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Light: Reflection and Refraction - Apply laws of refraction and refractive index

Grade 10CBSE

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

🔑Concepts

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

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

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

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First Law of Refraction: The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

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Second Law of Refraction (Snell's Law): The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for light of a given color and a given pair of media.

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The refractive index of a medium is a dimensionless quantity that represents how much the speed of light is reduced inside that medium compared to a vacuum.

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Optical density is not the same as mass density. For example, kerosene has a higher refractive index than water but a lower mass density.

📐Formulae

Snell’s Law: sin⁡isin⁡r=n21\text{Snell's Law: } \frac{\sin i}{\sin r} = n_{21}

Absolute Refractive Index: nm=cv\text{Absolute Refractive Index: } n_m = \frac{c}{v}

Relative Refractive Index: n21=v1v2=n2n1\text{Relative Refractive Index: } n_{21} = \frac{v_1}{v_2} = \frac{n_2}{n_1}

Speed of light in vacuum: c≈3×108 m/s\text{Speed of light in vacuum: } c \approx 3 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

Light enters from air to glass having refractive index 1.501.50. What is the speed of light in the glass? The speed of light in vacuum is 3×108 m/s3 \times 10^8 \text{ m/s}.

Solution:

Given: ng=1.50n_g = 1.50, c=3×108 m/sc = 3 \times 10^8 \text{ m/s}. Using formula ng=cvgn_g = \frac{c}{v_g}, we get vg=cng=3×1081.50=2×108 m/sv_g = \frac{c}{n_g} = \frac{3 \times 10^8}{1.50} = 2 \times 10^8 \text{ m/s}.

Explanation:

The speed of light decreases when entering a denser medium like glass. By dividing the speed of light in vacuum by the absolute refractive index of glass, we find the velocity in that specific medium.

Problem 2:

The refractive index of water is 1.331.33 and for glass is 1.501.50. Calculate the refractive index of glass with respect to water (ngwn_{gw}).

Solution:

Given nw=1.33n_w = 1.33 and ng=1.50n_g = 1.50. The relative refractive index ngw=ngnw=1.501.33≈1.127n_{gw} = \frac{n_g}{n_w} = \frac{1.50}{1.33} \approx 1.127.

Explanation:

The relative refractive index of medium 2 with respect to medium 1 is the ratio of the absolute refractive index of medium 2 to that of medium 1.