Question

Ray Optics 1 Problem Statement Light strikes a 5.0-cm thick sheet of glass at an angle of incidence in air of 50°. The sheet has parallel faces and the glass has an index of refraction 1.50. (a) What is the angle of refraction in the glass? (b) After traveling through the glass the light re-emerges into the air. What is the final angle of refraction in air? (c) As it leaves the glass, by what distance is the path of the ray is displaced from what it was before entering the glass? Visual Representation • Draw a ray diagram • Draw a line normal to the boundary • Show the ray bending in the correct direction • Label angles of incidence and refraction measured from the normal Mathematical Representation • Start with Snell's law and/or the law of reflection • Substitute in the known values • Solve for the desired unknown Check units and assess if your answer is reasonable

Answer 1

Solm- ng = 1 on 50 01-150° î O2 ng = = 1.5 t= 5cm Oz 11 ca) Let angle of refraction at A in glass is '®; So, Applying snell's law at A, na sind, = ng sind, 1* sin5o = 1.5sin Oz O2 = 30-710 Ang <b> Again applying snell's law at co Ng sin 02= na sino using ① & ② Masindi nasin as 03 - 01350° co let path is displaced by by distance, d'. Now, In A ACD, Ang Cos O2 = st ср AC AC Alan 5,815 cm 5. AC CS 30.71= 5 AC Now, In A ABC, Sina= AB d a me => Al Al d = 1.9264 / Ang Sin 19.29= 5.815 Ali o'c' &t Oro 3 = 02 02 = 50-36.71 - a 19.29°