Solve using snells law for the red light at the air

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Solve: Using Snell’s law for the red light at the air-glass boundary, air air red red sin sin n n θ θ = 1 air air red red sin sin n n θ θ = 1 1.0sin30 sin 19.30 1.513 ° = = ° From the geometry of the diagram, red violet red violet tan tan 10.0 cm 10.0 cm d d θ θ = = ( ) ( ) red 10.0 cm tan 19.30 3.502 cm d = ° = violet 3.502 cm 0.1 cm 3.402 cm d = = 1 1 violet violet 3.402 cm tan tan 18.79 10.0 cm 10.0 cm d θ = = = ° That is, white light is incident on a piece of glass at 30 ° , and it gets dispersed. The violet light makes an angle of 18.79 ° with the vertical. Using Snell’s law, violet violet air air sin sin n n θ θ = ( ) violet 1.0 sin30 1.552 sin18.79 n ° = = °
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23.56. Model: Use the ray model of light and the phenomena of refraction and dispersion. Visualize: Solve: Since violet light is perpendicular to the second surface, it must reflect at θ violet = 30 ° at the first surface. Using Snell’s law at the air-glass boundary where the ray is incident, air air violet violet sin sin n n θ θ = ( ) air air violet violet 1.0 sin50 sin 1.5321 sin sin30 n n θ θ ° = = = ° Since n violet = 1.02 n red , red 1.5021. n = Using Snell’s law for the red light at the first surface red red air air sin sin n n θ θ = 1 red 1.0sin50 sin 30.664 1.5021 θ ° = = ° The angle of incidence on the rear face of the prism is thus θ r glass = 30.664 ° 30 ° = 0.664 ° . Using Snell’s law once again for the rear face and for the red wavelength, red r glass air r air sin sin n n θ θ = red r glass 1 r air air sin sin n n θ θ = 1 1.5021sin0.664 sin 0.997 1.0 ° = = ° Because θ v air = 0 ° and θ r air = 0.997 ° , r air v air 0.997 1.00 . φ θ θ = = ° ≅ °
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23.57. Model: Use the ray model of light and the phenomenon of refraction. Visualize: Solve: (a) The critical angle θ c for the glass-air boundary is glass c air sin sin90 n n θ = ° 1 c 1.0 sin 41.81 1.50 θ = = ° For the triangle ABC, ( ) glass 1 c 120 90 180 θ θ + ° + ° − = ° ⇒ ( ) glass 1 180 120 90 41.81 11.81 θ = °− °− °− ° = ° Having determined θ glass 1 , we can now find θ air 1 by using Snell’s law: air air 1 glass glass 1 sin sin n n θ θ = 1 air 1 1.50 sin11.81 sin 17.88 1.0 θ × ° = = ° Thus, the smallest angle θ 1 for which a laser beam will undergo TIR on the hypotenuse of this glass prism is 17.9 ° . (b) After reflecting from the hypotenuse (face 3) the ray of light strikes the base (face 2) and refracts into the air. From the triangle BDE, ( ) ( ) glass 2 c 90 60 90 180 θ θ °− + ° + ° − = ° glass 2 90 60 90 41.81 180 18.19 θ = °+ °+ °− °− ° = ° Snell’s law at the glass-air boundary of face 2 is glass glass 2 air air 2 sin sin n n θ θ = glass glass 2 1 air 2 air sin sin n n θ θ = 1 1.50sin18.19 sin 27.9 1.0 ° = = ° Thus the ray exits 27.9 ° left of the normal.
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23.58.
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