The Wave Nature of Light

I. Interactions with Slits

I-a. Overview

	Single Slit Diffraction	Double Slit Interference	Diffraction Grating Interference
Physical Discription	A single slit with a size typically of about 1x10-6 m	Two parallel slits separated by a typical distance of 1x10-4 m	1000s of slits separated by a typical distance of 2x10-6 m
Typical pattern
Typical Size (on a screen 1 meter away)	1 to many meters	A few centimeters	A meter to many meters
Description (typical)	Distinct central bright space. Indistinct off center bright maxima. Reasonably distinct off center dark spaces.	Distinct central bright space. Reasonably distinct off center bright maxima. Very distinct off center dark spaces.	A very distinct center and off center bright maxima.
Relevant Equations D (slit width) d (slit separation) l (wave length) q (angle) L (screen distance) W (width/distance on screen) Note for a diffraction grating with N lines/cm d=1/(100N)	Angular width of central bright space q=sin-1(2l/D) Width on screen W=Ltanq Angle to the mth (m=1,2,...) dark space q=sin-1(ml/D) Screen distance (fom the middle of the central bright space) to the mth (m=1,2,...) dark space W=Ltanq	Angle to the center of the mth bright space q=sin-1(ml/d) Screen distance (fom the middle of the central bright space) to the middle of the mth (m=1,2,...) bright space W=Ltanq Angle to the center of the mth dark space q=sin-1((m+1/2)l/d) Screen distance (fom the middle of the central bright space) to the middle of the mth (m=0,1,2,...) dark space W=Ltanq	Angle to the center of the mth bright space q=sin-1(ml/d) Screen distance (fom the middle of the central bright space) to the middle of the mth (m=1,2,...) bright space W=Ltanq

Note that in all of the above it is assumed that the light is monochromatic (single wavelength. For the two interference patterens, and especially the diffraction grating. sometimes light with two or more wavelengths are used. The central bright space will be whatever the color mixture produces, but since light of different wavelengths is bent at different angles, each color will produce its own off center line pattern. This gives rise to a possible set of problems.

Interactive Examples

1. If light of 500 nm is incident on a slit of width 0.000005 what is the angular width to the nearest degree of the central maximum?
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2. If light of 500 nm is incident on a slit of width 0.000005 what is the width to the nearest centimeter of the central maximum on a screen 2 meters away?
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3. If light of 500 nm incident on a slit is to cover a whole screen, what is the maximum slit width to the nearest nm?
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4. If light of 500 nm is incident on double slits separated by 0.0001 meters, what is the distance from the center to the second bright space to the nearest mm on a screen 2 meters away.?
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5. If light of 500 nm is incident on double slits separated by 0.0001 meters, what is the distance from the center to the second dark space (m=2) to the nearest mm on a screen 2 meters away.?
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6. If light of 500 nm is incident on a grating with 5000 lines/cm, what is the distance from the center to the second bright space (m=2) to the nearest cm on a screen 2 meters away.?
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7. If light of 400 nm and light of 600 nm are incident on a N=5000 line/cm diffraction grating at what angle (if any) will the m+1 400 nm line overlap the m 600 nm line ?
   
   • For what value of m will this be true?
     

   • What is d?
     

   • What will be the angle to the nearest degree for this to occur?
     

   
   

   II. Thin Film Interference

   There are three main subtopics
   
   1. Interference caused by an air gap
      
      
      

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      Note: Upon reflection at a surface with higher index of reflection, a light wave undergoes a shift (phase shift) of a half of a wave length.

      ---

      
      
      • Light incident on figure I from above
        undergoes reflection at the upper glass-air interface without phase shift
        
        
      • undergoes reflection at the lower air-glass interface with phase shift
        
        
      • and the lower reflected ray travels an extra distance of 2t, where t is the thickness of the air gap at the point where the light strikes.
      
      
      If 2t = ml (m= 0, 1,2,...), due to the phase shift, the two rays will destructively interfere and a dark band results. Conversely, if 2t = (m+1/2)l, constructive interference occurs and a bright band results.
      
      

      Interactive Example

      Viewed head on what is the minimum air gap to the nearest micron ( mm ) that will produce constructive interference for light of 600 nm?
      

   2. Interference produced by a soap bubble
      
      
      The soap bubble problem is similiar to the air gap problem. Three differences:
      
      • Phase shift occurs at the first interface not the second, This does not change the nature of the problem.
        
        
      • The wave length in the film is not l but, instead, l/n, where n is the index of refraction of the film.
        
        
      • The incident light is generally white light, and the given is the color viewed (or seen), this corresponds to the wavelength which produces constructive interference. Thus the problem usually is given l what is t? The relevant equation is 2t =(m+1/2)l/n. Usually the problem is to find the thinnest layer and thus m=0.
      
      
      

      Interactive Example

      A soap bubble appears green ( l= 540 nm ) at a point on its front surface nearest the viewer. What is its minimum thickness to the nearest nm? Assume n=1.35
      

   3. Interference produced by a lens coating
      
   
   
   The purpose of a lens coating is to produce destructive interference (thus cutting down on reflective loss) in the mid spectrum region (550-600 nm). Since there is no phase shift at either interface (why?), distructive interference occurs when 2t= l/2n
   
   

   Interactive Example

   What is the thichness to the nearest nm of a lens coating ( n= 1.38 ) and is designed to eliminate reflected light at 550 nm when incident on glass ( n= 1.5 )?
   

   
   

   III. Polarization

   The other major class of problems in this chapter involves polarization. Normally light is unpolarized and when it goes through a polarizer, the component perpendicular to the axis of the polarizer is removed and the intensity is reduced by half. If that light, I_o, is now incident upon another polarized whose axis makes an angle q with the axis of the first polarizer, the light is further reduced by an amount
   

   I= I_o(cosq)²

   Interactive Example

   Unpolarized light of intensity 1000 W/m² is incident on two polarizers crossed at 37^o, to the nearest W/m² what is the intensity of the exiting light?
   

   
   
   
   Polarization occurs upon reflection, complete removal of the component perpendicular to the surface occurs when the refraced ray (green) makes a 90^o angle with respect to the reflected ray (red). This will occur when
   
   

   tanq_p= n₂/n₁

   
   where n₁ and n₂ are the indexes of refraction of the first and second media, respectively.
   

   Interactive Example

   At what angle to the nearest degree does incident sunlight on water (n=1.33) become fully polarized?
   

   
   finis
   © 1999 Carl Adler mailto:Carl@Image-ination.com