Question

QUESTION 7 A prism made of saphire with a refractive index of 1.762 is immersed in oil with a refractive index of 1.390. Light Initially traveling horizontally in the oil enters the prism as shown below. 60.0° a) What is the refracted angle inside the prism? b) What is the refracted angle when the light exits the prism?

Answer 1

angle of incidence 102 Of given reßactive index of Prism material h = 1.762 This prism immersed in Oil with refractive inder n = 1.390 Initially light travelling Horizontally in Oil and strives at incidence argle at oil - prism interface at O, C let say) Bom normal (hud to interface (a) from figure 09=309 60 ho nisino, Msino, Oil Oil 60 sino, chi sino = sino ₂₂ # sino 2 1₂ 0, = 23.23° (from normal n 2x 1.762 The angle of refraction from Horizontal = 30-23.23 = 6.770 The angle of refraction in prism is = 23. 23° ßom hoshal 6.77° from Horizontch using shell so low h₂ hi ne 20₂ sint sino ( 1.390 (*1.76 ha (5) Now The angle LOBA = 180 -190-8, +30) 30 LOBA 60 + O₂ So The argle 190-0, n, (normal na LCBO = 0= 90-60+0) 160 0 prism ofl oct 0g = 30-0, = 30 - 23.23° so light falls on angle of incident = 6.77 inter fare of Og = 6.77°

S and Up where Oy sinou ha using shells low again (npsino, = n sin or hasino ni sinou where he duantities from incident side angle of refraction in Oil ng and of Quantities from sin Oz refraktion side ni sinou = 1.762 si 16.77) = 0.14943 Esin! (0.14943) Sinoy = 0.14943 -8.594 The defaction angle in orl when say exists from frism is 04 - 8.59 8.59* from normalne ( Bom Horizontal also 1.390 Oy 8.59° Ou