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Physics - Refraction of Light through Plane Surfaces

Grade 10ICSE

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

🔑Concepts

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Refraction is the change in the direction of the path of light as it passes from one transparent medium to another due to a change in its speed.

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When light travels from an optically rarer medium to a denser medium, it bends towards the normal (i>ri > r), and its speed decreases.

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When light travels from an optically denser medium to a rarer medium, it bends away from the normal (i<ri < r), and its speed increases.

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The refractive index of a medium depends on the nature of the medium, the temperature, and the color (wavelength) of light. Refractive index decreases with an increase in wavelength: μviolet>μred\mu_{violet} > \mu_{red}.

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Principle of Reversibility: If the path of a ray of light is reversed, it follows exactly the same path in the opposite direction, implying aμg=1gμa_a\mu_g = \frac{1}{_g\mu_a}.

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Lateral Displacement is the perpendicular distance between the incident ray and the emergent ray when light passes through a parallel-sided glass slab.

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Factors affecting lateral displacement: (i) Thickness of the glass slab (tt), (ii) Angle of incidence (ii), and (iii) Refractive index of the glass (μ\mu).

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The phenomenon of Total Internal Reflection (TIR) occurs when light travels from a denser to a rarer medium at an angle of incidence greater than the critical angle (CC).

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Conditions for TIR: (i) Light must travel from a denser to a rarer medium, (ii) The angle of incidence must be greater than the critical angle (i>Ci > C).

📐Formulae

μ=sin⁡isin⁡r\mu = \frac{\sin i}{\sin r}

1μ2=v1v2=λ1λ2_1\mu_2 = \frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2}

μ=cv\mu = \frac{c}{v}

μ=Real DepthApparent Depth\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}

Shift=Real Depth×(1−1μ)\text{Shift} = \text{Real Depth} \times \left(1 - \frac{1}{\mu}\right) Rose by height

sin⁡C=1μ\sin C = \frac{1}{\mu}

δ=(i1+i2)−A\delta = (i_1 + i_2) - A

A=r1+r2A = r_1 + r_2

💡Examples

Problem 1:

A ray of light is incident on a glass slab of refractive index μ=1.5\mu = 1.5. If the speed of light in vacuum is 3×108 m s−13 \times 10^8 \text{ m s}^{-1}, calculate the speed of light in the glass slab.

Solution:

Given: μ=1.5\mu = 1.5, c=3×108 m s−1c = 3 \times 10^8 \text{ m s}^{-1}. Using the formula μ=cv\mu = \frac{c}{v}, we get v=cμ=3×1081.5=2×108 m s−1v = \frac{c}{\mu} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m s}^{-1}.

Explanation:

The speed of light in a medium is inversely proportional to its refractive index.

Problem 2:

A water tank appears to be 2.7 m2.7 \text{ m} deep when viewed from above. If the refractive index of water is 43\frac{4}{3}, find the actual depth of the tank.

Solution:

Given: Apparent Depth =2.7 m= 2.7 \text{ m}, μ=43\mu = \frac{4}{3}. Using μ=Real DepthApparent Depth\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}, Real Depth=μ×Apparent Depth=43×2.7=3.6 m\text{Real Depth} = \mu \times \text{Apparent Depth} = \frac{4}{3} \times 2.7 = 3.6 \text{ m}.

Explanation:

Refraction causes objects in a denser medium to appear closer to the surface than they actually are.

Problem 3:

Calculate the critical angle for a glass-air interface if the refractive index of glass is 2\sqrt{2}.

Solution:

Given: μ=2\mu = \sqrt{2}. The formula for critical angle is sin⁡C=1μ\sin C = \frac{1}{\mu}. Therefore, sin⁡C=12\sin C = \frac{1}{\sqrt{2}}. Since sin⁡45∘=12\sin 45^\circ = \frac{1}{\sqrt{2}}, we have C=45∘C = 45^\circ.

Explanation:

The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90∘90^\circ.