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If the refractive index of the second me...

If the refractive index of the second medium with respect to the first medium is `._(2)n_(1)` and that of the third medium with respect to the second medium is `._(3)n_(2)`, what and how much is `._(3)n_(1)` ?

Text Solution

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`._(3)n_(1)` is the refractive index of the third medium with respect to the first medium.
`._(2)n_(1)= ( v_(1))/(v_(2)) , ._(3)n_(2) = ( v_(2))/(v_(3)) , ._(3)n_(1)= ( v_(1))/(v_(3))= ( v_(1))/(v_(3)) xx(v_(2))/(v_(3))`
`:. ._(3)n_(1) = ._(2)n_(1) xx ._(3)n_(2)`
[Suppose medium 1`-=` air, medium 2 `-=` ice and medium 3 `-=` diamond . Then, `._(2)n_(1) =1.31,._(3)n_(2) = 1.847 ` `:. ._(3)n_(1) = ._(2)n_(1) xx ._(3)n_(2) = 1.31 xx 1.847 = 2.42` which is the refractive index of diamond with respect to air.]
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