Ploblem 1.4 (37.5 points) (Graded-index waveguide) The figure below shows a and and B, starting planar...

The following figure shows the ray diagram and the system. 18/2 -------y=0 (a) Calculate the distance d travelled by ray-A as follows: d=0A+AO Using trigonometric relations. O2 = cos OA 8/2 cos

Since, OA= AO Therefore, cos Critical angle for surface boundary at point-A is given as follows: Therefore, Cos 04 = V1-sin? 04 Ray-A hits the point-A at critical angle. Therefore, dacoso (01-19, 1m3) The velocity of light in medium-1 is v; =%.

Therefore, time taken by ray-A to cover distance d is given as follows:

(b) The distance d, covered by ray-B to reach point O' is given as follows: di = OB+BB'+B'B"+B"O' Since, OB=B"O BB'= B'B" Therefore, d. = 20B +2 BB Using trigonometric relations. 812 - coses OR OB = 8/2 ...... (1 cos Og Use Snell's law to get the following. n sin 0g = n, sin 03 Therefore, sin 0, -n sin Since 0g is critical angle at boundary of n, & nz.

Therefore, sin , = 12 sing 7,92 ho n Therefore, [email protected] = V1-sin oz Substitute cos 0, from above-mentioned equation to equation-1. 8/2 OB= V1-(ng/mL) Therefore, time taken by ray-A to cover the distance OB is given as follows: ty=- 18/2 ci-(ng/m2) Using the same approach as in part-A. The time taken to cover the distance BB' is given as follows:

128 c1-01, / , ) Therefore, the time taken to cover the distance d, is given as follows: t'= 2(t+t) col_120 9 8/2 Top1- (13/n,) cdi- (13/) 2n;8 + 18 |CV1- (13/n) -(13/1)

x=x-2169] Since, n=n, at y=- *:=x[1-2-()] Substitute 2 (A) =in to Substitute 24 = = in the above-mentioned equation. n} = n} [1-6] Therefore, |3x3 = "? (1-6) Now, v=re[1-24(97 Since, n= n; at y = 38/2. Substitute y = 38/2 and n=n; in the above-mentioned equation.

:=[1-20) =--=(2009) ) Substitute 2A ute 218 = & in the above-mentioned equation. = w[1–3” ] (d) The condition for which ray-A and ray-B arrive at same time on point O' is given as follows: t = t' Therefore, no - - = + - - 208 c(-(), (n)) Cl-( (n) no -(x / ) Rearrange as follows: - - -

(n-vor my a* -lo mog bor to my In part (c), following has been proved. n} = 1? (1-8) and n} = 1? (1–3" €) Therefore, * = VI-e) and = V1-5*) Substitute above-mentioned conditions in equation-1.

+ (11-(10-0)) (11-346) V-641-30) -(1-39 €) 2(1-E -E-(1-3 € 1 2(1-8) 1 (VE) 7 32-1)* 130€ Therefore, 2(1-E) 1 |(34 1 jaz+zaz-1=0 (e) The value of a can be calculated by solving equation - 2(1-€) a/2+ 1 2 -1=0 for a as follows: 2(1-E) 1 ...... (2) (30 – 1)arz tzaz -1=0 For 8 0, is given as follows:

2a = 0 Therefore, substitute E= 0 in equation-2. 2(1-0) 1 27 2012-1=0 Rearrange as follows: ...... (3) The above-mentioned equation is transcendental in nature and Its solution can be found 2 by graphical method by plotting 7 2,127 +d2 = y(@) and 1= y(@).

The intersection point of above-mentioned equations is value of a for which equation-3 is satisfied. 12 14/2+ 3 2012 3- 13 (2.607,1) y(a)=1 8 10 12 Therefore, ray-A and ray-B meet at same time if, la = 2.6071.