Consider a triangular piece of diamond surrounded by air, as shown below. The index of refraction of diamond is na light ray enters the diamond normal to the bottom surface (surface 1), as shown. Can this light refract out of the diamond when it reaches surface 2? Explain why it can or cannot refract out. 2.42 and the index of refraction of air is nair = 1.00. The %3D 60° 2 30 1
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Q: The index of refraction for violet light in silica flint glass is 1.66 and that for red light is…
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Q: = 1.0 cm and 7. A narrow beam of light passes through a plate of glass with thickness s an index of…
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- White light is incident on a 30o prism at the 40o angle shown in the figure. Violet light emerges perpendicular to the rear face of the prism. The index of refraction of violet light in this glass is 2.0% larger than the index of refraction for red light. At what angle ϕ does red light emerge from the rear face? Recall that nair = 1.0Current Attempt in Progress In the drawing, suppose that the angle of incidence is 0₁-27.1°, the thickness of the pane is 3.21 mm, and the refractive index of the pane is n₂ = 1.33. Find the amount (in mm) by which the emergent ray is displaced relative to the incident ray. Number i eTextbook and Media Save for Later Air (n3 = n1) Glass (₂) Air (₁) Units 102 8₁1 Incident ray jez 02 Emergent ray Displacement Attempts: 0 of 6 used Submit AnswerThe figure shows a refracted light beam in linseed oil making an angle of p = 30.6° with the normal line NN'. The index of refraction of linseed oil is 1.48. Air N Linseed oil Water (a) Determine the angle 0. (b) Determine the angle 0'.
- Needs Complete typed solution with 100 % accuracy.ou can determine the index of refraction of a substance by determining its critical angle. (a) What is the index of refraction of a substance that has a critical angle of 58.9° when submerged in ethanol, which has an index of refraction of 1.361?(b) What would the critical angle be for this substance in air? °(a) An opaque cylindrical tank with an open top has a diameter of 2.60 m and is completely filled with water. When the afternoon sun reaches an angle of 31.5° above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank (in m)? 4 Determine how the depth of the tank is related to the angle of refraction for the geometry described in the question. Drawing a diagram may be helpful. Then apply Snell's law to find the angle of refraction and use it to determine the depth of the water in the tank. m (b) What If? On winter solstice in Seattle, the sun reaches a maximum altitude of 19° above the horizon. What would the depth of the tank have to be (in m) for the sun not to illuminate the bottom of the tank on that day? m
- (a) The index of refraction for violet light in silica flint glass is 1.66, and that for red light is 1.62. What is the angular spread (in degrees) of visible light passing through a prism of apex angle 60.0° if the angle of incidence is 45.0°? See figure below. Visible light O Angular spread 0 Deviation of red light B V Screen R (b) What If? What is the angular spread (in degrees) of visible light passing through a prism of apex angle 60.0° if the angle of incidence is 90°?Needs Complete typed solution with 100 % accuracy.A very large piece of clear ice has a cave man frozen inside it. When viewed at an anglefrom the top of the ice the cave man appears to be 1.00 ft below the surface of the ice.What is the actual depth (in units of inches) of the cave man from the top of the ice?The index of refraction of the ice is 1.309 Do NOT use i for the angle of incidence or r for the refracted angle