An optical fiber consists of core of μ1 surrounded by a cladding of μ2<μ1. A beam of light enters from air at an angle α with axis of fiber. The highest α for which ray can be traveled through fiber is
A
cos−1√μ22−μ21
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
sin−1√μ21−μ22
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
tan−1√μ21−μ22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
sec−1√μ21−μ22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Dsin−1√μ21−μ22 Here the requirement is that i>c ⇒sini>sinc⇒sini>μ2μ1 ....(i) From Snell's law μ1=sinαsinr ....(ii) Also in △OBA r+i=90o⇒r=(90−i) Hence from equation (ii) sinα=μ1sin(90−i) ⇒cosi=sinαμ1 sini=√1−cos2i=√1−(sinαμ1)2 ....(iii) From equation (i) and (iii) √1−(sinαμ1)2>μ2μ1 ⇒sin2α<(μ21−μ22)⇒sinα<√μ21−μ22 αmax=sin−1√μ21−μ22