A parallel incident beam falls on a solid glass sphere at near normal incidence.Show that the image in terms of the index of refractive `mu` and the sphere of radius R is given by
`(R(2-mu))/(2(mu-1))`

A parallel incident beam falls on a solid glass sphere at normal incidenc. Prove that the distance of the fianll image after two refractions is at a distance (2-mu)//2(mu-1)a from the outer edge of the sphere. Refractive index of the sphere is mu and radius of the sphere is a.

A parallel incident beam falls on a solid glass sphere at normal incidenc. Prove that the distance of the fianll image after two refractions is at a distance (2-mu)//2(mu-1)a from the outer edge of the sphere. Refractive index of the sphere is mu and radius of the sphere is a.

A narrow parallel beam of light falls on a glass sphere of radius R and refractive index mu at normal incidence. The distance of the image from the outer edge is given by-

A parallel beam of light falls on a solid tranparent sphere. Q. If however thin beam is focussed at A, then find the refractive index of the sphere,

A parallel beam of light falls on a solid tranparent sphere. Q. For what value of refractive index mu , the thin beam can be focussed at centre of sphere.

A parallel beam falls on solid glass sphere of radius R and refractive index mu . What is the distance of final image after refractioin from two surfaces of sphere? What is the condition for the image to be real?