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Chapter 14 Refraction of Light Hr Physics
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Sorry this is late.
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Snell's Law n i sinΘ i = n r sinΘ r index of refraction of first medium x sine of the angle of incidence = index of refraction of second medium x sine of the angle of refraction The index of refraction will vary for different wavelengths (Table 1 gives values for wavelength 589 nm). This is how a prism works.
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Snell's Law Pracice Sunlight passes into a raindrop at an angle of 22.5˚ from the normal at one point on the droplet. What is the angle of refraction? (Hint: see Table 1 p.490)
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Snell's Law Practice 16.7˚
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Snell's Law Practice For an incoming ray of light in a vacuum wavelength 589 nm, fill in the unknown values: from(medium)to(medium)Θ i Θ r a. flint glasscrown glass25.0˚ ? b. air?14.5˚ 9.80˚ c. airdiamond31.6˚ ?
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Snell's Law Practice a. 27.5˚ b. glycerine c. 12.5˚
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Snell's Law Practice A ray of transparent light of vacuum wavelength 550 nm traveling in air enters a slab of transparent material. The incoming ray makes an angle of 40.0˚ with the normal, and the refracted ray makes an angle of 26.0˚ with the normal. Find the index of refraction of the transparent material. (Assume the index of refraction of air for light of wavelength 550 nm is 1.00).
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Total Internal Reflection Total internal reflection occurs for all angles greater than the critical angle. We can find the critical angle by sin Θ c = n r n i n i
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Critical Angle Practice Problems Glycerine is used to make soap and other personal care products. Find the critical angle for light traveling from glycerine (n = 1.473) into air.
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Critical Angle Practice Problems 42.8˚
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Solving Lens Problems As with mirrors M = h' = -d i h d r M is positive – image is upright and virtual M is negative – image is real and inverted
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Sign Conventions forLenses +- preal objectreal object in front ofin back of the lensthe lens qreal imagevirtual image in back of in front of the lensthe lens fconverging diverging
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Lenses Practice Problems An object is placed 20.0 cm in front of a converging lens of focal length 10.0 cm. Find the image distance and the magnification. Describe the image.
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Lenses Practice Problems 20.0 cm, M = -1.00; real, inverted image
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Lenses Practice Problems Sherlock Holmes examines a clue by holding his magnifying glass at arm's length and 10.0 cm away from an object. The magnifying glass has a focal length of 15.0 cm. Find the image distance and the magnification. Describe the image that he observes.
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Lenses Practice Problems -30.0 cm, M = 3.00; virtual, upright image
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Lenses Practice Problems An object is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification. Describe the image.
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Lenses Practice Problems -6.67 cm, M = 0.333; virtual, upright image
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