For a 645.0-nm light, calculate the critical angle for the material of the refractive index 1.47 surrounded by air. State your answer to nearest 0.1°.
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- Given: The light ray travels from the air to water under the angle of incidence of 40.0°. What is the angle of refraction of the light in water? The refraction index for air and water is 1 and 1.33, respectively. Submit a scanned copy of YOUR detailed and handwritten working in the Drag & Drop box field. Unclear, non-readable files or files sent via email will receive zero.For light of wavelength 589 nm, calculate the critical anglesfor the following substances when surrounded by a substance with an index of refraction of 1.341 sodium chloride. 74.53° 60.29° 1.23° 1.4° noneFrom air (n=1), white light enters a prism with an incidence angle of 75.00° as measured from the normal. With this information, find the angular separation of the dispersed colors - red with a wavelength of 660 nm (n=1.488) and blue with a wavelength of 470 nm (n=1.499). a. 0.358° b. 0.241° C. 0.12° d. 0.074°
- Light is incident along the normal on face AB of a glass prism of refractive index 1.61. What is the largest value the angle a can have without having any light reflected out of the prism at face AC if the prism is immersed in oil (noil = 1.47)? A Incident ray B А. 24° С. 39° В. 29° D. 66°Calculate 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 Answers: r = 8.2o , c = 78.4o , i = 81.8o light will propagateA laser beam is incident at an angle of 30.00to the vertical onto a solution of blood plasma in water. If the beam is refracted to 19.240 to the vertical: What is the index of refraction of the blood plasma solution? Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find: Its wavelength. Its frequency. Its speed in the solution. The critical angle for the blood plasma solution when surrounded by air
- Find the Brewster angle for an air-oil interface for light incident from the air. Take the oil refractive index to be n=1.79.For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by a substrance with an index of refraction of 1.435: (a) fused quartz 79.81º 1.53º 83.76º 1.39º (b) polystyrene 1.39º 74.38º 79.54º 1.39º (c) sodium chloride 68.34º 62.67º 1.39º 1.32ºLight is incident along the normal on face AB of a glass prism of refractive index 1.61. What is the largest value the angle a can have without having any light reflected out of the prism at face AC if the prism is immersed in oil (noil = 1.47)? A Incident ray B a A. 24° С. 39° В. 29° D. 66°
- The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660. A beam of white light in this glass strikes the glass-air interface at a 28.70° angle of incidence and refracts out into the air. What is the angular separation between the red and violet components of the spectrum that emerges from the glass? angular separation:A light beam strikes a piece of glass with an incident angle of 45.00°. The beam contains two colors: 450.0 nm and an unknown wavelength. The index of refraction for the 450.0-nm light is 1.482. Assume the glass is surrounded by air, which has an index of refraction of 1.000. Determine the index of refraction n, for the unknown wavelength if its refraction angle is 0.6950° greater than that of the 450.0 nm light.For 589-nm light, calculate the critical angle for the following materials surrounded by air. (a) cubic zirconia (n = 2.20) (b) gallium phosphide (n = 3.50) (c) polystyrene (n = 1.49)