The `xz` plane separates two media `A` and `B` with refractive indices `mu_(1)` & `mu_(2)` respectively. A ray of light travels from `A` to `B`. Its directions in the two media are given by the unut vectors, `vec(r)_(A)=a hat i+ b hat j` & `vec(r)_(B) alpha hat i + beta hat j` respectively where `hat i` & `hat j` are unit vectors in the `x` & `y` directions. Then :

Options

A. `mu_(1) a = mu_(2) alpha`
B. `mu_(1) alpha = mu_(2) a`
C. `mu_(1) b mu_(2) beta`
D. `mu_(1) beta = mu_(2) b`

Answer

A

Explanation

We have `cos theta =(vec(A).vec(B))/(ab)`
Similarly here
`cos i=(vec(r)_(A).hat(j))/(|r_(A).|.|1|)=((a hat(i)+b hat(j)).hat(j))/(sqrt(a^(2)+b^(2)))=(b)/(sqrt(a^(2)+b^(2)))`
and `cos r=(vec(r)_(B).hat(j))/(|r_(B)|.|1|)=((alpha hat(i)+beta hat(j)).hat(j))/(sqrt(alpha^(2)+beta^(2)))`
`= (beta)/(sqrt(alpha^(2)+beta^(2)))`
From shell's law
`mu_(1),sin i=mu_(2) sin r`
`rArr mu_(1) sqrt(1-cos^(2)i)=mu_(2) sqrt(1-cos^(2)r)`
`rArr mu_(1)(a)/(sqrt(b^(2)+a^(2)))=mu_(2) (alpha)/(sqrt(alpha^(2)+beta^(2)))`
`rArr mu_(1) sqrt(1-(b^(2))/(b^(2)+a^(2)))=mu_(2)sqrt(1-(beta^(2))/(alpha^(2)+beta^(2)))`
both `vec(r)_(A)` & `vec(r)_(B)` are unit vectors so magnitudes of both vectors is `1`
i.e. `sqrt(b^(2)+a^(2))=sqrt(alpha^(2)+beta^(2))=1`
So `mu_(1)a=mu_(2)alpha`.