b) In the diagram: identify the angle of incidence and the angle of reflection.
ID: 1462171 • Letter: B
Question
b) In the diagram: identify the angle of incidence and the angle of reflection.
angle of incidence: 2; angle of reflection: 4
angle of incidence: 2; angle of reflection: 3
angle of incidence: 1; angle of reflection: 4
angle of incidence: 1; angle of reflection: 2
angle of incidence: 3; angle of reflection: 1
angle of incidence: 3; angle of reflection: 4
angle of incidence: 4; angle of reflection: 3
angle of incidence: 4; angle of reflection: 1
angle of incidence: 1; angle of reflection: 3
angle of incidence: 3; angle of reflection: 2
angle of incidence: 4; angle of reflection: 2
c) In the diagram below: a ray of light traveling in air (n = 1.00) strikes the edge of a transparent rectangle (n = 1.57, width w = 41.2 cm) at point A. The angle of incidence of the ray is 1 = 50.2o. The ray passes through and exits through the opposite side at point B.
i. Find the vertical displacement from point A to point B. cm
ii. Calculate the time it took the ray to travel from point A to point B. seconds
Explanation / Answer
b)
angle of incidence: 2; angle of reflection: 3
c)
Apply Snells's law
sin(theta1)/sin(theta2) = n2/n1
sin(50.2)/sin(theta2) = 1.57/1
sin(theta2) = sin(50.2)/1.57
= 0.489
theta2 = sin^-1(0.489)
= 29.3 degrees
let y is vertical dispalcement from A to B,
tan(29.3) = y/w
y = w*tan(29.3)
= 41.2*tan(29.3)
= 23.12 cm <<<<<<<<<<-------------Answer
ii)
from fig, cos(theta2) = W/AB
AB = W/cos(theta2)
= 0.412/cos(29.3)
= 0.472 m
speed of light in the material, v = c/n2
= 3*10^8/1.57
= 1.91*10^8 m/s
so, time taken , t = AB/v
= 0.472/(1.91*10^8)
= 2.47*10^-9 s <<<<<<<<<<-------------Answer
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