When light passes through the glass block shown below, it is shifted laterally by distance d. The glass block has index of refraction nglass = 1.50. What is d? The image is attached below. Make sure to show all work.

Answer the following questions A. The critical angle for total internal reflection at a diamond-air interface is 25 degrees. Suppose the light falls at an angle to the normal. Then, total internal reflection will occur if the incident medium is: a) air and angle < 25 degrees b) air and angle > 25 degrees c) air and angle = 25 degrees d) diameter and angle > 25 degrees e) diamond and angle < 25 degrees B. A ray of light in water (index n1) hits its surface (with air) at a critical angle for total internal reflection. Now some oil (index n2) floats in the water. The angle between the ray in the oil and the normal is: note: sin -1 (1.00), it is read as the inverse sine of one. a) sin -1 (1/n1) b) sin -1 (1/n2) c) sin -1 (1.00) d) sin -1 (n1/n2) e) sin -1 (n2/n1) C. A ray of light passes obliquely (non-zero angle) through a glass plate that has parallel faces. The emerging ray: a) it leans more towards the normal than the incident ray b) is reflected internally toatl c) tilts…

Calculate the critical angle for the light traveling from glass (n glass= 1.54) to air (n=1.00). Express your answer in degrees. Normal format with 3 SF. air

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 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- nị