Question Video: Reflection | Nagwa Question Video: Reflection | Nagwa

# Question Video: Reflection Physics • Second Year of Secondary School

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What is the incident angle of the light ray shown in the diagram?

02:42

### Video Transcript

What is the incident angle of the light ray shown in the diagram?

We have our incident ray of light here that reflects off of our flat surface here. And then, this ray drawn in is the reflected ray. On our diagram, this 35-degree angle represents the angle here between a reflected ray and the surface. We want to know what is the incident angle of this light ray. Now, the first thing we can do is figure out which part of our figure represents that incident angle.

At this point, it’s important to remember that the incident angle as well as the reflected angle of a light ray, by convention, are always measured with respect to the normal line to the surface the ray reflects off of. That means that this dashed line here, the normal line to the surface that’s perpendicular to it, is a reference point for the incident as well as reflected angles of this ray. The ray’s incident angle is measured from this normal line to the ray itself. And we can represent this angle symbolically as 𝜃 sub 𝑖. And then, the reflected angle of the ray, likewise, is measured from this normal line. We start at the normal line and go to the reflected portion of that ray. And this angle we can represent symbolically as 𝜃 sub 𝑟, the angle of reflection.

Now, there’s an important relationship between 𝜃 sub 𝑖 and 𝜃 sub 𝑟 that’s worth recalling. The law of reflection tells us that the angle of incidence of a light ray is equal to its angle of reflection. Written down as an equation, we say that 𝜃 sub 𝑖 is equal to 𝜃 sub 𝑟. Looking back at our diagram then, we can see that if we can solve for 𝜃 sub 𝑟, then we’ll also have solved, by the law of reflection, for 𝜃 sub 𝑖. They’re equal. So what is 𝜃 sub 𝑟, the reflected angle of this ray?

If we look at the plane of the surface that the ray reflects off of, we know that this plane is at 90 degrees to the line we’ve drawn, called the normal line. That means that this angle here that we’ve drawn in must itself be equal to 90 degrees. And with that being the case, we can now write an equation for 𝜃 sub 𝑟 in terms of these two angles, 90 degrees and 35 degrees. Looking at the diagram, we can see that 𝜃 sub 𝑟, whatever that is, plus 35 degrees is equal to 90 degrees. We can see that this reflected angle, 𝜃 sub 𝑟, plus this 35-degree angle must sum up to 90.

So to solve for 𝜃 sub 𝑟, let’s subtract 35 degrees from both sides of this equation. When we do, that term cancels out with the positive 35 on the left-hand side. And we see that 𝜃 sub 𝑟 is equal to 90 degrees minus 35 degrees and that that is equal to 55 degrees. Now, remember that we wanted to solve for the incident angle but that by the law of reflection, that incident angle is equal to 𝜃 sub 𝑟. Therefore, we can write that 𝜃 sub 𝑖, the angle of incidence, is equal to 55 degrees. And that’s our final answer.

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