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Q.If the critical angle of water with respect to air is 48.75 and sin 48.75=0.75, cos 48.75=0.65 and tan 48.75=1.14 approximately, what will be the refractive index of water?

Assam AhsecAHSEC Higher Secondary (HS) Final Examination 2024Subjective· 1mImportance★★★★★
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Refractive index of water n=1/sin⁡C≈1.33n = 1/\sin C \approx 1.33.

At the critical angle CC, light travelling from the denser medium (water) to the rarer medium (air) refracts at 90∘90^\circ. By Snell's law applied at the water-air interface:

nsin⁡C=1×sin⁡90∘=1n \sin C = 1 \times \sin 90^\circ = 1

n=1sin⁡Cn = \frac{1}{\sin C}

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