I need help with a Pre lab about Optics

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Znex56

Engineering

Description

Please solve the following in the pre lab.

1- part (e) in question 1.

2- Question 2 ( All parts)

3- Question 3 ( All parts)

In question 3, please use group number 1 and 2 in solving the problem.

You can solve the problems on a separate sheet. If you choose to hand write it, please use clear hand writing.

I will attach couple lectures that may help in solving the problems. The pre lab sheet is attached.


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Laboratory #12 EE Science II Laboratory #12 Optics Pre-Laboratory Assignment Your name and group number: ___________________________________ Note: Please view Module 7 → Wave Reflection and Transmission Part 4 before completing this prelab. 1. Consider the geometry shown in Figure 1. A ray of red light traveling in air (Medium 1) is incident on an acrylic slab (Medium 2) of thickness t = 2 cm and εr = 2.25 at an angle θ1 = 60˚ as shown. A part of the ray reflects back to medium 1 at an angle θ2 while the rest propagates through medium 2 at a refracted angle θ3. This refracted ray re-emerges in air (Medium 3) on the side of the acrylic slab at an angle θ6 after a part of it reflects-off the acrylicair interface at angle θ5. Figure 1: Geometry for problem 1. a. Define index of refraction. Calculate the index of refraction in Medium 1, Medium 2, and Medium 3. Index of refraction in Medium 1, Medium 2, Medium 3 is n1 = _______, n2 = _______, and n3 = _______ respectively [1 Point] b. What is Snell’s law of reflection? [0.5 Points] c. What is Snell’s law of refraction? [0.5 Points] d. Use your answers to part b. and c. to calculate the following: [2 Point] e. θ1 = 60˚ θ4 = __________________ θ2 = __________________ θ5 = __________________ θ3 = __________________ θ6 = __________________ The acrylic slab displaces the ray of light by a distance d as shown in Figure 2. Calculate d using the equation given below: [1 Point]  University of South Florida 1 EE212-pre-180411.docx Laboratory #12  cos1 d  t sin 1 1   2 n2  sin 2 1    = _________________________   where, t is the thickness of the glass slab, θ1 = 60˚, and n2 is the refractive index of the acrylic slab. Figure 2: Displacement of light ray by a distance d due to the acrylic slab. 2. A prism is made of a dispersive material – one where the index of refraction is dependent on wavelength as shown in Figure 3 (a). As a result, a beam of white light incident on the prism splits into different colors since each color (wavelength) experiences a different refractive index and refracts with a different angle. You will analyze the behavior of the prism for three different colors or wavelengths – 400 nm, 550 nm, and 700 nm:  Determine which wavelength refracts the most and which one refracts the least and complete the blanks in Figure 3 (b). Show your work clearly for full credit. [3 Points] Hint: Use Snell’s Law to solve the problem. Figure 3: (a) Index of refraction of the prism as a function of wavelength (b) white light splits into different colors (wavelengths) in the prism 3. Consider the experimental set-up shown in Figure 4. A collimated laser beam of diameter h1 = 3mm is incident on a beam expansion lens set-up in which the focal point of both the lenses overlap. A collimated beam is one where light rays are parallel to each other and the beam does not diverge/converge. The first lens focuses the laser beam at its focal point, f1 = 30mm. Since the focal point of the second lens f2 is at the  University of South Florida 2 EE212-pre-180411.docx Laboratory #12 same location, a collimated beam is formed on its other side as shown. If f2 > f1, the diameter of the beam h2 will be greater than h1. The experiment is governed by the following equation M  f 2 h2  f1 h1 where, M is called the magnification and f2 depends on your group number as given below: Groups 1 and 2: f2 = 60 mm Groups 3 and 4: f2 = 100 mm Groups 5 and 6: f2 = 125 mm Group 7: f2 = 150 mm 1. 2. What is the distance between the two lenses (L) in Figure 4? [1 Point] Calculate h2. [1 Point] Figure 4: Beam expansion experimental setup.  University of South Florida 3 EE212-pre-180411.docx Plane Wave Reflection – Part 4 Electrical Engineering Science II – Electromagnetics Video Presentation 11/10/17 EE Science II - Electromagnetics 1 Outline • Plane wave reflection and transmission with oblique incidence 11/10/17 EE Science II - Electromagnetics 2 Plane Wave Oblique Incidence 11/10/17 EE Science II - Electromagnetics 3 Plane Wave Oblique Incidence: Reflection !" = ()" $" %" !' = 1 = $" %" 11/10/17 $' %' ()' EE Science II - Electromagnetics 1 = $' %' 4 Plane Wave Oblique Incidence: Transmission (or) Refraction !" = ()" $" %" !' = 1 = $" %" 11/10/17 $' %' ()' EE Science II - Electromagnetics 1 = $' %' 5 Snell’s Laws – other relations 11/10/17 EE Science II - Electromagnetics 6 Special Cases 11/10/17 EE Science II - Electromagnetics 7 Transmission and Reflection Coefficients: Oblique Incidence 11/10/17 EE Science II - Electromagnetics 8 Optics – Part 1 Electrical Engineering Science II – Electromagnetics Video Presentation 4/14/18 EE Science II - Electromagnetics 1 Outline • Dispersion • Optical Fibers 4/13/18 EE Science II - Electromagnetics 2 Electromagnetic spectrum 4/14/18 EE Science II - Electromagnetics 3 Wave Equation and Refractive Index From Maxwell’s equations, we can derive the wave equation (in free space) " ) # " ! # = %& '( )* Speed of propagation of wave: 1 1 += = 3×10 3/5 %&'& Refractive index is the ratio of speed of an EM wave in vacuum to that in matter + %' 6= = 7 %& '& 4/13/18 6 = %8 '8 ≈ '8 EE Science II - Electromagnetics 4 Spectral Dispersion of light Dispersion 6 = %8 '8 ≈ '8 • '8 dependents on the frequency/wavelength of the EM wave 4/13/18 EE Science II - Electromagnetics 5 Optical Fiber 4/13/18 EE Science II - Electromagnetics 6 Snell’s law of Reflection and Refraction Snell’s law of Reflection :; = :8 Snell’s law of Refraction 6< sin :; = 6" sin :@ When 6< > 6" and :; = :B No refraction :B = 4/14/18 C< sin 6" 6< EE Science II - Electromagnetics 7 Optical Fibers: Numerical Aperture Numerical aperture (EF) of a fiber defines the acceptance angle of light incident on the fiber; i.e., the maximum angle at which light can enter the fiber and undergo total internal reflection. Acceptance angle 6& :H 6" 6< :B EF = sin(:H ) EF = 4/13/18 " 6< " − 6" 6& • Depends on core and cladding indices • Limits the amount of light that can be coupled EE Science II - Electromagnetics 8 Optical Fibers: Temporal Dispersion 4/13/18 EE Science II - Electromagnetics 9 https://commons.wikimedia.org/wiki/File:Optical_fiber_types.svg Optics – Part 2 Electrical Engineering Science II – Electromagnetics Video Presentation 4/13/18 EE Science II - Electromagnetics 1 Outline • Lenses 4/13/18 EE Science II - Electromagnetics 2 Plane waves vs. Spherical waves 4/13/18 EE Science II - Electromagnetics 3 Lenses focus or diverge light https://commons.wikimedia.org/wiki/File:Lens_and_wavefronts_rotated.gif 4/13/18 EE Science II - Electromagnetics 4 Imaging with lenses 4/13/18 EE Science II - Electromagnetics 5
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Explanation & Answer

attached is my answer

Question 1:
With 1 = 60o , n2  1.5 , and t = 2 cm = 2 10−2 m , the distance d is calculated by:



cos 1

d = t sin 1 1 −
2
2


n

sin

2
1 


cos 60o
−2
o
 d = ( 2 10 m ) sin 60 1 −

1.52 − sin 2 60o 



0.5
 d = ( 0.02 )( 0.866 ) 1 −

1.52 − 0.8662 

 d = 0.01 m
Hence, the distance d is 0.01 m .
Question 2:
The indices of refraction of wavelengths 400 nm, 550 nm, and 700 nm are n1 = 1.47 , n2 = 1.46 ,
and n3 = 1.455 , respectively.
Let nair be the index of refraction of air and let  i be the incident angle.
By Snell’s law, we ha...


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