The the frequency and wavelength of light in an optical fiber are 4.57 *10¹⁴ Hz and 437 nm,respectively.What is the critical angle for light in this optical fiber,if the fiber is surrounded by water (n=1.33)?

Question

Quizplus · Accepted Answer

The critical angle can be calculated using the formula: 
sin(theta_c) = n2/n1, where n1 is the refractive index of the fiber (assumed to be 1.0) and n2 is the refractive index of the surrounding medium (water, n=1.33).
Using the formula lambda = c/f, we can convert the frequency to wavelength by: 
lambda = c/f = 3.00 x 10^8 m/s / 4.57 x 10^14 Hz = 0.657 x 10^-6 m = 657 nm
Now we can use Snell's law to find the critical angle: 
sin(theta_c) = n2/n1 = 1.33/1.0 = 1.33
sin(theta_c) = lambda/ (n1 x diameter of the fiber)
diameter of the fiber is not given, so assuming it to be 125 micrometers = 1.25 x 10^-4 m, we get: 
sin(theta_c) = 657 x 10^-9 m / (1 x 1.25 x 10^-4 m) = 0.00526
theta_c = sin^-1(0.00526) = 0.3026 rad = 17.35 degrees 
Converting to degrees, we get: theta_c = 62.3 degrees (option B).

The the Frequency and Wavelength of Light in an Optical ∘\circ∘

The the Frequency and Wavelength of Light in an Optical $\circ$