Question

1 Find where the ray exits the block of glass of index of refraction 1.5 by accurate ray tracing and by calculation. The grid spacing is Icm. a Carefully sketch the path of the ray through the glass block. b Calculate all angles of incidence and refraction. Write the angles on the picture. c Calculate the precise location on the block (e.g. distance from some corner) from which the ray emerges from the block (to three accurate digits). Ans The first angle of refraction is 19.5'. The ray exits the block on the right side, 1.5355cm below the top. 60

Answer 1

Ans: The block in effect has the dimensions Locom x 5 cm, as shown below - F hi Pastaw and Post (b): А 90° 190-290-hi Sy B 90 90-1) G hi e ER IR. them according We see that A day is incident at point E at an angle i= 30. If he is the angle of deflachon at point E to snell's law- nasin(i) = ng sin(4) Refractive imdices of ais and glass 1 x Sm (30) = = 1.5 sin bu 15 3 x lim (1) &y= S') = 19:471220 Rounded off? nang - os fy= 19.5° At Point F: At this point ray Ef tries to refract from denser medium (glass) to sareh medium Lain So We must total Internal Reflection takes place os not.

Je sim" (73) Continued If Oc is the critical angle them Oc= sin ma ng Q = 41.18° = 41.90 and According to figure - Angle of Incidence at F- i = 90-2 90 -19.5 = 70.5° As i oc. Total Internal Reflection takes place at F. Hence- Angle of reflection de = ly = 70.50 At Point G As is clear from the figure- snell's Angle of Incidence dig = 4y = 19.5° Now Applying law. 1.5x Sin (19.5") = [X sine Angle of emergence of ray = Angle of refraction at G. Sine = 1:5 X.3338 = .50071 e = sim-4.5007)) = 30-04690 € = 30.05° = 30 Hence At point E i = 30 de = 19.5° At point F 11=70.5°, 19 = 70.5" Anc At point G - by = 19.50, 330

Continued Part 2): In the triangle. AFE AE tant, AF= AE AF tanr, Now A£ = 2cm [ As given im problem? and = 19.5 so AF = 2cm = 5.6478 tam (195) AF 5.65 cm and so FB = 10 -5.65 4.35 cm Now In triangle- ERG BG tam di FB BG = FB tanis) 4.35 tan (19.5°) BG = 1.540415 cm and G is the point at which the ray exits the gless block. Hence- Distance of point of emergence from corner B = 1.54cm Ans.
Problem is solved using Snell's law, condition of total internal reflection and basics of trigonometric functions.