Optics

Question 1

The Electromagnetic Spectrum

Answer 1

• Through a vacuum, all electromagnetic waves travel at a fixed speed: c= 3.00 x 10⁸ m/s; regardless of their frequency
• Electromagnetic waves can be categorized by their frequency (or wavelength)
  
  • The full range of waves is called the electromagnetic spectrum
    
    • Types of waves: radiowaves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays
    • MEMORIZE ORDER OF VISIBLE SPECTRUM
      
      • In order of increasing wave frequency: red, orange, yellow, gren, blue, and violet (Mneomic: ROYGBV)
        
        • The wavelengths of colors in the visible spectrum are usually expressed in nanometers

Question 2

Interference

Answer 2

• Waves experience interference when they meet
  
  • Whether they interfere constructively or destructively depnds on their relative phase
    
    • If they meet in phase (crest meets crest), they combine constructively
    • If they meet out of phase (crest meets trough), they combine destructively
      
      • If waves that have the same wavelength meet, the difference in the distances they’ve traveled determine whether they are in phase
        
        • Assuming that the waves are coherent (phase difference remains constant over time and does not vary)
          
          • If the difference in their path lengths, Δlength, is a whole number of wavelengths (0, plus or minus λ, plus or minus 2λ, etc) they’ll arrive in phase at meeting
          • If difference is a whole number plus one half wavelength (plus or minus 1/2λ, plus or minus (1 + 1/2λ), etc.) then they’ll arrive exactly out of phase
• Constructive Interference: Δlength= mλ
• Destructive Interference: Δlength= (m + 1/2)λ
  
  • m is an integer

Question 3

Young’s Double Slit Interference Experiment

Answer 3

• When a wave encounters a slit with a width that’s comparable with its wavelength, the wave will fan out after it pases through which is known as diffraction
  
  • The waves will diffract through the slits and spread out and interfere as they travel toward the screen
    
    • The screen shows the results of this interference: bright bands (fringes) centered at points at which waves interfere constructively, alternating with dark fringes, where the waves interfere destructively
      
      • To locate position of say bright fringes on screen use: y_m= mλL/d
        
        • y measrues the vertical displacement along the screen from the center of the screen (y= 0, the point directly across from the midpoint of the slits)
          
          • The bright fringe directly opposite the midpoint of the slits- the central maximum- will have the greatest intensity, the bright fringes with m= plus or minus 1 will have lower intensity, those with m= plus or minus 2 will be fainter and so on
            
            • If more than two slits cut in the barrier, the interference pattern becomes sharper, and the distinction between dark and bright fringes becomes more pronounced
              
              • Barriers containing thousands of tiny slits per centimeter= called diffraction gratings- are used specifically for this purpose

Question 4

Single-Aperture Diffraction

Answer 4

• A diffraction pattern will also form on the screen if the barrier contains only one slit
  
  • The central maxima will be very pronounced, but lower-intensity maxima will also be seen because of interference from waves arriving from different locations within the slit itself
  • The central maximum will become wider as the width of the slit is decreased

Question 5

Reflection

Answer 5

• Light directed toward a smooth transparents surface when hit, some of its energy will be reflected off the surface and some will be transmitted into the new medium
  
  • The directions of the reflected and transmitted beams can be calculated using the angles that the beams make with the normal to the interface
    
    • The normal is a line perpendicular to the interface
    • The angle that the incident beam makes with the normal is called the angle of incidence or θ₁
    • The angle that the reflected beam makes with the normal is called the angle of reflection, θ^’₁
    • The angle that the transmitted beam makes with the normal is called the angle of refraction, θ₂
    • The incident, reflected, and transmitted beams of light all lie in the same plane

Question 6

Law of Reflection

Answer 6

Relationship between θ₁ and θ^’₁ is called the law of reflection

• θ₁= θ^’₁

Question 7

Index of Refraction

Answer 7

• When light travels through empty space (vacuum), its speed is 3.00 x 10⁸ m/s but when a light travels through a material medium (e.g. water), it’s constantly absorbed and re-emitted by the atoms of the material and, as a result, its apparent speed, v, is some fraction of c
  
  • The reciprocal of this fraction is called the medium’s index of refraction
    
    • n= c/v
      
      • n has no units; and is never less than 1

Question 8

Snell’s Law

Answer 8

Equation relating θ₁ and θ₂ involves the index of refraction of the incident medium (n₁) and the index of refraction of the refracting medium (n₂)

• n₁sinθ₁= n₂sinθ₂
  
  • If n₂ > n₁ (e.g. when the light slows down in n₂) then Snell’s law tells us that θ₂ < θ₁ meaning:
    
    • the beam will bend (refract) toward the normal as it enters the medium
  • If n₂ < n₁ (e.g. when the light speeds up in n₂), then θ₂ > θ₁, and the beam will bend away from the normal

Question 9

Dispersion of Light

Answer 9

• When light travels through a material medium, it displays dispersion, which is a variation in wave speed with frequency (or wavelength)
  
  • Index of refraction should be acompanied by a statement of the frequency of the light used to measure v, since different frequencies have different speeds and different indices
    
    • In general, higher frequency waves have higher indicies of refraction
    • When white light (which is combination of all colors of the visible spectrum) hits a glas prism, the beam is split into its component colors as each color has its own index
      
      • Snell’s law dictates that each color will have its own angle of refraction, therefore each color emerges from the prism at a slightly different angle, so the light disperses into its component colors

Question 10

Total Internal Reflection

Answer 10

• When a beam of light strikes the boundary to a medium that has a lower index of refraction, the beam bends away from the normal
  
  • As the angle of incidence increases, the angle of refraction becomes larger
    
    • At some point, when the angle of incidence reaches a critical angle, θ_c the angle of refraction becomes 90^o, which means the refracted beam is directed along the surface
    • For angles of incidence that are greater than θ_c, there is no angle of refraction; the entire beam is reflected back into its original medium
      
      • This phenomenon is called total internal reflection
        
        • Occurs when n₁ > n₂ AND θ₁ > θ_c, where θ_c= sin^-1(n₂/n₁)
          
          • Total internal reflection cannot occur if n₁ < n₂, if n₁ > n₂ then total internal reflection is a possibility; it will occur if the angle of incidence is greater than the critical angle, θ_c

Question 11

Mirror

Answer 11

• A mirror is an optical device that forms an image by reflecting light
• Flat mirrors are called plane mirrors

Question 12

Plane Mirrors

Answer 12

• An image is said to be real if light rays actually focus at the image
  
  • A real image can be projected onto a screen
  • For a flat mirror, light rays bounce off the front of the mirror, therefore no light focuses behind it
    
    • Therefore the images produced by a flat mirror are not real; they are virtual
  • Flat mirrors produce upright images
  • Images formed by a flat mirror is niether magnified nor diminished (minified) relative to the size of the object
  • Plane mirror can be considered a spherical mirror with an infinite radius of curvature (infinte focal length) if f= infinity then 1/f= 0 and the mirror equation becomes 1/s₀ + 1/s_i= 0 therefore s_i= -s₀

Question 13

Spherical Mirror

Answer 13

• A spherical mirror is a mirror that’s curved in such a way that its surface forms part of a sphere
  
  • The center of this imaginary sphere is the mirror’s center of curvature
  • The radius of the sphere is called the mirror’s radius of curvature, R
  • Halfway between the mirror and the center of curvature, C, is the focus (or focal point), F
    
    • The intersection of the mirror’s optic axis (its axis of symmetry) with the mirror itself is valled the vertex, V
      
      • The distance from V to F is called the focal length, f, which is one-half of the radius of the curvature (f= R/2)
  • If the mirror had a parabolic cross-section, any ray parallel to the axis would be reflected by the mirror through the focal point
    
    • Spherical mirrors do this for incident light rays near the axis (paraxial rays) because in the region of the mirror that’s close to the axis, the shapes of a parabolic mirror and a spherical mirror are nearly identical

Question 14

Concave vs. Convex Mirror

Answer 14

• Concave mirror: mirror whose reflective side is caved in toward the center of curvature
• Convex mirror: mirror which has a reflective side that curves away from the center of curvature
• To distinguish mathematically between concave and convex mirrors
  
  • Focal length f for concave mirrors and negative f for convex mirrors

Question 15

Ray Tracing for Mirrors

Answer 15

• Ray tracing: representative rays of light are sketched in a diagram that depicts the object and the mirror
  
  • The point at which the reflected rays intersect (or appear to intersect) is the location of the image
• Rules governing rays:
  
  • Concave Mirrors:
    
    • An incident ray parallel to the axis is reflected through the focus
    • An incident ray that passes through the focus is reflected parallel to the axis
    • An incident ray that strikes the vertex is reflected at an equal angle to the axis
  • Convex Mirrors:
    
    • An incident ray parallel to the axis is reflected away from the virtual focus
    • An incident ray directed toward the virtual focus is reflected parallel to the axis
    • An incident ray that strikes the vertex is reflected at an equal angle to the axis

Question 16

Mirror Equation

Answer 16

1/s₀ + 1/s_i= 1/f

• s₀ is the object’s distance from the mirror, s_i is the image’s distance from the mirror, and f is the focal length of the mirror (applies to lenses as well)
  
  • The value of s₀ is always positive for a real object, but s_i can be positive or negative
    
    • The sign of s_i indicates whether the image is real or virtual
      
      • If s_i is positive, the image is real
      • If s_i is negative, the image is virtual

Question 17

Magnification Equation

Answer 17

m= -s_i/s₀

• Equation gives the magnification (applies to lenses as well)
• The height of the image, h_i is |m| times the height of the object, h₀
  
  • If m is positive, then the image is upright relative to the object
  • If m is negative, it’s inverted
• Because s₀ is always positive:
  
  • If s_i is positive, then m is negative
    
    • Real images are always inverted
  • If s_i is negative, then m is positive
    
    • Virtual images are always upright

Question 18

Four questions about image formed by mirror

Answer 18

1. Where is the image?
2. Is the image real or is it virtual?
3. Is the image upright or is it inverted?
4. What is the height of the image (compared with that of the object)?

• The mirror equation answers questions 1 and 2
• The magnification equation answers questions 3 and 4

Question 19

Lenses

Answer 19

• A lens is an optical device that forms an image by refracting light
  
  • Converging lens
    
    • Converges parallel paraxial rays of light to a focal point on the far side (lens could be bi-convex: both of its faces are convex, converging lens all have at least one convex face)
      
      • Light rays actually focus at F, this point is called a real focus, its distance from the lens is the focal length, f
      • Only converging lenses can produce real images (if s₀ > f)
  • Diverging lens
    
    • Causes parallel paraxial rays of light to diverge away from a virtual focus, F, on the same side as the incident rays (Diverging lenses all have at least one concave face)
    • Diverging lenses can only produce virtual images
• Equations for mirrors can be used to analyze images of lenses
  
  • Nature of the image (whether it’s real or virtual) is determined by the side of the lens upon which the image is formed
    
    • If the image is formed on the side of the lens that’s opposite the object, then the image is real
      
      • If the image is formed on the same side of the lens as the object, then it’s virtual
  • Everything the same except the reversal of the physical locations of real versus virtual images

Question 20

Ray Tracing for Lenses

Answer 20

• Representative rays of light can be sketched in a diagram along with the object and the lens; the point at which the reflected rays intersect (or appear to intersect) is the location of the image
• Rules that govern these rays:
  
  • Converging lenses:
    
    • An incident ray parallel to the axis is refracted through the real focus
    • Incident rays pass undeflected through the optical center, O (the central point within the lens where teh axis intersects the lens)
  • Diverging lenses:
    
    • An incident ray parallel to the axis is refracted away from the virtual focus
    • Incident rays pass undeflected through the optical center, O