16. A light ray is incident on the interface between a diamond and air from with

Question

16. A light ray is incident on the interface between a diamond and air from within the diamond. What is the critical angle for total internal reflection n(diamond) = 2.42 and nair) = 1)? A) 65.6 B) 24.4 C) 33.3 D) 48.8% E) 413% A converging lens has a focal length of +20 cm. An object 5 cm in height is placed 40 em in front of the lens (applies to problem 17 & 18) 17. Where is the image located? A) There is no image. B) There is a real image 13.33 cm behind the lens. c) There is a virtual image 1333 cm in front of the lens. D) There is a real image 40 cm behind the lens. E) There is a virtual image 40 cm behind the lens. 18. What is the height of the image? A) The image is 5 cm high and inverted. B) No image can be seen. c) The image is 5 cm high and erect. D) The image is 1.67 cm high and erect. EB) The image is 1.67 cm high and inverted. 19. An object 3 cm high is placed 10 cm in front of a mirror. What type of mirror and what radius of curvature is needed for an inverted image that is 6 cm high? A convex, R - 20 cm B) convex, R 20 em C concave, R - 13.3 cm D) conve, R in 40 em E conve, R - 40 em

Explanation / Answer

16)

here,

refractive index of air , na = 1

n(diamond) = 2.42

the crictical angle , theta = arcsin(na/n)

theta = arcsin(1/2.42)

theta = 24.4 degree

the critical angle is B) 24.4 degree

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