P-2 Please solve this problem very clearly and briefly with step by step explanation. NOTE: but take the index of refraction n = 1.50.
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P-2 Please solve this problem very clearly and briefly with step by step explanation.
NOTE: but take the index of refraction n = 1.50.
![48. (II) Light is incident on an equilateral glass prism at a 45.0°
angle to one face, Fig. 32-52. Calculate the angle at which
light emerges from the
opposite face. Assume
that n = 1.54.
45.0°
FIGURE 32-52
Problems 48 and 65.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F475feb7f-02d2-4072-b49b-ad6dcf51cdba%2F03e4d8ae-8e12-494d-a2f5-3d2ae8c3e4fe%2Fmmlj8op_processed.png&w=3840&q=75)
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- A goldfish is swimming inside a spherical bowl of water having an index of refraction n = 1.333. Suppose the goldfish is p 10.4 cm from the wall of a bowl of radius |R| = 15.8 cm, as in the figure below. Neglecting the refraction of light caused by the wall of the bowl, determine the apparent distance of the goldfish from the wall according to an observer outside the bowl. 1.07 Your response differs from the correct answer by more than 10%. Double check your calculations. cm behind the glassCalculate the angle of refraction at the air/core interface, r . critical angle, c and incident angle at the core/cladding interface, i . Will this light ray propagate down the fiber? You have the following data: nair = 1, ncore = 1.46, ncladding =1.43, incident =12o r = 8.2o, c = 78.4o, i = 81.8olight will propagate Formula summery Index of Refraction n= n sin 01= n, sin 0, o = sin1(2) Snell's Law Critical Angle Acceptance angle a= sin1 Numerical Aperture NA = sin a = ni- nịLight rays in a material with index of refrection 1.39 can undergo total internal reflection when they strike the interface with another material at a critical angle of incidence. Find the second material's index of refraction n when the required critical angle is 73.3°. n =
- w9-10 A diamond in air is illuminated with white light. On one particular facet, the angle of incidence is 25.70°. Inside the diamond, red light (λ = 660.0 nm in vacuum) is refracted at 10.88° with respect to the normal; blue light (λ = 470.0 nm in vacuum) is refracted at 10.13°. How would a diamond look if there were no dispersion? a. The diamond would look white. b. The diamond would look red. c. The diamond would look blue. d. The diamond would be clear.A goldfish is swimming inside a spherical bowl of water having an index of refraction n = 1.333. Suppose the goldfish is p = 10.7 cm from the wall of a bowl of radius |R| = 19.2 cm, as in the figure below. Neglecting the refraction of light caused by the wall of the bowl, determine the apparent distance of the goldfish from the wall according to an observer outside the bowl.cm behind the glassWhite light is incident onto a 30° prism at the 40° 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 of red light. (Figure 1) Figure White light 40⁰ 30° 1 of 1 Part A At what angle does red light emerge from the rear face? Express your answer with the appropriate units. 6 = ☐ O μA Value Submit Request Answer < Return to Assignment Units Provide Feedback ?
- The index of refraction n for water is 1.33. What is the speed of light v in water? U = m/sA ray of light is incident on an air/water interface. The ray makes an angle of θ1 = 29 degrees with respect to the normal of the surface. The index of the air is n1 = 1 while water is n2 = 1.33. Write an expression for the reflection angle ψ, with respect to the surface.Light traveling through medium 3 (n3 = 2.4) is incident on the interface with medium 2 (n2 = 2.0) at angle θ. If light does enter into medium 2 but no light enters into medium 1 (n1 = 1.6), what can we conclude about the range of values for θ?