At the interface between two materials having refractive indices n1 and n2, the critical angle for reflection of an em wave is 1C. The n2 material is replaced by another material having refractive index n3 such that the critical angle at the interface between n1 and n3 materials is 2C. If n3 > n2 > n1; n2n3 = 25 and 2C - 1C = 12, then 1C is :
- -1 ( 16n1 )
- -1 ( 13n1 )
- -1 ( 56n1 )
- -1 ( 23n1 )
Answer
Solution
Since n2 > n1 and n3 > n1, total internal reflection takes place for light going from n2 (or n3) into n1:
1C = n1n2, 2C = n1n3 = n1n2 n2n3 = 25 1C
2C - 1C = - 35 1C, which is negative and can never equal + 12.
Taking the magnitude, 35 1C = 12 1C = 56, i.e. 1C = -1 ( 56 ), which does not depend on n1.
None of the given options is correct; the question was awarded as bonus.