A ray of light enters a diamond from the air. If the refractive index of diamond is 2.42, by what percent does the speed of the light reduce on entering the diamond.?
Answers
Answer:
242% of the speed of the light reduced on entering the diamond.
Step-by-step explanation:
Given, Refractive Index = 2.42
we know that, speed of light = 3×10^8
Hence, light reduced = 3×10^8/2.42 = 123966942
Percentage = (3×10^8 / 123966942 × 100)%
= 242 % (answer).
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Explanation:
Refractive index (μ) = c(speedoflightinvaccum)v(speedoflightinanymedium)c(speedoflightinvaccum)v(speedoflightinanymedium) Given μ = 2.42, c = 3 × 108. Putting the value in the above formula, we get ⇒ 2.42 = 3×108v3×108v ⇒ v = 3×1082.423×1082.42 ⇒ v = 1.23 × 108 m/s. Speed of light in diamond is 1.23 × 108 m/s % of speed in diamond = 1.23×1083×1081.23×1083×108x 100 = 0.41 x 100 = 41% Thus there is 41% decrease in the speed of light in diamond than in air./951231/light-enters-diamond-from-refractive-index-diamond-what-percentage-speed-light-diamond