A optical fiber consists of a cylindrical core n=1.46 surrounded by a concentric cylindrical jacket cladding n=1.43. The cladding in turn, is surrounded by air n=1.00. An incoming ray of light has a angle alpha to the central axis of the core. For what range of angles for alpha will the ray experience total internal reflection at the core-cladding interface?
Then what will the sketch look like until it reaches the right side of the optical fiber?

I do not know where to even begin.

madam, i do not know what core/ clads are but my calculation shows that between 0 and 90 degrees, alpha has to be less than 17degrees .( on both sides, ie. a cone of 17 degree semi vertical angle)
if this matches answers, then i will tell u to just draw the figure neatly, apply snell law at two surfaces, one at air -core interface when ray enters core and the other at core - clad interface and apply total internal reflection conditions.if u need further explanation, please reply.
the relevant equations are 1 sin(alpha)=1.46 sin theta eq1 ......snell law at core air interface
so angle of incidence at core clad interface=90 - theta
critical angle = sin inverse(1.43/1.46)
so for TOTAL INT REFLECSION, 90 - theta>sin inverse(1.43/1.46)
so theta<90- sin inverse (1.43/1.46), take sine on both sides
sin theta<cos(sin inverse 1.43/1.46)
ie. sin theta<0.201677
now use eq 1 to replace sin theta, sin alpha/1.46<.201677
or, sin alpha<0.294448637
taking inverse
alpha< 17.1244 degrees.

I've attached a drawing of the Optic Fiber in my Physics problem for your review. Can you help with the 'rays' from here?

Attached Files

Sample Drawing.doc (24.0 KB, 5 views)