A ray hits the y-axis making an angle `theta` with y-axis as shown in the figure. The variation of refractive index with x-coordinate is `mu=mu_0 (1-(x)/(d))` for `0lt=xlt=d(1-(1)/(mu_0))` and `mu=mu_0` for `x lt 0`, where d is a positive constant. The maximum x-coordinate of the path traced by the ray is

If mu_0 be the permeability and k_0 , the dielectric constant of a medium, its refractive index is given by

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.

A thin oil layer floats on water. A ray of light making an angle of incidence of 40^(@) shines on oil layer. The angle of refraction of light ray in water is [mu_(oil) = 1.45, mu_(water) = 1.33]

A thin oil layer floats on water. A ray of light making an angle of incidence of 40^(@) shines on oil layer. The angle of refraction of light ray in water is (mu_(oil)=1.45, mu_("water")=1.33)

The relation between refractive indices u, mu_(1),mu_(2) . if the behaviour of light ray is as shown in figure

In the figure shown, if friction coefficient of block 1 kg and 2kg with inclined plane is mu_(1) = 0.5 and mu_(2) = 0.4 respectively, then

A light ray enters into a medium whose refractive index varies along the x-axis as n(x) = n_(0) sqrt(1+(x)/(4)) where n_(0)=1 . The medium is bounded by the plates x =0,x =1 & y =0 . It the ray enters at the origin at an angle 30^(@) with x-axis