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Ray Optics and Optical Instruments - Refraction through a Prism

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A prism is a transparent refracting medium bounded by two plane surfaces inclined at an angle called the Angle of Prism (AA).

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When a ray of light passes through a prism, it undergoes refraction at two surfaces. The total deviation produced is the Angle of Deviation (δ\delta).

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The relation between the angle of incidence (ii), angle of emergence (ee), angle of prism (AA), and angle of deviation (δ\delta) is given by δ=i+e−A\delta = i + e - A.

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The sum of the angles of refraction at the two faces is equal to the angle of prism: r1+r2=Ar_1 + r_2 = A.

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At a specific angle of incidence, the deviation is minimum (δm\delta_m). At this position, the ray passes symmetrically through the prism, such that i=ei = e and r1=r2=A2r_1 = r_2 = \frac{A}{2}.

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For thin prisms (where AA is very small), the angle of deviation is nearly independent of the angle of incidence: δ≈(n−1)A\delta \approx (n - 1)A.

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Dispersion occurs because the refractive index (nn) of the material depends on the wavelength (λ\lambda) of light, causing different colors to deviate by different amounts.

📐Formulae

A=r1+r2A = r_1 + r_2

δ=i+e−(r1+r2)=i+e−A\delta = i + e - (r_1 + r_2) = i + e - A

n=sin⁡(A+δm2)sin⁡(A2)n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

δ=(n−1)A (For thin prisms)\delta = (n - 1)A \text{ (For thin prisms)}

i=A+δm2 (At minimum deviation)i = \frac{A + \delta_m}{2} \text{ (At minimum deviation)}

💡Examples

Problem 1:

A ray of light is incident on an equilateral glass prism (n=1.5n = 1.5) such that the angle of incidence is 45∘45^\circ. Calculate the angle of deviation if the light passes through the prism symmetrically.

Solution:

Given A=60∘A = 60^\circ (equilateral prism) and i=45∘i = 45^\circ. Since the light passes symmetrically, it is in the position of minimum deviation, meaning i=ei = e. Using the formula δm=i+e−A\delta_m = i + e - A, we get δm=45∘+45∘−60∘=30∘\delta_m = 45^\circ + 45^\circ - 60^\circ = 30^\circ.

Explanation:

Symmetric passage of light implies the condition of minimum deviation where the angle of incidence equals the angle of emergence.

Problem 2:

Calculate the refractive index of the material of an equilateral prism for which the angle of minimum deviation is 60∘60^\circ.

Solution:

Given A=60∘A = 60^\circ and δm=60∘\delta_m = 60^\circ. Using the prism formula: n=sin⁡(A+δm2)sin⁡(A2)n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}. Substituting values: n=sin⁡(60∘+60∘2)sin⁡(60∘2)=sin⁡(60∘)sin⁡(30∘)=3/21/2=3≈1.732n = \frac{\sin\left(\frac{60^\circ + 60^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(60^\circ)}{\sin(30^\circ)} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1.732.

Explanation:

The prism formula relates the refractive index to the angle of the prism and the angle of minimum deviation.