Reflection and Refraction (Lateral Displacement)by Sarah K Tags: lateral displacemnt, reflection, refraction 

#1
Sep1808, 12:59 AM

P: 1

1. The Problem
A thin beam of white light is directed at an angle α = 36.4° with respect to the horizontal, onto the top of a square block of silicate flint glass near its top corner. The perpendicular length of the block (or height is 1cm). The block’s refractive index n is 1.66 for blue light (λ0 = 400 nm) and 1.61 for red light (λ0 = 700 nm). The block is surrounded by vacuum. Find the lateral displacement l between the two rays leaving the bottom surface of the block, as shown. (Note that l is measured perpendicularly to the rays.) 2.The relevant equation: n1sin(theta1)=n2sin(theta2) I do realize that I need use this formula twice but I am not exactly sure what LATERAL DISPLACEMENT means :S if someone can explain what is it I am supposed to find, I would appreciate it. Hints are more than welcome. Thanks 



#2
Sep1808, 01:22 AM

HW Helper
P: 4,442

In the rectangular block the incident ray and refracted ray are parallel to each other. The perpendicular distance between them is called lateral displacement.
The formula for lateral displacement S = t*sin( theta1 theta2)/cos(theta2) where t is the height of the block. Find the lateral displacement for red and blue, and find the difference between them. 



#3
Dec1808, 05:10 PM

P: 1

what is the relationship between the width of a glass block and angel of incidence and the lateral displacement if the width of a glass block is kept constant while performing a simple refractive lab??




#4
Aug1009, 07:17 AM

P: 1

Reflection and Refraction (Lateral Displacement)"A beam of light travelling horizontally in the air goes through a titled rectangular piece of glass. The piece of glass has a thickness w = 32.0mm, a refractice index n2= 1.54, and is titled by an angle alpha = 26 degrees with respect to the horizontal. As a result, the outgoing beam is displaced laterally with respect to the incoming direction." 



#5
Aug1009, 12:37 PM

P: 128

Would it not be simpler (i.e more manageable) to work out the angle of refraction for each beam then work out using trig where each ray would end up on the other face.Tthe difference is the lateral displacement the way I read it. That equation posted does not seem appropriate to difficulty of the question. It seems like something you would learn at degree level.
To the last poster I presume you are asked to find out how far below the incident ray the refracted ray exits the block. Again Snell's law and a bit of trig is required. Draw a diagram and work out the angle of refraction. Then use trig to find out where the ray end up. Also you may O.K for you to solve the problem using a scale diagram on graph paper or similar. 



#6
Aug1009, 11:48 PM

HW Helper
P: 4,442

n = [tex]\frac{sin i}{sin r}[/tex] In the problem angle of incident ray θ ιs given with respect to the horizontal. There fore the i = 90  θ. Find r. A simple geometry and trigonometry will give a relation between the width of a glass block and angel of incidence and the lateral displacement. So the lateral shift S = t[tex]\frac{sin( i  r )}{cos r}[/tex] where t is the width of the glass slab. Since the angler of incidence for red and blue is the same, find lateral shift for red and blue and find the difference. 



#7
Aug1109, 12:04 AM

HW Helper
P: 4,442

If you draw the ray diagram you can see that the angle of incidence is 90^{o}  26^{o} = 54^{o} Find the angle of refraction r. Substitute in the formula and find the lateral shift. In this problem thickness and the height is the same and it is the distance between the incident surface and the refracted surface. 


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