King Saud University
College of Applied Medical Sciences
Optometry Department
OPTO-356: Second Mid-Term Exam
Nov. 12th, 2003 |
Time Allowed: Two Hours
Student’s Name:
Student’s Number:
Course Instructor: Dr. Idriss M. Bedja
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THEORY:
QUESTION 1:
List the main differences between the laser source and incandescent lamp source.
Laser source Incandescent lamp source
Stimulated emission | Spontaneous emission
Coherent light | Incoherent light
Monochromatic, and confined | Multi-wavelnegths and
parallel Beam | divergent beam
QUESTION 2:
a) Why can laser be safely used for medical surgery? Because specific laser wavelength is being absorbed by a specific type of biological tissue.
b) Describe briefly one example of medical application: In Glaucoma we have high fluid pressure which may destroy the optic nerve. Use a laser to open a tiny hole to retrieve a destructive pressure .
PROBLEM 1:
Measurements are made of the intensity distribution in a Young’s interference pattern. At a particular value of y (distance from the center of the screen), it is found that I/I0 = 82% when light of wavelength 5890 A is used.
a) A) What wavelength of light should be used in order to reduce the relative intensity at the same location to 55%?
b) B) What is the shift (variation) in minimum phase differences in radians between sources accompanied with this change in wavelengths?
c) What is the new deviation angle θ attributed to the new wavelength?
Solution:
Data: Requests:
Iav/Io = 0.82 l2= ? (if new Iav/Io = 0.55) at same y
l1= 5890 A° DF = F(l2) - F(l1) = ?
q (l2) = ?
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a)- Same location y, also same |πδ/l2= cos-1(0.742) = 42.1°
Path difference δ, for both | l2= 351.1 nm
equations. |
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Iav/Io = cos2 (πδ/l2) = 0.55 | b)- DF = F(l2) - F(l1) = ?
We need so to get δ, from | F(l1)/2= πδ/l1 ----- F(l1)= Iav/Io = 0.82 | 50.2°
0.82 = cos2 (πδ/l1) | F(l2)/2= πδ/l2 ----- F(l2)= 84.2°
cos (πδ/l1) = Ö0.82 = 0.91 |
πδ/l1= cos-1 (0.91) = 25.1 ° | DF = 84.2° - 50.2° = 34°
δ = 25.1 °x l1 = 82.13 nm
180°
cos (πδ/l2) = Ö0.55
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PROBLEM 2:
A soap bubble of index of refraction 1.33 reflects both red and green colors in white light. What minimum thickness of soap bubble will allow this to happen? (In air, λred =700 nm, λgreen=500 nm.)
Solution: n=1.33 tmin= ?
lr=700 nm
lg=500 nm
The system is air /layer/air ------ low index/high index/ low index ---so we use formula of constructive interference
2t=(m + ½)/l/n
For Red | For Green
t=(m + ½)/l/2n | t=(m + ½)/l/2n
m=0 t= 131.6 nm | m=0 t= 94 nm
m=1 t= 394.7 nm | m=1 t= 282 nm
m=2 t= 657.2 nm | m=2 t= 470 nm
| m=3 t= 658 nm
So min thikness which allow both colors is
t= 657.2 nm