A parallel beam of light travelling in x direction is incident on a glass slab of thickness t. The refractive index of the slab changes with y as `mu = mu_(0) ( 1 – ( y2) /( y_(2)^( 0) ) ) `where `mu_(0)` is the refractive index along x axis and `y_(0)` is a constant. The light beam gets focused at a point F on the x axis. By using the concept of optical path length calculate the focal length f. Assume` f gtgt t` and consider y to be small.

Options

A. `Z_(0)^(2)//(2mu_(0)t)`
B. `Z_(0)^(2)//(mu_(0)t)`
C. `mu_(0)Z_(0)^(2)//(2t)`
D. None

Answer

A

Explanation

On comparing optical path we get
`tmu_(0)(1-Z^(2)/Z_(0))+sqrt(Z^(2)+F^(2)) = tmu_(0)+F`
`Rightarrowsqrt(Z^(2)+F^(2)) = (mu_(0)tZ^(2))/Z_(0)^(2)+F`
`RightarrowZ^(2)+F^(2)=mu_(0)^(2)t^(2)(Z^(2)/Z_(0)^(2))^(2)+F^(2)+2mu_(0)tZ^(2)/Z_(0)^(2)F`
since is very small
`Rightarrow Z^(2)=2mu_(0)tZ^(2)/Z_(0)^(2)F Rightarrow F=Z_(0)^(2)//(2mu_(0)t)`