04 - Optics - Prisms and Critical Angles

8/17/2019 04 - Optics - Prisms and Critical Angles

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Lesson 4:

Prisms andCritical Angles


2/14

Optics HW #4 due NEXT TUESDAY!!! ◦

Note the change…


3/14

IV.C.1. Reflection and Refraction

Students should understand the principles ofreflection and refraction, so they can:

b) Show on a diagram the directions of reflected andrefracted rays.c) Use Snell’s Law to relate the directions of the incidentray and the refracted ray, and the indices of refraction ofthe media.

Students will be able to

1. use Snell’s law to predict the ray path of light througha prism.

2. use Snell’s law to determine the critical angle for agiven boundary.

L e s s

o n

4 :

P r i s m

s a n d C r i t i c a l A n g l e s


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Whiteboard optics


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Light enters a prism as shown, and passes through theprism.a) Complete the path of the light through the prism, and show the

angle it will make when it leaves the prism.b) If the refractive index of the glass is 1.55, calculate the angle of

refraction when it leaves the prism.c) How would the answer to b) change if the prism were immersed

in water?

30o

60oglass

air


6/14

Light enters a prism made of air from glass.a) Complete the path of the light through the prism, and show the

angle it will make when it leaves the prism.

b) If the refractive index of the glass is 1.55, calculate the angle of

refraction when it leaves the prism.

30o

60o

glass

air


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The smallest angle of incidence for which

light cannot leave a medium is called thecritic l ngle of incidence.

If light passes into a medium with a

l ss r refractive index than the originalmedium, it bends w y from the normaland the angle of refraction is greater

than the angle of incidence. If the angle of refraction is > 90o,

the light cannot leave the medium.


8/14

This drawing reminds usthat when light refractsfrom a medium with alarger n into one with asmaller n, it bends w yfrom the normal.

n2

n1

n1 > n2


9/14

This shows light hitting aboundary at the criticalangle of incidence,

where the angle ofrefraction is 90o.

No refraction occurs!

n2

n1

n1 > n2

c

r = 90o

Instead of refraction,total internal reflectionoccurs when the angleof incidence exceedsthe critical angle.


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n1sin(1) = n2sin(90o)

n1sin(c) = n2sin(90o)sin(c) = n2/ n1c = sin-1(n2/n1)


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Binoculars use a

combination of

prisms that reflect

the incoming light.

As long as the

incident angles

exceed the criticalangle, the light will

be reflected.


12/14

Light enters the fiber

optic tube at an angle

above the critical angleand is thus totally

reflected down the

‘light pipe’ to the other

end.

For commercial use,

two different glasses

are used, wrapped in a

protective cover. Which

must have the greater

index of refraction, thecore or the cladding?


13/14

A. What is the critical angle of incidence for a gemstonewith refractive index 2.45 if it is in air?

B. If you immerse the gemstone in water (refractive index

1.33), what does this do to the critical angle ofincidence?


14/14

The glass core of an optical fiber has an index ofrefraction of 1.60. The index of refraction of thecladding is 1.48.

What is the maximum angle a light ray can make

with the wall of the core if it is to remain inside thefiber?

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04 - Optics - Prisms and Critical Angles