Practice: What is the distance, d, between the incoming and outgoing rays?
|Ch 01: Intro to Physics; Units||1hr & 22mins||0% complete|
|Ch 02: 1D Motion / Kinematics||4hrs & 13mins||0% complete|
|Ch 03: Vectors||2hrs & 43mins||0% complete|
|Ch 04: 2D Kinematics||2hrs||0% complete|
|Ch 05: Projectile Motion||2hrs & 57mins||0% complete|
|Ch 06: Intro to Forces (Dynamics)||3hrs & 20mins||0% complete|
|Ch 07: Friction, Inclines, Systems||2hrs & 43mins||0% complete|
|Ch 08: Centripetal Forces & Gravitation||3hrs & 47mins||0% complete|
|Ch 09: Work & Energy||1hr & 58mins||0% complete|
|Ch 10: Conservation of Energy||2hrs & 49mins||0% complete|
|Ch 11: Momentum & Impulse||3hrs & 45mins||0% complete|
|Ch 12: Rotational Kinematics||3hrs & 3mins||0% complete|
|Ch 13: Rotational Inertia & Energy||7hrs & 4mins||0% complete|
|Ch 14: Torque & Rotational Dynamics||2hrs & 10mins||0% complete|
|Ch 15: Rotational Equilibrium||4hrs & 8mins||0% complete|
|Ch 16: Angular Momentum||3hrs & 6mins||0% complete|
|Ch 17: Periodic Motion||2hrs & 16mins||0% complete|
|Ch 19: Waves & Sound||3hrs & 25mins||0% complete|
|Ch 20: Fluid Mechanics||4hrs & 31mins||0% complete|
|Ch 21: Heat and Temperature||3hrs & 15mins||0% complete|
|Ch 22: Kinetic Theory of Ideal Gases||1hr & 44mins||0% complete|
|Ch 23: The First Law of Thermodynamics||1hr & 28mins||0% complete|
|Ch 24: The Second Law of Thermodynamics||3hrs & 9mins||0% complete|
|Ch 25: Electric Force & Field; Gauss' Law||3hrs & 34mins||0% complete|
|Ch 26: Electric Potential||1hr & 55mins||0% complete|
|Ch 27: Capacitors & Dielectrics||2hrs & 2mins||0% complete|
|Ch 28: Resistors & DC Circuits||3hrs & 20mins||0% complete|
|Ch 29: Magnetic Fields and Forces||2hrs & 34mins||0% complete|
|Ch 30: Sources of Magnetic Field||2hrs & 30mins||0% complete|
|Ch 31: Induction and Inductance||3hrs & 38mins||0% complete|
|Ch 32: Alternating Current||2hrs & 37mins||0% complete|
|Ch 33: Electromagnetic Waves||1hr & 12mins||0% complete|
|Ch 34: Geometric Optics||3hrs||0% complete|
|Ch 35: Wave Optics||1hr & 15mins||0% complete|
|Ch 37: Special Relativity||2hrs & 10mins||0% complete|
|Ch 38: Particle-Wave Duality||Not available yet|
|Ch 39: Atomic Structure||Not available yet|
|Ch 40: Nuclear Physics||Not available yet|
|Ch 41: Quantum Mechanics||Not available yet|
|Ray Nature Of Light||11 mins||0 completed|
|Reflection Of Light||12 mins||0 completed|
|Refraction Of Light||32 mins||0 completed|
|Total Internal Reflection||13 mins||0 completed|
|Ray Diagrams For Mirrors||36 mins||0 completed|
|Mirror Equation||20 mins||0 completed|
|Refraction At Spherical Surfaces||10 mins||0 completed|
|Ray Diagrams For Lenses||23 mins||0 completed|
|Thin Lens And Lens Maker Equations||25 mins||0 completed|
Concept #1: Reflection of Light
Hey guys, in this video we're going to talk about the reflection of light off of a boundary between two media, let's get to it. Now remember when a wave encounters a boundary it can do one of two things, it can either reflect off of that boundary or it can transmit through that boundary and propagate in the new media in reality waves are going to do a little bit of both now for light boundary's are typically referred to or media I should say are typically referred to as one of two things they can either be reflective like the surface of a mirror and mainly only allow reflection or they can be transparent like glass and mainly allow transmission. For now we want to talk about reflective surfaces alright light reflects off of a boundary in the same manner as a ball undergoing an elastic collision with a wall, if we have a ball at some point with some momentum right here and some direction when it hits the wall it's going to conserve that momentum but it's going to go in the opposite direction, and it's going to enter with some angle and it's going to leave with some other angle light is going to do the same thing if I draw a ray of lights encountering a boundary between two media that ray is going to leave at some angle we have something called the Law of reflection which holds for the elastic collision just like it holds for reflection of light that states that that reflected angle is actually the same as the incident angle the angle with which it hits the boundary. Now just as a side note this isn't going to be important right now but this is going to be very important later whenever light encounters a boundary we always measure the angle relative to the normal direction, relative to some line that is perpendicular to the boundary right, this is the normal just like the normal force is perpendicular, the normal direction is perpendicular to a surface.
Let's do an example, we want to find the missing angle theta using the law of reflection, so this light ray is incident at 65 degrees it undergoes reflection right here which means that the outgoing angle is also going to be 65 degrees that's just what the law of reflection says. Now notice this is the normal to this surface which means that it's at a 90 degree angle from the surface so this angle right here is going to be the complementary angle to 65 degrees which is 25 degrees. Right because they have to add up to 90, now notice we have a triangle right here that looks like this here is our 25 degree angle, here's our 125 degree angle and here is an unknown angle. Let me minimise myself, remember that the sum of all the internal angles within a triangle has to be 180 so 25 degrees plus 120 degrees plus this unknown angle has to be 180 degrees. That means that this unknown angle is actually 35 degrees. So that tells us what this angle is this angle is 35 degrees. Now once again this line right here is the normal to the second surface which means that its perpendicular to the second surface so this angle right here is going to be the complimentary angle to 35 degrees right, the angle that when added to 35 degrees equals 90 and that is 55 degrees. So what is theta have to be? Well the law of reflection says it has to be the same as that incident angle which on the second surface is 55 degrees so theta therefore is 55 degrees. Alright guys that wraps up our discussion on the reflection of light and the law of reflection. Thanks for watching.
Example #1: Hanging Mirror
Hey guys lets do an example, a flat mirror hangs 0.2 meters off the ground. If a person 1.8 meters tall stands 2 meters from the mirror, which we called X that can be seen in the mirror? The geometry for this problem is already set up all we have to do is use the law of reflection to figure it out. So this light ray is coming up off the ground from this point encountering the mirror at its lowest points and then leaving in this direction to your eyes which are going to be here, what we have to do is use the law of reflection to figure out what angle properly adjust that light ray so that meets you at exactly 1.8 meters off the ground. Here is the normal because we always use the normal when measuring angles so this is our incident angle theta 1 and this is our reflected angle theta 1 prime and remember that those two are equal now notice if I were to continue this normal line right here we form a triangle right the triangle that we form is 2 meters wide, how tall is it ? well it's not 1.8 meters tall because the mirror the bottom of this point right here is 0.2 meters off the ground so it's actually 1.8 minus 0.2 which is 1.6 meters and this angle right here is theta 1 prime, which is the angle that we're interested in finding. So clearly we can just use trigonometry to find this we can use the tangent and we consider the tangent of theta 1 prime is going to be the opposite edge which is 1.6 meters divided by the adjacent edge which is 2 and that tells us that theta 1 prime is just 51.3 degrees. So now we know that this angle is 51.3 degrees so this angle too is 51.3 degrees.
So we have a new triangle right here, let me minimise myself we have a new triangle now this lower triangle, where this angle the incident angle is 51.3 degrees this height is 0.2 meters and this length is X. What's this angle going to be? Well this is what's known as an alternate interior angle to 51.3 and all alternate interior angles are the same so this is going to be 51.3 degrees. Alright so once again we can use the tangent to find what X should be we can just say then that the tangents of 51.3 degrees equals the opposite which is 0.2 meters divided by the adjacent which is X or that X is 0.2 over the tangent of 51.3 degrees and finally that X is just 0.25 meters. Just using geometry and trigonometry we can answer this question the crux of the physics is that V these angles right here are the same and that's the law of reflection. Alright guys thanks for watching.
Practice: What is the distance, d, between the incoming and outgoing rays?
Join thousands of students and gain free access to 55 hours of Physics videos that follow the topics your textbook covers.
Enter your friends' email addresses to invite them: