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(a) A coin kept inside water [µ=43\dfrac{4}{3}] when viewed from air in a vertical direction appears to be raised by 3.0 mm. Find the depth of the coin in water.

(b) How is the critical angle related to the refractive index of a medium?

Refraction Plane Surfaces

ICSE 2023

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Answer

(a) Given, µ=43\dfrac{4}{3}

Shift = 3.0 mm

We know,

refractive index = real depthapparent depth\dfrac{\text{real depth}}{\text{apparent depth}}

43\dfrac{4}{3} = xapparent depth\dfrac{\text{x}}{\text{apparent depth}}

apparent depth = 3x4\dfrac{\text{3x}}{4}

Shift = real depth - apparent depth

3=x−3x4⇒3=4x - 3x4⇒3=x4⇒x=4×3=123 = \text{x} - \dfrac{\text{3x}}{4} \\[1em] \Rightarrow 3 = \dfrac{\text{4x - 3x}}{4} \\[1em] \Rightarrow 3 = \dfrac{\text{x}}{4} \\[1em] \Rightarrow \text{x} = 4 \times 3 = 12

∴ Depth of coin in water = 12 mm

(b) Critical angle [C] = sin-1 1μ\dfrac{1}{μ}

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