Concepts

Class 10
Chapter 10 Class 10 - Light - Reflection and Refraction

There are 2 main laws of Refraction of Light

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## Law 1

The Incident Ray,Refracted Ray and Normal Ray all lie on the same plane

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## Law 2

The Ratio of sin of Angle of Incidence and Angle of Refraction is constant for a particular medium

This constant is called Refractive Index

It is actually called Refractive Index of the second medium with respect to the first medium.

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The second law of refraction is also called Snell's Law

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Let's do some examples

## A beam of light passes from air into a substance X. If the angle of incidence be 72Β° and the angle of refraction be 40Β°, calculate the refraction index of substance X. (Given: sin 72Β° = 0.951 and sin 40Β° = 0.642 )

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### Transcript

Example Suppose for a ray of Light. Angle of Incidence = 37Β° and Angle of Refraction = 24Β°. Find Refractive Index (Given: sin 37Β° = 0.6 and sin 24Β° = 0.4) We know that Refractive Index = (π ππ (π΄ππππ ππ πππππππππ))/(π ππ (π΄ππππ ππ πππππππ‘πππ)) Refractive Index = (π ππ (37Β°))/(π ππ (24Β°)) Refractive Index = 0.6/0.4 Refractive Index = 6/4 Refractive Index = 3/2 Refractive Index = 1.5 Example A beam of light passes from air into a substance X. If the angle of incidence be 72Β° and the angle of refraction be 40Β°, calculate the refraction index of substance X. (Given: sin 72Β° = 0.951 and sin 40Β° = 0.642 ) Given, Angle of incidence = β i = 72Β° Angle of Refraction = β r = 40Β° We have to find Refraction index of substance X By second law of refraction, Refractive index = sinβ‘π/sinβ‘π = sinβ‘γ72Β°γ/sinβ‘γ40Β°γ = 0.951/0.642 = 951/642 = 317/214 = 1.48 Refractive index of substance X is 1.48