**Geometric Optics
**So..

Law of reflection, the angle of incidence equals the angle of reflection. | |

The focal length is half the radius of curvature. | |

This is called the mirror equation. f denotes the focal length, d_o is the object distance, d_i is the image distance. Make sure you realize that these values can be negative as they are displacement from the mirror. | |

m is the lateral magnification of the mirror. h_i is the height of the image, h_o is the height of the object. d_i is the image distance, d_o is the object distance. | |

n, the index of refraction of a material is equal to the speed of light in a vacuum (approximately 3.00x10^8m/s) over the speed of light in the respective material. | |

This equation is known as Snell's Law. The left side of the equation with subscript 1 corresponds to one medium and its angle of incidence. The right side with subscript 2 corresponds to another medium and the angle of refraction. | |

Theta_c, is the critical angle. This is the incident angle that of which the angle of refraction will be 90 and the refracted ray will skim the surface of a material. |

For a plane mirror, light is reflected at the same angle that of which it strikes the mirror. The ray of light that strikes the mirror is referred to as the incident ray. The ray of mirror that reflects off the mirror is called the reflected ray. The angle of incidence and the angle of reflection, named respectively, are equivalent when taken from the normal. The normal is the line that bisects the full angle such that the angle of incidence and angle of reflection are equivalent. This normal is perpendicular to the plane mirror as well. The virtual image in a plane mirror is skewed proportionally to the actual image. The size of the virtual image depends on the distance between the actual image and the plane mirror. The further away the actual image is from the plane mirror, the smaller the virtual image appears. The closer the actual image is from the plane mirror, the more similar in size the virtual image is to that of the actual image. To apply the law of reflection to surfaces that are not flat, treat the point that the incident ray strikes as a plane mirror. From here, draw the tangent to that point and the plane mirror. The reflected ray can be obtained because of the law of reflection that states the angle of incidence is equivalent to the angle of reflection.

For a concave mirror, when the object is located beyond the center of curvature, there will be a real image which is smaller, inverted and it will appear in front of the mirror. When the object is located at the center of curvature, there will be a real image, inverted, same size as object, located at the center of curvature which is in front of the mirror. When the object is located between the center of curvature and the focal point, there will be a real image, which is inverted and larger, in front of the mirror. When the object is located at the focal point, there will be no image because the reflected rays are parallel. When the object is located between the focal point and the mirror, there will be a virtual image which is erect, beyond the mirror, and larger.

For a convex mirror, the images formed will always be virtual, erect, smaller, and beyond the mirror.

Refraction is the bending of light. When a ray of light is incident on an interface between two different mediums, a generalization can be made. If the index of refraction of the medium the incident ray travels through is greater than the index of refraction of the medium the emergent ray travels through, then the angle made by the incident ray to the normal will be less than the angle made by the emergent ray to the normal. If the index of the medium through which the incident ray travels through is less than that of the medium of which the emergent ray travels through, then it is just the opposite. Light tends to bend towards the normal when it passes from a material of lower index of refraction to that of a higher index. It tends to bend away from the normal as it passes from material of higher index to lower index. The index of refraction is a constant for a given material, which is the ratio c/v, where c is the speed of light in a vacuum and v is the velocity of light in that respective material.

**A Summary.**

**1.** The ray model of light suggests that light travels in straight
lines under a variety of circumstances. Geometric optics is the study of straight
line rays at various angles.

**
2.** The law of reflection states that the angle of incidence equals
the angle of reflection. On a flat surface, the normal perpendicular to the
surface is that of which the angle of incidence and reflection are measured
from.

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**Problem 1**

An object stands 30 cm in front of a thin converging lens with a focal length
of 15 cm. Describe the image formed.

Since the lens is converging, f is positive. Use the lens equation, f^-1 =
d_0^-1 + d_i^-1 to find that d_i is +30cm. m = -d_i/d_o, so m = -1.

Since d_i is +, the image is 30cm -behind- the lens and -real-.

Since m is -, the image is inverted.

Since m is 1, they're the same size.

**A little about the eye..**

**1.** The eye focuses on objects at different distances by the
process of accommodation. To focus on a nearby object, the muscles contract,
causing the center of the lens to be thicker, which then shortens the focal
length. For a far away object, the muscle relaxes so that the lens is thinner.
This increases the focal length.

**2. **A nearsighted eye has the problem that the eyeball is too
long, or sometimes the curvature of the cornea is too great. The objects are
focused in front of the retina then. This can be corrected with a diverging
lens, since it causes the light to diverge then be refocused at the retina.
A far sighted eye is just the opposite. The eyeball is too short, or the curvature
of the cornea isn’t curved enough. This can be corrected with a converging
lens.

**Problem 1
**A ray of monochromatic light enters an equilateral triangle made of
crown glass ( n = 1.52 ) at 45 degrees to the normal, at what angle does it
exit the triangle?

**The Wave Nature of Light**

2t=(2m+1)*lambda/2n | minimum thickness of soap bubble skin formula. m is the order, n is the index of refraction. |

m is the order of the fringe, lambda is the wavelength of the light beam, d is the distance between two slits. |

**1.** The interference of waves depends on the amplitudes, phase
of the waves and wavelength. There will be constructive interference when the
paths of the two waves differ by one wavelength, or any whole number of wavelengths.
Destructive interference occurs when the wavelengths differ by a half. Diffraction
is affected by slit width and wavelength. Larger the slit, the more bands there
will be. Longer the wavelength, the more bands there will be.

**
2.** Evidence that shows light has wave properties are that light waves
interfere with each other constructively and destructively, as well as show
diffraction patterns.

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We'll start off with an appetizer.

**Problem 1**

Monochromatic light falls on two very narrow slits 0.040mm apart. Successive
fringes on a screen 5.00 m away are 5.5cm apart near the center of the pattern.
What is the wavelength and frequency of the light?

This is a rather tricky question, but once you understand it, it makes all
the following problems easier.

Well, start off by drawing a picture, one of the best methods of problem solving
(in my opinion). Since we know d = 0.040 x 10^-3 m, and the distance between
the slits and the screen is 5.00m, then drawing a triangle reveals theta to
equal the arctangent of (5.5x10^-2)/5. Now that we have theta, substituting
things in is easy.

You should come up with lambda equaling 440 nanometersif you let m, the order,
equal 1.

Finding the frequency can be achieved by c = lambda*nu, where nu is the frequency
and c is the speed of light in a vacuum.

nu = c/lambda.

3.0*10^8/(440*10^-9) => 6.82 x 10^14 Hz.

**Problem 2
**A parallel beam of light from an angry duck with a wavelength of 656
nm falls on two very very narrow slits, 0.050 mm apart. How far apart are the
fringes in the center of the pattern if the scren is 2.6 m away?

Using the triangle again, the arctangent of 656*10^-9/(0.050*10^-3) results
in theta. Now we can use the triangle with screen distance and inteference fringe
length once again.

tan theta = x/2.6 ( x is the interference fringe length.)

3.4*10^-2 m = d.

**Problem 3**

Light of wavelength 680 nm falls on two narrow slits and produces an interference
pattern of chickens in which the fourth-order fringe is 48 mm from the central
fringe on a screen 1.5 m away. What is the seperation of the slits?

Here, we want to find "d."

Using the triangle.. (hey look lots of triangles) theta = arctangent(48*10^-3/1.5)

Now that we have theta.. substitute into m*lambda = d sin theta.

(4*680*10^-9)/Sin[ArcTan[(48*10^-3)/(1.5)]] = d

8.5*10^-5 m = d.

**Problem 4
**Light of wavelength 400 nm in air falls on two slits 5.00 * 10^-2 mm
apart. The slits are immerse in chicken blood (index of refraction 1.33), as
is a viewing screen 40.0 cm away. How far apart are the fringes on the screen?

Mmmm (^)< QUACK QUACK QUACK chicken blood.

Anyway.. We solve for theta not using the triangle starting off.

m*lambda = d sin theta.

400*10^-9/(5*10^-5) = sin theta.

Now that we have theta, we know that there is refraction.. so use Snell's law.
n1 sin theta 1 = n2 sin theta 2.

Now we have theta 2 which is approx .3446.. and Hey it's another triangle with
the viewing screen.

tan theta2 = x/(40*10^-2). (where x is the distance between the fringes)

x = 2.406*10^-3 m for the distance from fringe to fringe.

**Problem 5**

EM waves from a line source pass through a narrow slit and form an interference
pattern on a screen. As the slit narrows, what happens to the width of the central
maximum?

Since m*lambda = d sin theta, you notice that d is inversely proportional to
the sin of theta. Therefor, as d decreases, the width of the central maximum
will increase.

**Problem 6**

A grating has 7000 lines/cm. How many spectral orders can be seen when it is illuminated by white light?

m*lambda = d sin theta.

d = (700000 lines/cm )^-1 = 0.0001428571429 cm/line * 1 m/100cm

d = 1.428571429 x 10^-7

m = ( 1.428571429 x 10^-7 sin 90 )/(700x10^-9m)

m = .2040816327

m = ( 1.428571429 x 10^-7 sin 90 )/(400x10^-9m)

m = .3571428572