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Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.
What is Limiting Angle of Incidence of a Ray?
A ray of light incident on a refracting surface of a prism may not emerge from the second refracting surface. It depends on the refracting angle of the prism. Every prism has a limiting value of its refracting angle. Light can emerge from the prism if angle of the prism is equal or less than this critical value, otherwise no light can emerge from the prism.
Let ABC be the principal section of a prism [Fig.]. PQRS is the path of a ray through the prism placed in air where ray RS grazes along the second face AC.
Let the angles of incidence and refraction at the face AB be i1 and r1 respectively and the corresponding angles at the face AC be r2 and i2,where i1 = 90°.
So, r2 = θc, the critical angle between glass and air.
If the refracting angle of the prism A is equal to the limiting angle, then the ray Incident at an angle of incidence i1 to the face AB of the prism makes a grazing emergence along the second refracting surface AC.
A = r1 + r2
For refraction at Q,
sin i1 = µsin r1 or, r1 = sin-1\(\left(\frac{\sin i_1}{\mu}\right)\)
For refraction at R,
sin 90° = µsin r2 or, r2 = sin-1\(\left(\frac{1}{\mu}\right)\)
or θc = sin-1\(\left(\frac{1}{\mu}\right)\)
From equation (1) we get,
A = sin-1\(\left(\frac{\sin i_1}{\mu}\right)\) + sin-1\(\left(\frac{1}{\mu}\right)\) ….. (2)
Special cases:
i) Limiting angle of the prism for normal incidence on the first face: When ray, PQ is incident on the face AB normally, then i1 = 0. In this case, if the emergent ray grazes along the surface AC then from equation (2) we get,
A = sin-1 \(\left(\frac{\sin 0}{\mu}\right)\) + sin-1 \(\left(\frac{1}{\mu}\right)\) = sin-1 \(\left(\frac{1}{\mu}\right)\) = θc
Hence, the ray can emerge from the prism through its second surface till the refracting angle of the prism remains less than its critical angle. But as the refracting angle of the prism becomes greater than its critical angle, no ray
emerges from the surface AC. Then the face AC acts as a total reflecting surface.
ii) Limiting angle of the prism for grazing incidence on the first face: For grazing incidence on the face AB, i1 = 90°
Then from equation (2) we get,
A = sin-1\(\left(\frac{\sin 90^{\circ}}{\mu}\right)\) + sin-1\(\left(\frac{1}{\mu}\right)\)
= sin-1\(\left(\frac{1}{\mu}\right)\) + sin-1\(\left(\frac{1}{\mu}\right)\) = 2 sin-1\(\left(\frac{1}{\mu}\right)\) = 2θc
So, if the refracting angle of the prism is greater than 2θc and if a ray is incident on the face AB grazing the surface, then it will be totally reflected from the face AC. It means, the ray will not emerge in air through the face AC. Hence,
no emergent ray will be obtained.
Thus, from the above discussions we conclude that for any incidence no ray can emerge from the prism if the angle of the prism is greater than twice the critical angle for the material with respect to the surrounding medium.
Limiting Angle of Incidence for No Emergent Ray from a Given Prism
Just like a prism has a limiting refracting angle for no emergent ray, a prism with a definite refracting angle also possesses limiting angle of incidence for no emergent ray from it. 1f the angle of incidence i1 becomes less than this limiting incident angle, then there will be no corresponding emergent ray.
Let ABC be the principal section of a prism [Fig.]. The ray of light PQ is incident at Q on the face AB. After refraction through the prism, the emergent ray RS grazes the second face AC of the prism.
Let the angles of incidence and refraction at the face AB be i1 and r1 and the corresponding angles at the face AC be r2 and i2 respectively. Here, i2 = 90°.
Now, angle of prism, A = r1 + r2 = constant.
From, A = r1 + r2; we get, r2 = A – r1
r1 reduces with the decrease of i1. Again, r2 increases with the decrease of r1.
Now, if r2 is greater than θc the ray QR is totally reflected back from the face AC inside the prism and it does not emerge in air.
So when r2 = θc then i1 = limiting angle of incidence.
If µ is the refractive index of the material of the prism then,
This is the limiting angle of incidence. If the angle of incidence is less than this limiting angle, no ray will emerge from the second face of the prism.
Numerical Examples
Example 1.
To get emergent ray from a right-angled prism its refractive index should not exceed \(\sqrt{2}\)—prove it. [AIEEE ’04]
Solution:
The condition of getting emergent ray from a prism is that the refracting angle of the prism should be equal to or less than twice the value of the critical angle θc.
Here the angle of the prism, A = 90°
∴ A ≤ 2θc or, 90° ≤ 2θc or, θc ≥ 45°
∴ sin θc ≥ sin 45° or, sin θc ≥ \(\frac{1}{\sqrt{2}}\)
∴ sin θc = \(\frac{1}{\mu}\)
∴ \(\frac{1}{\mu}\) ≥ \(\frac{1}{\sqrt{2}}\) or, µ ≤ \(\sqrt{2}\)
Example 2.
The refractive Index of a prism having refracting angle 75° is \(\sqrt{2}\). What should be the minimum angle of incidence on a refracting surface so that the ray will emerge from the other refracting surface of the prism?
Solution:
According to the question, the emergent angle is = 90°.
So for refraction at the second face of the prism,
µ = \(\frac{\sin i_2}{\sin r_2}\) = \(\frac{\sin 90^{\circ}}{\sin r_2}\)
or, sinr2 = \(\frac{1}{\mu}\) = \(\frac{1}{\sqrt{2}}\) = sin 45° or, r2 = 45°
know, A = r1 + r2
∴ 75° = r1 + 45° or, r1 = 30°
For refraction at the first face,
µ = \(\frac{\sin i_1}{\sin r_1}\) or, \(\sqrt{2}\) = \(\frac{\sin i_1}{\sin 30^{\circ}}\)
or, sin i1 = \(\sqrt{2} \times \frac{1}{2}\) = \(\frac{1}{\sqrt{2}}\) = sin 45° or, i1 = 45°
∴ The required angle of incidence = 45°.
Example 3.
Find the value of the limiting angle of incidence if the refractive index of the material of prism is 1.333 and angle of prism is 60° . [WBCHSE Sample Question]
Solution:
Here, refractive index of the material of the prism, µ = 1.333; angle of prism, A = 60°
Example 4.
The refracting angle of a prism is 60° and its refractive index is \(\sqrt{\frac{7}{3}}\). What should be the minimum angle of incidence on the first refracting surface so that the ray can emerge somehow from the second refracting sur-face?
Solution:
Let i be the limiting angle of incidence, then sini = \(\sqrt{\mu^2-1}\) ᐧ sinA – cos A
= \(\sqrt{\frac{7}{3}-1}\) ᐧ sin60° – cos 60°
= \(\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2}\) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) = sin 30°
∴ i = 30°
Example 5.
The refractive index of the material of a prism is \(\sqrt{\frac{3}{2}}\) and the refracting angle is 90°. Calculate the angle of minimum deviation and the corresponding angle of incidence. Show that the limiting angle of incidence for getting emergent ray is 45°.
Solution:
Here, refractive index of the material of the prism, µ = \(\sqrt{\frac{3}{2}}\); angle of prism, A = 90°.
∴ For minimum deviation, angle of incidence = 60°
To obtain the emergent ray let i be the limiting angle of incidence. Then,
sin i = \(\sqrt{\mu^2-1}\)sin A – cos A
= \(\sqrt{\frac{3}{2}-1}\) ᐧ sin 90° – cos 90° = \(\frac{1}{\sqrt{2}}\) = sin 45°
∴ i = 45°
Example 6.
The refractive index of a prism is \(\sqrt{2}\). A ray of light is incident on the prism grazing along one of its refracting surface. What should be the limiting angle of the prism for no emergent ray from the other face?
Solution:
If the limiting angle of the prism for no emergent ray is A, then
A = 2sin-1\(\frac{1}{\mu}\) = 2sin-1\(\frac{1}{\sqrt{2}}\) = 2 × 45° = 90°