BHS 120.1 – Optics I
Notetaker: Lara Match
Date: 09/25/2012, 1st hour
Page#
Friday’s Exam:
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Will have Critical angle problems on exam
Will have regular prism that you’ll do 4 steps for, laid out as four different question
Will have 4-5, conceptual prism questions
Will have to distinguish what’s important and what’s not important info (Good to start now determining
what’s important and what’s not because that’s what you’ll do in clinic taking case histories)
Won’t have to calculate internal reflection, just say if it will happen or not
Friday problems will say, “You have a prism SET for minimum deviation, P = , nprism = , nsurround =, find the
deviation”
Words of Wisdom [WoW]
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All prism problems in geometric optics can be solved with the four steps. (Special case formulas are not
necessary)
All equilateral triangles are isosceles but not vice versa.
As soon as you see a problem, solve for the critical angle at the second surface so you have a logic
reference point for the rest of your answers.
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Critical Angle
Condition: only happens when going from a dense to a rarer medium - look for this first when reading exam
questions carefully!
Incident angle > Critical angle = internal reflection
Incident angle < Critical angle = refraction
Incident angle = Critical angle = grazing angle of 90˚
Note on geometric optics versus real world behavior of light:
Geometric optics says 100% reflected – not true in real world. Geometric optics pretends there is no absorption,
that it’s either totally reflected or refracted. When light hits surface in real world, you get refraction, reflection
and absorption. All three are happening all the time.
Example Critical Angle Problem
Given:
 Isoceles prism, BR (base right angle) is 60˚
 i1 = 50˚
 nprism = 1.50
 nsurround = 1.00
What is i’? Work backwards. 1.00sin(50) = 1.50sin(x)
What’s the angle of refraction at second surface ( i1’)? No angle because you’ll have internal deviation.
What is the critical angle of the second surface? If I find it, and see i 1 is greater than that, then I know there will be
no i1’, exiting ray. Any internal angle > 41.33 will give internal reflection. Look at a problem, first thing just solve
for critical angle.
BHS 120.1 – Optics I
Notetaker: Lara Match
Date: 09/25/2012, 1st hour
Page#
Internal Reflection/Deviation
[WoW] From the baby triangle on the right (made from the second inclined surface and the internally deviated ray)
you can get angle incident on third surface measured from the interface of the third surface, subtract that from 90 to
get the incident angle on the third surface measured from the normal of the third surface!
3 characteristics of prisms that likely lead to internal deviation:
 i ≤ 10˚
 P ≤ 70˚
 n – n1 ≥ 0.7
Steps to solve for deviation for an internal reflecting prism
1. Draw
2. Critical angle
3. Snell’s law at first surface
4. P – I’ = i1
5. Snell’s law at second interface
6. Deviation, d = i + i1’ – P, to calculate internal deviation, you need to know BL angle.
Clinically - Internally Deviating Prisms
The eye will point toward apex of prism so back part of eye can catch base-deviated light (for ophthalmic prisms)
Conceptual Trends
The smaller the i the larger i1 (Think P = i’ + i1)
Larger i makes larger i’ makes smaller i1 makes smaller i1’ (the two surfaces are inversely proportional)
Example of Prism set for Minimum Deviation
Because this is minimum deviation, internal angles are equal, external angles are equal. What happens at the first
surface happens in reverse at the second surface. When min deviation is indicated, all you need is angle of refraction
(you can get that through the prism angle).
Given:
 Isoceles prism, P = 40
 Nprism = 1.50
 Nsurround = 1.00
Solve for the two external angles 180 – P = 140 / 2 = 70
Solve for the two internal angles. P is the sum of the two internal angles. P divided by 2 gives each internal angle.
i’ = P / 2 = 40 / 2 = 20
i1 = P / 2 = 40 / 2 = 20
What is i?
i = 30.87˚ *CHECK! Is that logical? i’ < i when going from rare to denser medium? Yes!
Deviation = 20.87. Any angle greater or less than 30.87 will give greater deviation
Clinically
You solve for minimum deviation when prescribing prism.
Special Case formula for minimum deviation
nprism / nsurround = sin[(P + d) / 2] / sin(P / 2)
OR you could still do it with the four steps!
BHS 120.1 – Optics I
Notetaker: Lara Match
Date: 09/25/2012, 1st hour
Page#
Right Angle Prisms (Acute Prisms)
Given:
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BR = 60
Light enters perpendicular to surface (Perpendicular incidence goes straight through)
nprism = 1.50
nsurround = 1.00
Can you find total deviation? Yes! 4 steps
What is P? 180 – BR60 – BL90 = 30˚
P – i’ = 30 – 0 = 30˚
i’ = 48.50˚
Deviation using Logic (A and B)
<a = 0 (no deviation at first surface)
<b = 18.59˚
d=a+b
d = 18.59˚
Deviation using Formula (i, i1’ and P)
d = i + i1’ – P
d = 0 + 48.59 – 30
d = 18.59˚
Special case formula for perpendicular incidence
np / ns = sin(P + d) / sin(P)
4 steps is still easier!!!