Derive the optical law of reflection. Hint: Let light go from the point A = (x1, Y1) to B = (x2, Y2) via an arbitrary point P = (x, 0) on a mirror along the x axis. Set dt/dx = (n/c)dD/dx = 0, where D = distance APB, and show that then 0 = 4. 1. B
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- Suppose a ray comes from a point (2, 8) and hits at mirror point (6, 6). Assume that perpendicular vector a point (6, 6) is ?̅= (0, 3). Find the reflection emitting point from (6, 6).Using the direction of the reflected ray formula.Let C be a black circular disk in front of a white background. The circular disk is parallel to the image plane. This disk is projected on the image plane through a pinhole. What would be the shape of the disk’s projection on the image plane ? [X,Y,Z]T with (X-X0) 2 +(Y-Y0) 2 =R2 , Z=Z0, where [X0,Y0,Z0] T is the center of the disk and R is its radiusA light ray is incident normal to one face of a triangular block and strikes point P on the other side. The block is made of material with index of refraction of 1.71. The block is immersed in oil, having index of refraction of 1.10. What is the maximum angle y for which total internal reflection occurs at P? P O Not possible to have total reflection at P. O 30° O 60° O 50° O 40° O 45°
- Q.7 A retroreflector is any optic that reflects an incident ray, such that the exiting ray is parallel (but opposite) to the incoming ray. One version of a "cat's eye" retroreflector uses a thin positive lens of focal length fand a flat mirror as shown below. (a) Determine the matrix of retroreflector shown in Fig. Q7. (b) Use the matrix method to prove that it is, indeed, a retroreflector. Fig. Q7= O For refraction across an air-glass interface where ng 1.63, make a plot of the refracted angle as a function of the incident angle, with 0; = 0 to 90°.Consider scenarios A to F in which a ray of light traveling in material 1 is incident onto the interface with material 2. (Figure 1) Material 1 (n1) Material 2 (n2) A air (1.00) water (1.33) В water (1.33) air (1.00) diamond (2.42) air (1.00) D air (1.00) quartz (1.46) E benzene (1.50) water (1.33) F diamond (2.42) water (1.33) Part A For which of these scenarios is total internal reflection possible? List all correct answers in alphabetical order. For example, if scenarios A and E are correct, enter AE. • View Available Hint(s)
- When a light ray crosses from water into glass, it emerges at an angle of 30° with respect to the normal of the interface. What is its angle of incidence? I used Snell's Law to find the angle of incidence as 34.8 degrees (rounded to 3 sig fig), using index of refraction for water as 1.333 and for glass as 1.52. However, the answer was marked as wrong. Could you please explain steps for solving to get theta as approximately 42 degrees?o A ray of light strikes a flat block of glass (n= 1.50) of thickness 2.00 cm at an angle of 30.0° with the normal. Trace the light beam through the glass, and find the angles of incidence and refraction at each surface.A ray of light strikes a surface at an angle of 72.2° with respect to the normal. Use your knowledge on reflection and refraction of light to draw the reflected and/or refracted rays through the crown glass and the materials A (nA=1.6) and B (nB=1.362). Justify your answers Note: your illustration must include the significant angles such as the angles of reflection and refraction.)
- Consider an incident ray striking the surface of a material at an angle Ѳ1 = 65° with respect to the normal. The ray undergoes refraction. What is the angle Ѳ2 ? Given that n1 = 1.1 and n2 = 1.47.The critical angle for total internal reflection at a turpentine-air interface is 42.8°. A ray traveling in the liquid has an angle of incidence of 32.0° at the interface. What angle does the refracted ray in air make with the normal? O51.3° O 14.7° O 53.0° O 23.8° here to search L6 Light rays enter the plane surface of a glass hemisphere of radius 5 cm and refractive index 1.5. a. Using the system matrix representing the hemisphere, de- termine the exit elevation and angle of a ray that enters parallel to the optical axis and at an elevation of 1 cm. b. Enlarge the system to a distance x beyond the hemi- sphere and find the new system matrix as a function of x. c. Using the new system matrix, determine where the ray described above crosses the optical axis.