Figure P22.47 shows the path of a beam oflight through several layers with diffe

Question

   Figure P22.47 shows the path of a beam oflight through several layers with different indices of refraction.(n4 = 1.11) IMAGE: http://s1002.photobucket.com/albums/af149/jakeloftin2186/?action=view&current=6.jpg (a) If 1 = 23.0°, what is the angle2 of the emerging beam?
1°
(b) What must the incident angle 1 be inorder to have total internal reflection at the surface between themedium with n = 1.20 and the medium withn4 = 1.11?
2°    Figure P22.47 shows the path of a beam oflight through several layers with different indices of refraction.(n4 = 1.11) IMAGE: http://s1002.photobucket.com/albums/af149/jakeloftin2186/?action=view&current=6.jpg (a) If 1 = 23.0°, what is the angle2 of the emerging beam?
1°
(b) What must the incident angle 1 be inorder to have total internal reflection at the surface between themedium with n = 1.20 and the medium withn4 = 1.11?
2° (a) If 1 = 23.0°, what is the angle2 of the emerging beam?
1°
(b) What must the incident angle 1 be inorder to have total internal reflection at the surface between themedium with n = 1.20 and the medium withn4 = 1.11?
2°

Explanation / Answer

at the interface between the materials having refractive indices1.6 and 1.4 1.6 sin23=1.4sinr1 from this calculate r1 this r1 will become angle of incidence for the materials havingrefractive indices 1.4 and 1.2 again 1.4sinr1=1.2 sinr2 calculate r2 finally 1.2sinr2=1.11sin2 calculating we will get the value of 2 the angle emergence. part b for total internal reflection the formula is 1.2/1.11=1/sinr2 calculating r2 and moving backwards we will get the value ofincident angle 1

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