2. The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

Explanation:

Using the Snell's law formula to solve this question which states that the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. This constant is known as the refractive index for the given pair of media. Mathematically,

n = sin(i)/sin(r) where;

i is the angle of incidence

r is the angle of refraction.

n is the refractive index.

Given the refractive index of the optical fibre n₁ = 1.60 and that of cladding n₂ = 1.48

n₂/n₁ = sin(i)/sin(r)

The light ray can make with the wall of the core when its angle of refraction is 90⁰. The angle of incidence at this maximum point is known as the critical angle.

On substitution:

1.48/1.60 = sin(i)/sin90

1.48/1.60  = sin(i)/1

sin(i) = 1.48/1.60

sin(i) = 0.925

i = sin⁻¹0.925

i = 67.66⁰

Hence the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber is 67.66⁰.

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

b.a device that inputs heat into a cool object