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What is the Refractive Index of a Thin Prism?
Definition: The prism, whose refracting angle is very small (not more than 10°), is called a thin prism.
Deviation produced by a thin prism: ABC is a thin prism [Fig.]. A ray PQ is incident on the refracting face AB nearly normally. For nearly normal incidence, i1 \(\simeq\)0 ∴ i2\(\simeq\)0 .
If µ is the refractive index of the material of the prism, then
µ = \(\frac{\sin i_1}{\sin r_1}\) = \(\frac{i_1}{r_1}\) or, i1 = µr1
and µ = \(\frac{\sin i_2}{\sin r_2}\) = \(\frac{i_2}{r_2}\) or, i2 = µr2
So the deviation of the ray,
δ = i1 + i2 – A = µr1 + µr2 – A = µ(r1 + r2) – A
= µA – A [∵ r1 + r2 = A]
=(µ – 1)A
Again, if the ray PQ is incident on the face AB normally, then i1 = r1 = 0. So, A = r2.
Therefore, the deviation of the ray,
δ = i1 + i2 – A = µr2 – A = µA – A = (µ – 1)A
So, for normal and nearly normal incidence, the deviation of a ray in a thin prism,
δ = (µ – 1)A
Thus it is seen that for normal or nearly normal incidence, the deviation of a ray in a thin prism depends only on the refracting angle of the prism and the refractive index of its material but not on the angle of incidence. So if the angle of the incidence is small, the deviation of a ray in case of a thin prism remains constant.
Numerical Examples
Example 1.
A very thin prism deviates a ray of tight through 5°. If the refractive Index of the material of the prism is 1.5, what is the value of the angle of prism?
Solution:
Angle of deviation for a thin prism,
δ = (µ – 1)A …… (1)
Here, δ = 5° and µ = 1.5
Therefore from equation (1) we get,
5° = (1.5 – 1) or, A = 10°
Example 2.
A prism having refracting angle 4° is placed in air Calculate the angle of deviation of a ray incident normally or nearly normally on it. The refractive index of the material of the prism = \(\frac{3}{2}\) [HS ‘94]
Solution:
The refracting angle of the prism, A = 4°. So it is a thin prism. We know that the deviation of a ray in a thin prism for normal or nearly normal incidence is given by,
δ = (µ – 1)A ∴ δ =(\(\frac{3}{2}\) – 1) × 4° = 2°
Example 3.
A thin prism with refracting angle 5° and having refractive index 1.6 is kept adjacent to another thin prism having refractive index 1.5 such that one is Inverted with respect to the other. An Incident ray fall-ing vertically on the first prism passes through the second prism without any deviation. Calculate the refracting angle of the second prism.
Solution:
According to the condition,
(µ1 – 1)A1 = (µ2 – 1)A2
or, (1.6 – 1) × 5° = (1.5 – 1)A2 or, A2 = \(\frac{0.6 \times 5}{0.5}\) = 6°
So the refracring angle of the second prism = 6°.