Left over from Lecture 13 and Chapter 33 1)Show electric and magnetic fields are out of phase in demo. 2)Reflection and Refraction 1)Red or green laser through smoky block 2)Show maximum bending angle 3)Total internal reflection 1) Show it with block 1)Dispersion through prism 2)Brewsters Law 1)Show using polaroid paper
Dispersion: Different wavelengths have different velocities and therefore different indices of refraction. This leads to different refractive angles for different wavelengths. Thus the light is dispersed. The frequency does not change when n changes.
Why is light totally reflected inside a fiber optics cable? Internal reflection
9 Equilateral prism dispersing sunlight late afternoon. Sin θ refr = 0.866n λ
How much light is polarized when reflected from a surface? Polarization by Reflection: Brewsters Law
Fresnel Equations The parallel and perpendicular components of the electric vector of the reflected and transmitted light wave is plotted below when light strikes a piece of glass. Now the intensity of polarized light is proportional to the square of the amplitude of the oscillations of the electric field. So, we can express the intensity of the incoming light as I 0 = (constant) E 2. E
What causes a Mirage 1.06 1.09 1.08 1.07 1.08 1.09 sky eye Hot road causes gradient in the index of refraction that increases as you increase the distance from the road Index of refraction
Geometrical Optics: Study of reflection and refraction of light from surfaces The ray approximation states that light travels in straight lines until it is reflected or refracted and then travels in straight lines again. The wavelength of light must be small compared to the size of the objects or else diffractive effects occur.
Fermat’s Principle Using Fermat’s Principle you can prove the Reflection law. It states that the path taken by light when traveling from one point to another is the path that takes the shortest time compared to nearby paths. JAVA APPLET Show Fermat’s principle simulator http://www.phys.hawaii.edu/~teb/java/ntnujava/index.html
Two light rays 1 and 2 taking different paths between points A and B and reflecting off a vertical mirror 1 2 A B Plane Mirror Use calculus - method of minimization
Write down time as a function of y and set the derivative to 0.
Plane Mirrors Where is the image formed Mirrors and Lenses
Plane mirrors Normal Angle of incidence Angle of reflection i = - p Real side Virtual side Virtual image eye Object distance = - image distance Image size = Object size
Problem: Two plane mirrors make an angle of 90 o. How many images are there for an object placed between them? object eye 1 2 3 mirror
pocket Assuming no spin Assuming an elastic collision No cushion deformation d d Using the Law of Reflection to make a bank shot
What happens if we bend the mirror? i = - p magnification = 1 Concave mirror. Image gets magnified. Field of view is diminished Convex mirror. Image is reduced. Field of view increased.
Rules for drawing images for mirrors Initial parallel ray reflects through focal point. Ray that passes in initially through focal point reflects parallel from mirror Ray reflects from C the radius of curvature of mirror reflects along itself. Ray that reflects from mirror at little point c is reflected symmetrically Lateral magnification = ratio of image height/object height
One simple way to measure focal length of a concave mirror?
Spherical refracting surfaces Using Snell’s Law and assuming small Angles between the rays with the central axis, we get the following formula:
In Figure 34-35, A beam of parallel light rays from a laser is incident on a solid transparent sphere of index of refraction n. Fig. 34-35 (a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere? (b) What index of refraction, if any, will produce a point image at the center of the sphere? Enter 'none' if necessary. Chapter 34 Problem 32
Thin Lens Equation Lensmaker Equation What is the sign convention? Lateral Magnification for a Lens
Sign Convention p Virtual side - V Real side - R i Light Real object - distance p is pos on V side (Incident rays are diverging) Radius of curvature is pos on R side. Real image - distance is pos on R side. Virtual object - distance is neg on R side. Incident rays are converging) Radius of curvature is neg on the V side. Virtual image- distance is neg on the V side. r2r2 r1r1
Rules for drawing rays to locate images from a lens 1) A ray initially parallel to the central axis will pass through the focal point. 2) A ray that initially passes through the focal point will emerge from the lens parallel to the central axis. 3) A ray that is directed towards the center of the lens will go straight through the lens undeflected.
Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays... F1F1 F2F2 p Virtual side Real side Example
Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays. Image is real, inverted... F1F1 F2F2 p Virtual side Real side Example
Given a lens with the properties (lengths in cm) r 1 = +30, r 2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Virtual side Real side r1r1.. F1F1 F2F2 p r2r2 Example
Given a lens with the properties (lengths in cm) r 1 = +30, r 2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Image is virtual, upright. Virtual side Real side r1r1.. F1F1 F2F2 p r2r2 Example
A converging lens with a focal length of +20 cm is located 10 cm to the left of a diverging lens having a focal length of -15 cm. If an object is located 40 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. Treat each lens Separately. f1f1 f1f1 Lens 1Lens 2 f2f2 f2f2 10 40 +20-15 Example
f1f1 f1f1 Lens 1Lens 2 f2f2 f2f2 10 40 +20-15 Ignoring the diverging lens (lens 2), the image formed by the converging lens (lens 1) is located at a distance 40 This image now serves as a virtual object for lens 2, with p 2 = - (40 cm - 10 cm) = - 30 cm. 30 Since m = -i 1 /p 1 = - 40/40= - 1, the image is inverted
Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i 2 < 0). f1f1 f1f1 Lens 1Lens 2 f2f2 f2f2 10 40 +20-15 40 30 The magnification is m = (-i 1 /p 1 ) x (-i 2 /p 2 ) = (-40/40)x(30/-30) =+1, so the image has the same size orientation as the object.
Chapter 34 Problem 91 Figure 34-43a shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Figure 34-43b). A "normal" eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Figure 34-43c).
(a) Suppose that for the parallel rays of Figure 34-43a and Figure 34-43b, the focal length f of the effective thin lens of the eye is 2.52 cm. For an object at distance p = 48.0 cm, what focal length f' of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f'?