# Quiz 8

```November 14, 2008
PHY2054 Discussion-Fall ‘08
Quiz 8 (Chapter 21.8-21.13, 22.1-22.7)
Name:
UFID:
*1. (5pts) Assume that the intensity of solar radiation incident on Earth is 1 340 W/m2. The Earth is
1.50×1011 m away from the Sun. Calculate the intensity of the solar radiation incident on Venus,
which is 1.08×1011 m away from the Sun.
The wave front of solar radiation is a sphere centered at the Sun. Thus the total power of the solar
radiation is given by intensity times area of the sphere.
P = IE×4πrE2 ( = 1340×4×π×(1.50×1011)2 = 3.79×1026 W)
At the surface of the Venus, the intensity is
IV= P/(4πrV2) = IE×4πrE2/(4πrV2) = IE×(rE/rV)2 = 1340×(1.5/1.08)2 = 2.58×103 W/m2
*2. (5pts) A beam of light both reflects and refracts at the surface between air and glass. If the index
of refraction of the glass is 1.55, find the angle of incidence in the air that would result in the
reflected ray and the refracted ray being perpendicular to each other.
Since the angle of reflection equals the angle of incidence and a straight line
makes 180º, we have
θ1+90º+θ2 = 180º ⇒ θ2 = 90º- θ1,
where θ2 is the angle of refraction.
Applying Snell’s law, we solve the equation for θ1.
sinθ1 = ngsinθ2 ⇒ sinθ1 = ngsin (90º-θ1) = ngcosθ1
⇒ tanθ1 = ng ⇒ θ1 = tan-1(ng) = tan-1(1.55) = 57.2º
***3. (5pts) An optical fiber consists of a transparent core surrounded by cladding, which is a
material with a lower index of refraction than the core. The index of refraction of the cladding is
1.15. What is the minimum index of refraction of the core if all the incoming rays are transmitted in
the optical fiber? Assume the optical fiber is surrounded by air.
If a light ray strikes the cladding with an angle of incidence larger
than the critical angle, the ray transmits in the fiber. Therefore,
θ>θc ⇒ sinθ > sinθc = n1/n2,
where n1 is the refractive index of the cladding and n2 is that of the
core. φ2 and θ are related as θ = 90º-φ2, and φ1 and φ2 satisfies
Snell’s law: sinφ1 = n2sinφ2. Using these two equations, we express θ in terms of φ1.
n1/n2,< sinθ = sin(90º-φ2) = cosφ2 = √(1-sin2φ2) = √(1-sin2φ1/n2)
⇒ (n1/n2)2 < 1-sin2φ1/n2 ⇒ n2 > √(n12+ sin2φ1)
The maximum angle of incidence at the left surface is 90º, thus for all the ray to transmit, n2 needs to
be larger than
n2 > √(n12+ sin290º) = √(1.152+12) = 1.52
**4. (5pts) A prism with an apex angle of 55.0º has an index of refraction of 1.650 for blue light of
wavelength 430 nm and 1.615 for red light of wavelength 680 nm. A beam containing these two
colors is incident from air onto the left surface of the prism and only the red light goes out of the
right surface of the prism. Find the range of the allowed values of the incident angle at the left
surface.
Since only the red light traverses the prism, only the blue light undergoes
the total internal reflection at the right surface. When the angle of
incidence is critical at the right surface, the angle of incidence at the left
surface is calculated as follows:
sinθRc = 1/nR ⇒ θRc= sin-1(1/nR) = sin-1(1/1.615)= 38.3º
θR2 = 90-[180-φ-(90- θRc)] = φ- θRc = 55-38.3 = 16.7º
sinθR1 =nRsinθR2 ⇒ θR1 = sin-1(nRsinθR2) = sin-1(1.615×sin(16.7º)) = 27.7º for the red light
θBc = sin-1(1/nB) = sin-1(1/1.65)= 37.3º
θB2 = φ- θBc = 55-37.3 = 17.7º
θB1 = sin-1(nBsinθB2) = sin-1(1.65×sin(17.7º)) = 30.1º for the blue light
The angle of incidence at the right surface gets larger when the angle of incidence at the left surface
gets smaller, thus the range of the allowed angles of incidence is
27.7º < θ < 30.1º.
```
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