Light is incident from air on oil at an angle of 30 degrees after moving through oil one and another medium oil 2 it enters glass and then it enters water. If the refractive indices of Glass and water are 1.5 and 1.3 respectively find the angle which is the Ray makes with normal in water.

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Solution

As we know
μ sin i= (constant) ⇒ μ sin iair = μwater sin rwater
sin 30 = 1.3 sin r

sin r = 12 × 1.3 = 12.6; r=sin^-1 (12.6)

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Light is incident from air on oil at an angle of 30 degrees after moving through oil one and another medium oil 2 it enters glass and then it enters water. If the refractive indices of Glass and water are 1.5 and 1.3 respectively find the angle which is the Ray makes with normal in water.

As we know μ sin i= (constant) ⇒ μ sin iair = μwater sin rwater sin 30 = 1.3 sin r sin r = 12 × 1.3 = 12.6; r=sin^-1 (12.6)

As we know
μ sin i= (constant) ⇒ μ sin iair = μwater sin rwater
sin 30 = 1.3 sin r

sin r = 12 × 1.3 = 12.6; r=sin^-1 (12.6)