Show the relation sin i c = 1 μ d r

When angle of incidence becomes equal to critical angle i = i c . By definition of critical angle if angle of incidence in the denser medium is greater than critical angle light ray will reflect back to the same medium. In the diagram ray A O is the incident ray and ray O B is the reflected ray. Also, the angle of refraction becomes 90 ° . By snell's law we can write. μ d sin i = μ r sin r . ∴ sin i sin r = μ r μ d . for i = i c r = 90 ° . ∴ sin i c sin 90 ° = μ r d . Now, by relative refractive index, we can write μ r d = 1 μ d r . Also, sin 90 ° = 1 ∴ sin i c = 1 μ d r .

E M B I B E

S CHAND'S ICSE PHYSICS BOOK 2>Chapter 4 - Refraction at Plane Surfaces>QUESTIONS>Q 21

EASY

10th ICSE

IMPORTANT

Earn 100

Apparent depth of a swimming pool is $1.5 m$ . What is the real depth of the pool? $μ_{w} = 1.33$

Important Questions on Refraction at Plane Surfaces

HARD

10th ICSE

IMPORTANT

S CHAND'S ICSE PHYSICS BOOK 2>Chapter 4 - Refraction at Plane Surfaces>QUESTIONS>Q 22

A bird is flying at the height of

3 m

above the river surface while a fish is

4 m

below the surface. At what depth would the fish appear to the bird? At what height the bird would appear to the fish? Given refractive index of water with respect to air

({}_{a}μ_{w}) = 4 / 3

HARD

10th ICSE

IMPORTANT

S CHAND'S ICSE PHYSICS BOOK 2>Chapter 4 - Refraction at Plane Surfaces>QUESTIONS>Q 23

Figure shows two unmixable transparent liquids refractive indices $μ_{1}$ and $μ_{2}$ filled in a container. The depth of each liquid is $d$ . Calculate apparent depth of a coin placed at bottom of container.

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