# Question Video: Calculating the Critical Angle Using Distance and Time Values Physics • 9th Grade

Light travels a distance quantified by 3π₯ meters in 5π‘ seconds in air. If light travels a distance of π₯ meters in π‘ seconds in another medium, then the critical angle of this medium is οΌΏ degrees.

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### Video Transcript

Fill in the blank. Light travels a distance quantified by three π₯ meters in five π‘ seconds in air. If light travels a distance of π₯ meters in π‘ seconds in another medium, then the critical angle of this medium is blank degrees. (A) 53.13, (B) 30.96, (C) 36.87, (D) 59.04.

This question is asking us to calculate the critical angle for some unknown medium. Letβs start by reminding ourselves about the critical angle.

We can imagine a ray of light traveling through a medium and then reaching the boundary of a second medium. When this happens, the light ray is often refracted so that it passes through the second medium like this. However, if a light ray reaches the new material at a particular angle of incidence, called the critical angle, which weβll label π sub c, the light ray wonβt be transmitted through the second medium. Instead, itβll travel along the boundary, like this. This happens when the angle of refraction is equal to 90 degrees.

To figure out the critical angle, we need to recall Snellβs law of refraction. π sub i times the sin of π sub i equals π sub r times the sin π sub r, where π sub i is the refractive index of the first medium. π sub r is the refractive index of the second medium. And π sub i and π sub r are the angles of incidence and refraction.

Letβs replace the angle of incidence with the critical angle π sub c and substitute in π sub r equals 90 degrees. This leaves us with π sub i times the sin of π sub c equals π sub r times the sin of 90 degrees. The sin of 90 degrees equals one. So this just becomes π sub i times the sin of π sub c equals π sub r. Since weβre interested in the critical angle, we can rearrange this to make the sin of π sub c the subject. sin of π sub c equals π sub r over π sub i. So the sine of the critical angle is equal to the refractive index of the second medium divided by the refractive index of the first medium.

Recall that the refractive index of a medium is equal to the speed of light in vacuum π divided by the speed of light in the medium, which weβll call π£. Letβs find an expression for the refractive index of the first medium π sub i. In this case, weβre told that the medium is air. We know that light travels a distance of three π₯ meters in a time of five π‘ seconds when in this medium. Using the formula speed equals distance over time, we can find that the speed of light π£ in this medium is equal to three π₯ over five π‘ meters per second. So the refractive index π sub i is equal to π divided by three π₯ over five π‘. We can rewrite this slightly more simply as π sub i equals five π times π‘ over three π₯.

Now letβs find an expression for the refractive index of the second medium, π sub r. We know that light travels a distance of π₯ meters in π‘ seconds in this medium. So, the speed of the light in this medium is equal to π₯ over π‘ meters per second. To find the refractive index, we simply divide this into the speed of light in a vacuum to get π sub r equals π divided by π₯ over π‘ or π times π‘ over π₯.

Next, we can substitute these expressions into our formula for the sine of the critical angle: sin of π sub c equals π sub r over π sub i. Doing this, we find that the sine of the critical angle is equal to π times π‘ over π₯ divided by five π times π‘ over three π₯. We can see that the factors of π times π‘ over π₯ cancel out, leaving us with a value of three divided by five. This is equal to the sine of the critical angle.

To find the critical angle itself, we need to take the arcsine of both sides of this equation. This leaves us with the expression π sub c equals the arcsin of three divided by five. Running this through our calculator, we can find that the arcsin of three-fifths is equal to 36.8698 and so on degrees. When rounded to two decimal places, this gives us the answer 36.87 degrees. This corresponds to option (C). So the correct answer to this question is (C). The critical angle of this medium is 36.87 degrees.