Question

Chapter 25 Part 3

1. You can determine the index of refraction of a substance by determining its critical angle. What is the index of refraction of a substance that has a critical angle of 46.7º when submerged in water (n = 1.333)?

2. What magnification will be produced by a lens of power –4 D (such as might be used to correct myopia) if an object is held 28.3 cm away?

3. A doctor examines a mole with a 16.7 cm focal length magnifying glass held 11.6 cm from the mole What's the magnification?

4. Components of some computers communicate with each other through optical fibers having an index of refraction n = 1.622. What time in nanoseconds is required for a signal to travel 0.109 m through such a fiber?

Answer 1

1) theta = 46.7 degrees, n = 1.333
from snells formula
n1 sin(theta) = n2 sin(theta2)
n1*sin(46.7) =1.333*sin(90)
n1 = 1.832

2) P = -4D
f = 1/P = -1/4 = -1/400
from lens formula
1/f= 1/v -1/u
-1/400 = 1/v -1/28.3
v = 30.45 cm
magnification = v/u = 30.45/28.3
m = 1.076

3) f = 16.7 cm , u =11.6 cm
from lens formula 1/f =1/v - 1/u
1/16.7 =1/v - (1/11.6)
v = 6.845 cm
magnification = v/u = 6.845/11.6
m = 0.59

4) n = 1.622, d = 0.109 m
n = c/v
v = d/t
1.622 = 3*10^8*t/0.109
t = 5.89*10^-10 s