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What is Meant by Minimum Deviation in the Case of a Prism? Write Down the Condition for Minimum Deviation
Prism : Some Definitions
Prism: A prism is a portion of a transparent medium confined using two plane faces inclined to each other [Fig.].
In the figure DEFGHK is a prism and it is confined by the two planes DEHK and DFGK.
Refracting face: The two plane faces inclined to each other at some angle by which the prism is bound are called the refracting faces. In Fig., DEHK and DFGK are the refract-ing faces.
Edge: The line along which the two refracting faces meet is called the edge of the prism. In Fig., the line DK is the edge of the prism.
Refracting angle or angle of the prism: The angle included between the two refracting faces is called the refracting angle or simply the angle of the prism. In the figure, ∠EDF is the angle of the prism.
Side face and base: In general, besides of two refracting surfaces a prism further on is enclosed by three more surfaces. Among these surfaces, two surfaces are triangular and are placed perpendicular to the edge of the prism. In figure these two surfaces are DEF nd KHG. These are the side faces of the prism. Another one is rectangular and situated perpendicular to the side face. In figure the surface is EFGH. It is called base of the prism.
Principal section: The triangular cross section cut by a perpendicular plane at right angles to the edge of a prism is called a principal section of the prism. In Fig., ABC is a principle section of the prism.
Light can enter or emerge from the prism through its refracting surfaces as those surfaces are plain and smooth. In some cases, base and side faces of a prism are made rough so that no light passes through it.
A prism is usually represented by its principal section.
Refraction of Light along the Principal Section of a Prism
In Fig., ABC is the principal section of a prism. AB and AC are the refracting faces and SC is the base of the prism. A ray PQ is incident on the face AB at Q where NQO is the normal. The prism is supposed to be an optically denser medium with
respect to its surroundings. So, after refraction on plane AB, the refracted ray QR bends towards the normal NQO .
The refracted ray QR is then incident on the face AC at R where N’RO is the normal. The emergent ray RS bends away from the normal N’RO after refraction.
So PQRS is the whole path of the ray of light. The angle between the direction of the incident ray and the direction of the emergent ray gives the angle of deviation. In Fig., the angle of deviation, δ = ∠MTS.
Expression for angle of deviation: Let the angle of the prism = ∠BAC = A . For refraction on the face AB, the angle of incidence = ∠PQN = i1 and angle of refraction = ∠RQO = r1 and for refraction on the face AC, the angle of incidence of QR = ∠QRO = r2 and angle of refraction = ∠N’RS = i2
Now we get from the ΔQRT, angle of deviation,
δ = ∠MTR = ∠TQR + ∠TRQ
= (i1 – r1) + (i2 – r2) = i1 + i2 – (r1 + r2)
Now, from the quadrilateral AQOR, as NO and N’O are nor-mals on AB and AC respectively, so A + ∠QOR = 180°.
Again, from the triangle QOR,
∠QOR + r1 + r2 = 180° ∴ A = r1 + r2 ……. (1)
and δ = i1 + i2 – A …… (2)
So the angle of deviation δ depends on the angle of incidence. From the principle of reversibility of light ray it follows that, if a ray enters the face AC along SR at an angle of incidence i2, it will emerge along QP at an angle i1 and suffer the same deviation. In other words, the same deviation occurs for two values of i.
Angle of Minimum Deviation
The angle of deviation of a ray of light passing through a prism depends on the angle of incidence because in the relation
δ = i1 + i2 – A, A is constant and i2 depends on i1. A graph is drawn taking the angle of incidence i as abscissa and the angle of deviation δ as ordinate [Fig.]. The graph indicates that the deviation decreases at first with the increase in the angle of incidence and then attains a minimum (δm) for a particular value of the angle of incidence i0. Then with further increase in the angle of incidence, the deviation also increases.
So for every prism there is a fixed angle of incidence for which the deviation suffered by a ray of light traversing the prism becomes minimum. This angle of deviation is called the angle of minimum deviation of a prism. The position of the prism for which the value of the deviation becomes minimum is called the position of minimum deviation of the prism. This position of the prism is unique i.e., for only one position of the particular prism, the deviation becomes minimum.
Analysis of (i-δ) graph and condition of minimum deviation: Aline AB parallel to i -axis is drawn [Fig.]. The coordinates of the points, A and B are respectively (i1, δ) and (i2, δ). The angle of deviation is δ for angle of incidence i1 or i2. We have seen that for angle of incidence i1 the emergent angle is i2. The length of the line AB = (i2 – i1). As the line AB moves parallel to itself downwards, the value of δ decreases and the length of the line (i2 – i1) also decreases gradually. At the point C, the length of the line AB becomes zero and δ also becomes minimum. Then i2 – i1 = 0 or i2 = i1.
So we can say that during refraction through a prism, when the angle of incidence and the angle of emergence are equal, the angle of deviation becomes minimum.
Condition of minimum deviation by calculus: We know for refraction through a prism, the angle of deviation of a ray of light,
δ = i1 + i2 – A …… (1)
and angle of the prism,
A = r1 + r2 …….. (2)
The angle of deviation δ depends on the angle of incidence i1. For minimum deviation δ,
\(\frac{d}{d i_1}(\delta)\) = 0 ∴ \(\frac{d}{d i_1}\)(i1 + i2 – A) = 0
or, 1 + \(\frac{d i_2}{d i_1}\) = 0 [∵ angle of the prism, A = constant]
∴ \(\frac{d i_2}{d i_1}\) = -1 ….. (3)
Differentiating equation (2) we have,
\(\frac{d}{d r_1}(A)\) = \(\frac{d}{d r_1}\left(r_1+r_2\right)\)
or, 0 = 1 + \(\frac{d r_2}{d r_1}\) or, \(\frac{d r_2}{d r_1}\) = -1 ….. (4)
For refraction at Q and R [Fig.], according to Snell’s law we have, sin i1 = µ sinr1 and sini2 = µ sinr2; where, µ = refractive index of the material of the prism.
Differentiating the above equations we have,
So the deviation is minimum when the angle of incidence (i1) is equal to the angle of emergence (i2).
Brightness of a ray of light at minimum deviation: From the graph of the angle of incidence i and angle of deviation δ [Fig.], it is found that generally if the angle of incidence is different, the angle of deviation is also different. But if the angle of deviation attains its minimum value δm, then it is observed that for a wider range of angle of
incidence (from i1 to i2 vide [Fig.]), the angle of deviation of the ray becomes almost equal to the angle of minimum devi-ation δm, i.e., all the rays within this range emerge from the prism making minimum angle of deviation δm. So it can be said that in case of minimum deviation the brightness of the emergent ray increases considerably. For this characteristic, minimum deviation of a prism is of great importance.
Path of Ray through a Prism for Minimum Deviation
Suppose a ray of light passes through a prism with minimum deviation along the path PQRS [Fig.]. According to the condition of minimum deviation i1 = i2.
If the refractive index of the material of the prism is R, p then
µ = \(\frac{\sin i_1}{\sin r_1}\) = \(\frac{\sin i_2}{\sin r_2}\)
∴ r1 = r2
Since ∠AQR = 90° – r1 and ∠ARQ = 90° – r2
So ∠AQR = ∠ARQ [∵ r1 = r2]
So the triangle AQR ¡s an isosceles triangle having AQ = AR. So, for minimum deviation of a ray, the point of incidence, Q and the point of emergence, R are equidistant from the vertex A of the prism. So it is evident that when the deviation of a ray in a prism is minimum, the path of the ray through the prism becomes symmetrical.
Suppose, for the prism, AB = AC
∵ AQ = AR ∴ \(\frac{A Q}{A B}\) = \(\frac{A R}{A C}\)
i.e., the line QR is parallel to BC.
So, for an isosceles prism, when the deviation is minimum the path of the ray through the prism becomes parallel to the base of the prism.
Again the deviation of the ray dueto refraction at AB = (i1 – r1) and the deviation of the ray due to refraction at AC = (i2 – r2).
At the position of the minimum deviation of the prism, i1 = i2 and r1 = r2. So the deviations stated earlier becomes equal.
Therefore, we can say that at the minimum deviation position of the prism the total deviation is divided equally between the two refracting faces of the prism.
Refractive Index and Angle of Minimum Deviation
We know that in case of refraction through a prism, the angle of deviation of a ray is δ = i1 + i2 – A and the angle of the prism, A = r1 + r2.
For minimum deviation, i1 = i2. Again when i1 = i2 then r1 = r2
So, angle of minimum deviation,
δm = i1 + i1 – A = 2i1 – A or, i1 = \(\frac{A+\delta_m}{2}\)
Again, A = r1 + r1 = 2r1 or, r1 = \(\frac{A}{2}\)
Now considering refraction at the face AB we have, angle of incidence = i1 and angle of refraction = r1.
If the refractive index of the material of the prism is µ then,
µ = \(\frac{\sin i_1}{\sin r_1}\) = \(\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}\)
So, if we know the values of the angle of the prism A, and the angle of minimum deviation δm, we can determine the value of the refractive index of the material of the prism.
Numerical Examples
Example 1.
The refractive index of the material of a prism is \(\sqrt{\frac{3}{2}}\) and the refracting angle is 90°. What are the values of the angle of minimum deviation of the prism and the angle of incidence at the minimum deviation position?
Solution:
Here A = 90°; µ = \(\sqrt{\frac{3}{2}}\). Let the angle of the minimum deviation be δm.
Example 2.
The refractive index of the material of a prism is \(\sqrt{2}\) and the angle of minimum deviation is 30°. Calculate the value of the refracting angle of the prism. [HS ’03]
Solution:
Here µ = \(\sqrt{2}\); δm = 30°
Let the refracting angle of the prism be A
Example 3.
The angle of minimum deviation is the same as the angle of a glass prism of refractive index µ = \(\sqrt{3}\). What is the angle of the prism? [HS(XI) ’06; WBJEE ’01]
Solution:
Example 4.
What will be the angle of emergence of a ray of light through a prism for an angle of incidence 45° ? The angle of the prism = 60°; refractive index of the prism = \(\sqrt{2}\).
Solution:
Here A = 60°; µ = \(\sqrt{2}\)
For refraction at the first face,
µ = \(\frac{\sin i_1}{\sin r_1}\) or, \(\sqrt{2}\) = \(\frac{\sin 45^{\circ}}{\sin r_1}\)
or, r1 = 30°
We know, A = r1 + r2
∴ r2 = A – r1 = 60° – 30° = 30° = r1
Example 5.
A ray of light is incident at an angle of incidence 40° on a prism having refractive index 1.6. What should be the value of the angle of the prism for minimum deviation? Given sin 40° = 0.6428; sin 23°42′ = 0.4018.
Solution:
Here µ = 1.6.
The angle of incidence at the first face = i1 = 40°
Let the angle of emergence at the second face = i2
For minimum deviation, i1 = i2; so, i2 = 40°
Example 6.
The refracting angle of a glass prism is 60° and the refractive index of glass is 1.6. If the angle of incidence of a ray of light on the first refracting surface is 45°. Calculate the angle of deviation of the ray. Given that sin 26° 14′ = 0.4419, sin 33° 46′ = 0.5558 and sin 62°47′ = 0.8893.
Solution:
Here A = 60°; µ = 1.6
Angle of incidence on the first face = i1 = 45°
For refraction on the first face, µ = \(\frac{\sin i_1}{\sin r_1}\)
∴ sin r1 = \(\frac{\sin i_1}{\mu}=\frac{\sin 45^{\circ}}{1.6}\)
= \(\frac{1}{\sqrt{2} \times 1.6}\) = 0.4419 = sin 26.23°
or, r1 = 26.23°
We know, A = r1 + r2
∵ r2 = A – r1 = 60° – 26.23° = 33.77°
For refraction at the second face,
μ = \(\frac{\sin i_2}{\sin r_2}\) or, 1.6 = \(\frac{\sin i_2}{\sin 33.77^{\circ}}\)
or, sin i2 = 1.6 × sin 33.77°
= 1.6 × 0.5559 = 0.8894 = sin62.8c
∴ i2 = 62.8°
So, the angle of deviation,
δ = i1 + i2 – A = 45° + 62.8° – 60° = 47.8°
Example 7.
If the refracting angle of a prism is A, the refractive index of its material is µ and the angle of deviation of a ray of light incident normally on the first refracting
face is δ, then prove that µ = \(\frac{\sin (A+\delta)}{\sin A}\).
Solution:
Angle of deviation for refraction in the prism,
δ = i1 + i2 – A and A = r1 + r2
For normal incidence i1 = 0 and r1 = 0
∴ δ = i2 – A or, i2 = A + δ
and A = r2
Refractive index of the prism,
µ = \(\frac{\sin i_2}{\sin r_2}\) = \(\frac{\sin (A+\delta)}{\sin A}\)
Example 8.
A ray of light is incident at an angle of 60° on a face of a prism with refracting angle 30°. If the ray emerges from the other face and makes an angle of 30° with the incident ray then, show that the emergent ray passes perpendicularly through the refracting surface. Deter-mine the refractive index of the material of the prism.
Solution:
According to question, δ = 30°, i1 = 60° and A = 30°.
We know, δ = i1 + i2 – A
∴ 30° = 60° + i2 – 30° or, i2 = 0
∴ The emergent ray is perpendicular to the refracting surface.
Again, A = r2 + r1
As, i2 = 0, so angle of incidence of the second face, r2 = ∴ r1 = A = 30°
∴ Refractive index of the material of the prism,
µ = \(\frac{\sin i_1}{\sin r_1}\) = \(\frac{\sin 60^{\circ}}{\sin 30^{\circ}}\) = \(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\) = \(\sqrt{3}\)
Example 9.
A glass prism of refracting angle 60° and of refractive index 1.6, is immersed in water (refractive index is 1.33). What is its angle of minimum deviation? [sin36.87° = 0.6]
Solution:
Here, A = 60°; aµg = 1.6; aµw = 1.33
Example 10.
A ray of light passes through an equilateral prism in such a way that the angle of incidence becomes equal to the angle of emergence and each of these angles are \(\frac{3}{4}\) th of the angle of deviation. Determine the angle of deviation. [HS(XI) ’08]
Solution:
Here, A = 60° and i1 = i2 = \(\frac{3}{4} \delta\)
Now, δ = i1 + i2 – A = \(\frac{3}{4} \delta\) + \(\frac{3}{4} \delta\) – 60° = \(\frac{3}{2} \delta\) – 60°
or, \(\frac{1}{2} \delta\) = 60° or, δ = 120°
∴ Angle of deviation = 120°
Example 11.
The angle of minimum deviation of a glass prism of refracting angle 60° is 30° . If velocity of light in vac-uum is 3 × 108m ᐧ s-1, then determine its velocity in glass. [HS(XI) ’08]
Solution:
Refractive index of material of the prism,