The refractive index of the core of an optical fibre is `mu_(2)` and that of the cladding is `mu_(1)` .The angle of incidence on the face of the core so that the light ray just under goes total internal reflection at the cladding is

If refractive index of core & cladding are mu_(1)&mu_(2) respectively then :

If refractive index of core & cladding are mu_(1) & mu_(2) respectively then :

A light rays is incident on a prism of prism angle A and refractive index mu . Show that the smallest value of the angle of incidence for which the reflected ray does not suffer total internal reflection at the second face, i.e. just comes out of it, is given by i=sin^-1[(mu^2-1)^(1//2)sinA-cosA] .

An optical fibre is made of glass fibre of refractive index 1.68. The outer coating of the glass fibre is made of material of refractive index 1.44. What is the range of angles of the incident rays with the axis of the pipe for which total internal reflection inside the optical fibre takes place?

A light ray going from air is incident (as shown in figure) at one end of a optical fibre used for communication purpose (refractive index of core mu=1.5 ) making an incidence angle of 60^(@) on the lateral surface. so that it undergoes a total internal reflection. how much time would it take to traverse the straight fibre of length 1 km?

Light is incident at point A on one of the faces of a diamond crystal (mu = 2.0) . Find the maximum allowed value of angle of incidence theta so that light suffers total internal reflection at point B.

A ray of light travelling in a transparent medium f refractive index mu , falls on a surface separating the medium from air at an angle of incidence of 45^(@) . For which of the following value of mu the ray can undergo total internal reflection ?

A ray of light travelling in transparent medium of refractive index mu falls on a surface separating the medium from air at an angle of incidence of 45^(@) . For which of the following value of mu the ray can undergo total internal reflection ?