Light travels in a line of optical fiber by total internal reflection between a core of refractive index 1.48 and a cladding 1.46 refractive index.

The critical angle for total internal reflection to occur inside the core of the fiber is ...

80.6º
48.8º
43.2º
42.5º
24.4º

I keep getting this one wrong too. Please help

For critical reflection, the angle of refraction in the less dense medium is 90 degrees.

Snell's Law says that, in this case,
1.46 sin 90 = 1.46 = 1.48 sin theta
theta = sin^-1 1.46/1.48 = sin^-1 0.9865
Look up the angle. It is one of your choices.

inverse sin--> =

Sin^-1(1.46-1.48)= 80.57 deg
90-80.57 = The Max_ang = 9.43 deg

To find the critical angle for total internal reflection to occur inside the core of the fiber, we need to use Snell's Law and the concept of total internal reflection.

Snell's Law states that the ratio of the sine of the angle of incidence (θi) to the sine of the angle of refraction (θr) is equal to the ratio of the refractive indices of the two media involved:

n₁sin(θi) = n₂sin(θr)

In this case, the light is traveling from the core (medium 1) to the cladding (medium 2). We are given the refractive indices of the core and the cladding as 1.48 and 1.46, respectively.

To find the critical angle, we need to determine the angle of incidence for which the angle of refraction becomes 90 degrees. At this angle, any increase in the angle of incidence would result in no refraction and complete reflection within the core.

When the angle of refraction is 90 degrees, sin(θr) is 1. Therefore, the equation becomes:

n₁sin(θc) = n₂ (θc is the critical angle)

Solving for θc, we have:

sin(θc) = n₂ / n₁

θc = arcsin(n₂ / n₁)

Plugging in the refractive indices, we get:

θc = arcsin(1.46 / 1.48)

Now, you can use a scientific calculator or an online calculator to find the arcsin of (1.46 / 1.48).

Calculating it, we get:

θc ≈ 42.5 degrees.

Therefore, the correct answer is 42.5º.