Question

You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Based on the specifications, you know that the greatest angle that the light will make with the horizontal is no greater than 25⁰. Assuming you will be using the scope in the body which has the same index of refraction of water (n = 1.33). What is the minimum index of refraction n2 required for the design to be functional?

Answer 1

Answer:

Explanation:

 For entry of light into tube of unknown refractive index

sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube

r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.

sin ( 90 - 25 ) / sin( 90 - C)  = μ

sin65 / cos C = μ

sinC = 1.33 / μ  , where 1.33 is the refractive index of body liquid.

From these equations

sin65 / cos C = 1.33 / sinC

TanC = 1.33 / sin65

TanC = 1.33 / .9063

TanC = 1.4675

C= 56°

sinC = 1.33 / μ

μ = 1.33 / sinC

= 1.33 / sin56

= 1.33 / .829

μ = 1.6   Ans