Electromagnetic Waves

Transcription

1 Welcome Back to Physics 1308 Electromagnetic Waves James Clerk Maxwell 13 June November 1879

2 Announcements Assignments for Tuesday, November 20th: - Reading: Chapter Watch Videos: - Lecture 22 - Images and Mirrors Homework 12 Assigned - due before class on Tuesday, November 20th. Midterm Exam 3 will be in class on Thursday, November 15th. It will cover material corresponding to that covered through chapter 29 in your textbook, and all associated lecture, homework and in class material.

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5 Review Question As more identical resistors R are added to the parallel circuit shown here, the total resistance between points P and Q A) increases B) decreases C) remains the same Answer: B decreases The potential difference is the same across each resistor, so the same current flows through each resistor as would flow through an isolated resistor R. The sum of the currents through all the parallel resistors (Req) must equal the input current. Thus, the total current increases as resistors are added the tot tab resistance decreases.

6 Review Key Concepts Electric charge exists in two kinds, positive and negative. These exert influence on each other via a field of force. The electric field is a conservative force field, with associated concepts of potential energy and the new concept of electric potential. Fields accelerate charges, and this is the basis of circuits. There are relationships between fields and the energy stored and release in circuits (e.g. in capacitors and resistors). Magnetic fields, fields of force exerted by certain materials such as magnetite and iron, also affect electric charges. However, they do so at right angles to the field lines and the direction of motion of the charges. Moving electric charge creates magnetic field. Changing magnetic field penetration in a conductor causes voltage, and thus electric current. A conductor that encloses its OWN magnetic field possesses selfinduction. Magnetic fields store energy, just like electric fields. Light is an electromagnetic wave.

7 Key Concepts Law of Reflection: A reflected ray lies in the plane of incidence and has an angle of reflection equal to the angle of incidence (both relative to the normal). Law of Refraction: A refracted ray lies in the plane of incidence and has an angle of refraction θ2 that is related to the angle of incidence θ1 by

8 (a)case n2 is equal to n1: θ2 is equal to θ1 and refraction does not bend the light beam, which continues in the undeflected direction. (b)case n2 is greater than n1: θ2 is less than θ1 and refraction bends the light beam away from the undeflected direction and toward the normal. (c)case n2 is less than n1: θ2 is greater than θ1 and refraction bends the light beam away from the undeflected direction and away from the normal.

9 Chromic Dispersion of White Light: The blue component is bent more than the red component. (a) Passing from air to glass, the blue component ends up with the smaller angle of refraction. (b) Passing from glass to air, the blue component ends up with the greater angle of refraction

10 Question 1 Complete the following statement: Chromatic dispersion occurs in some materials because different wavelengths of light (choose all that are correct) A) propagate at different speeds. B) are reflected at different angles. C) are transmitted in differing amounts. D) have differing indices of refraction.

11 Question 2 You are standing directly above a fish in an aquarium. The actual depth of the fish is one-half the distance from the surface to the bottom. As you look down at it, where does the fish appear to be? A) I see it at its actual depth. B) I see it at the surface of the water. C) I see it below its actual depth, but above the bottom of the aquarium. D) I see it above its actual depth, but below the surface of the water. E) I see it at the bottom of the aquarium.

12 Demonstrations of Reflection and Refraction

13 Instructor Problem: Safety Glasses Safety glass ( Laminated Glass ) is a shatterproof glass made from three layers of clear glass, then plastic, then glass. The plastic keeps the glass from shattering when it is cracked. Consider a light ray entering such a window from air outside. It makes an angle θ = 37.0 degrees with respect to the normal to the airglass boundary. Given the indices of refraction of the materials involved, determine the angles of refraction at the 1st air-glass boundary, then at the 1st glass-plastic boundary.

14 Given: n air =1.000 n glass =1.468 n PVB =1.488 First consider the light ray striking the air-glass interface. We are given the angle of incidence with respect to the normal to the air-glass interface. The ray begins in air and strikes the boundary. We expect some of the light to refract. From Snell s Law we can calculate the angle of refraction inside the glass.

15 What about the next refraction, at the glass-plastic boundary? Again, we need Snell's Law. We need the angle of incidence of the light ray, now traveling in glass, as it strikes the glass-plastic boundary. We know the angle with respect to the normal to the air-glass interface. Since the air-glass interface is parallel to the glass-plastic interface, it is also making an angle of θglass,2 = θglass,1with respect to the glass-plastic interface Applying Snell's Law again

16 You might then wonder why safety glass is safe at all, if such de flections of light occur inside the material. As an additional exercise, you could keep going, running the light from the plastic to the last glass layer (you'll end the angle of refraction there is = 24.38) and then at last at the final glass-air boundary. You'll end that this last refraction yields an angle with respect to the normal to the last glass-air boundary of = 37. The key here is that we have the same material at the end (air) as we had the beginning (also air), and so there is no noticable distortion of the world outside the safety glass window because, at the end of light's journey, it leaves the safety glass at the same angle it entered in the same material. Indeed, the true positions of objects outside the glass will be slightly shifted overall, but the glass is so thin that the e ffect is not noticable with the human eye. However, the key thing is that relative dimensions are not distorted by the refractions.

17 Student Problem: Reflect and Refract! Light traveling horizontally in air strikes a plane mirror tilted at a 30-degree angle to the horizontal. The reflected light then strikes a piece of glass, n= A) What is the angle of refraction in the glass? B) What is the critical angle at the glass-air boundary on the other side of the glass? Will the refracted ray be totally internally reflected?

18 Part A: The fi rst step is to determine the angle that the refl ected ray makes when it strikes the airglass boundary. We given that the incident ray enters horizontally before striking the mirror and that the mirror is tilted up from the floor at a 30-degree angle. We must find the angle, with respect to the normal, that the incident ray makes. The normal makes a 90-degree angle to the mirror surface. The mirror surface makes a 90-degree angle to the horizontal on its right side. Thus the angle between the normal and the horizontal on its right side is 120-degrees, making the angle to the horizontal on its left side 60-degrees. Thus, it follows that angle between the normal and the incident ray is 60-degrees. The ray enters from the left, and since it, too, is horizontal then it makes a 60-degree angle with respect to the normal.

19 The law of reflection states that the reflected angle equals the incident angle, thus our reflected ray is also 60 degrees. Now, we need to employ Snell s Law to find the refracted ray. But we should be careful with our angles. The angle made with respect to the normal of the glass interface is 30 degrees.

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21 The End for Today xkcd: 1469:UV