A plastic optical fiber has a polystyrene core (n=1.495) and a layer of fluorina

Question

A plastic optical fiber has a polystyrene core (n=1.495) and a layer of fluorinated polymer cladding with a slightly smaller index of refraction (n=1.402). (a) What is the speed of light in the polystyrene core? (b) If a yellow light with a wavelength of 550 nm in vacuum is sent through the fiber, what is the wavelength of the light in the fiber? (c) What is the smallest angle to the normal that the light can strike the core-cladding interface and still be totally internally reflected back into the core?<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

Explanation / Answer

a)Light in in fibre travels through core.

speed= c/1.495=2.0*1068 m/s

b) frequency of light remains same

f=c*wavelength

(3*10^8)*550*10^-9=(2*10^8)*(new wavelength)

new wavelength= 825 nm

c)Use internal reflection

Directly using the formula

sine(smallest angle)=1.402/1.495

smallest angle=69.68 degrees with normal

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.