theory of the Prism Spectrometer". www.ukessays.com.
9. ^ "Experimental set up for the measurements of angle of minimum deviation using prism spectrometer". ResearchGate.
10. ^ "Measurement of the dispersion of glass with a prism spectrometer". studylib.net.
11. ^ "Determination of Minimum Deviation For Given Prism". BYJU'S.
12. ^ "ISRO", SpringerReference, Berlin/Heidelberg: Springer-Verlag, 2011, doi:10.1007/springerreference_222294, retrieved 2021-10-22
13. ^ "Rainbow". www.schoolphysics.co.uk.
14. ^ "Halo 22°". HyperPhysics.

External links

• Minimum Deviation Part 1 and Part 2 at Khan Academy
• Refraction through a Prism in NCERT Tectbook
• Minimum Deviation by Prism by Mark A Peterson, Mount Holyoke College

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