Friday, June 19, 2015

We See You Spidey!

No, a screen in front of an object is not going to make the object invisible.  To be invisible, all light rays from around the object must bend around to avoid the object and emerge in the same direction on the opposite side.  Consider a sphere of decreasing refractive index along the radial line towards the center,


Light rays from one side of the sphere are bent by decreasing refractive index and emerge parallel to the incidence ray on the opposite side.  Rays can still be blocked from within such a sphere, but what if instead \(n(r)\) is made less dense than the surrounding and then made to increase progressively toward the center of the sphere.


In this case an object at the center of such a sphere can avoid the light rays round it.   But first \(n(r=r_o)\) must be made less then \(1\).  How can that be done?

We know from Snell's Law,

\(\cfrac{\psi_{n\,i}}{\psi_{n\,r}}=\cfrac{sin\,\theta}{sin\,\alpha}=n\)

If \(\psi_{n\,r}\) were to increase at the point of incidence,

\({\psi_{n\,r}}\gt{\psi_{n\,i}}\)

then it would be as if,

\(\cfrac{\psi_{n\,i}}{\psi_{n\,r}}=n\lt1\)

There after the ray is refracted progressively by smaller refractive angle as \(n(r)\) increases.  In effect the ray bends concave around the center of the sphere.

How then to increase \({\psi_{n\,r}}\) at the surface of the sphere?