Chapter 14Chapter 14Refraction of LightRefraction of Light

Hr PhysicsHr Physics

Sorry this is late.Sorry this is late.

Snell's LawSnell's Lawnnii sinsinΘΘii = n= n

rr sinsinΘΘrr

index of refraction of first medium index of refraction of first medium x x sine of the angle of incidence = sine of the angle of incidence =

index of refraction of second index of refraction of second medium x sine of the angle of medium x sine of the angle of refractionrefraction

The index of refraction will vary for The index of refraction will vary for different wavelengths (Table 1 gives different wavelengths (Table 1 gives values for wavelength 589 nm).values for wavelength 589 nm).

This is how a prism works.This is how a prism works.

Snell's Law PraciceSnell's Law Pracice

Sunlight passes into a raindrop at an Sunlight passes into a raindrop at an angle of 22.5angle of 22.5˚̊ from the normal at from the normal at one point on the droplet. What is one point on the droplet. What is the angle of refraction? (Hint: see the angle of refraction? (Hint: see Table 1 p.490)Table 1 p.490)

Snell's Law PracticeSnell's Law Practice

16.716.7˚̊

Snell's Law PracticeSnell's Law Practice

For an incoming ray of light in a For an incoming ray of light in a vacuum wavelength 589 nm, fill in vacuum wavelength 589 nm, fill in the unknown values:the unknown values:from(medium)from(medium) to(medium)to(medium) ΘΘii ΘΘrr

a. a. flint glassflint glass crown glasscrown glass 25.0˚25.0˚ ? ?

b.b. air air ?? 14.5˚ 9.80˚14.5˚ 9.80˚

c. c. airair diamonddiamond 31.6˚ 31.6˚ ? ?

Snell's Law PracticeSnell's Law Practice

a. 27.5a. 27.5˚̊

b. glycerineb. glycerine

c. 12.5c. 12.5˚̊

Snell's Law PracticeSnell's Law PracticeA ray of transparent light of vacuum A ray of transparent light of vacuum wavelength 550 nm traveling in air wavelength 550 nm traveling in air enters a slab of transparent enters a slab of transparent material. The incoming ray makes material. The incoming ray makes an angle of 40.0an angle of 40.0˚̊ with the normal, with the normal, and the refracted ray makes an and the refracted ray makes an angle of 26.0˚ with the normal. Find angle of 26.0˚ with the normal. Find the index of refraction of the the index of refraction of the transparent material. (Assume the transparent material. (Assume the index of refraction of air for light of index of refraction of air for light of wavelength 550 nm is 1.00).wavelength 550 nm is 1.00).

Total Internal Total Internal ReflectionReflection

Total internal reflection occurs for all Total internal reflection occurs for all angles greater than the critical angles greater than the critical angle.angle.

We can find the critical angle byWe can find the critical angle by

sin sin ΘΘcc = = nnrr

nnii

Critical Angle Practice Critical Angle Practice ProblemsProblems

Glycerine is used to make soap and Glycerine is used to make soap and other personal care products. Find other personal care products. Find the critical angle for light traveling the critical angle for light traveling from glycerine (from glycerine (nn = 1.473) into air. = 1.473) into air.

Critical Angle Practice Critical Angle Practice ProblemsProblems

42.842.8˚̊

Solving Lens ProblemsSolving Lens Problems

As with mirrorsAs with mirrors

M = M = h' h' = = -d-dii

hh d drr

M is positive – image is upright and M is positive – image is upright and virtualvirtual

M is negative – image is real and M is negative – image is real and invertedinverted

Sign Conventions Sign Conventions forLensesforLenses

++ --

pp real objectreal object real objectreal object

in front ofin front of in back ofin back of

the lensthe lens the lensthe lens

qq real imagereal image virtual imagevirtual image

in back of in back of in front ofin front of

the lensthe lens the lensthe lens

ff converging converging divergingdiverging

Lenses Practice Lenses Practice ProblemsProblems

An object is placed 20.0 cm in front An object is placed 20.0 cm in front of a converging lens of focal length of a converging lens of focal length 10.0 cm. Find the image distance 10.0 cm. Find the image distance and the magnification. Describe the and the magnification. Describe the image.image.

Lenses Practice Lenses Practice ProblemsProblems

20.0 cm, M = -1.00; real, inverted 20.0 cm, M = -1.00; real, inverted imageimage

Lenses Practice Lenses Practice ProblemsProblems

Sherlock Holmes examines a clue by Sherlock Holmes examines a clue by holding his magnifying glass at holding his magnifying glass at arm's length and 10.0 cm away from arm's length and 10.0 cm away from an object. The magnifying glass has an object. The magnifying glass has a focal length of 15.0 cm. Find the a focal length of 15.0 cm. Find the image distance and the image distance and the magnification. Describe the image magnification. Describe the image that he observes.that he observes.

Lenses Practice Lenses Practice ProblemsProblems

-30.0 cm, M = 3.00; virtual, upright -30.0 cm, M = 3.00; virtual, upright imageimage

Lenses Practice Lenses Practice ProblemsProblems

An object is placed 20.0 cm in front An object is placed 20.0 cm in front of a diverging lens of focal length of a diverging lens of focal length 10.0 cm. Find the image distance 10.0 cm. Find the image distance and the magnification. Describe the and the magnification. Describe the image.image.

Lenses Practice Lenses Practice ProblemsProblems

-6.67 cm, M = 0.333; virtual, upright -6.67 cm, M = 0.333; virtual, upright imageimage