Question

Light of wavelength 425.0nm enters the end of an optical fiber from air at an angle of 10.25 with respect to the normal. Its wavelength inside the fiber is 314.5nm. i. What is the index of refraction inside the fiber? ii. What is the angle between the light ray and the normal inside the fiber? iii. What is the angle between the light ray and the surface when the light reaches the upper edge? iv. If the index of refraction outside the upper edge of the fiber is 1.44, what is the angle between the light and the normal to the surface as it exits the upper edge?

Answer 1

Answer:

The correct answer was given: Brain

answer:

{i}^{2} r

explanation:

power = v×i

= ir×i (ohms law )

= i^2r

Answer 2

Explanation:

thanks for points mate ............