====== Problems ======
===== - Speed of light in a medium =====
Calculate the speed of light in zircon, a material used in jewelry to imitate diamond. Index of refraction of zircon is $n_z=1.923$. Compare the speed with the speed in diamond (index of refraction $n_d=2.419$).
Speed of light in zircon $v_z = c/n_z$ where speed of light in vacuum $c=300$ Mm ($300\times 10^6$ m). So $v_z = 156$ Mm.
Speed of light in diamond $v_d = c/n_d = 124$ Mm.
So the speed is lower in diamond because it has higher index of refraction.
===== - Determining index of refraction =====
Light enters a medium from air with an angle of incidence $\theta_1=30^\circ$ and is refracted at an angle $\theta_2=22^\circ$. What is the index of refraction of the medium?
Snell's law or the law of refraction states $n_1 \sin\theta_1 = n_2 \sin\theta_2$ where $n_1=1$ is the index of refraction of air and $n_2$ that of the other medium.
So the index of refraction of the medium
$$ n = \frac{\sin\theta_1}{\sin\theta_2} = \frac{\sin 30^\circ}{\sin 22^\circ} = 1.33. $$
Note that light bends more in a medium with larger index of refraction.
===== - Determining angle of refraction =====
Light enters from air to a medium at an angle of $30^\circ$ with the normal to the surface. Calculate the angle of refraction in the medium if the medium is (a) diamond ($n=2.419$) and (b) zircon ($n=1.923$).
Snell's law relates the incident angle $\theta_1=30^\circ$ with the angle of refraction $\theta_2$. Index of refraction of air $n_1=1$ and index of the other medium $n_2\equiv n$.
According to the law $\sin\theta_1 = n\sin\theta_2$. So the angle $\theta_2 = \sin^{-1}(\sin\theta_1/n)$.
If the second medium is diamond, $\theta_2 = \sin^{-1}(\sin 30^\circ/2.419) = 11.9^\circ. $
If the second medium is zircon, $\theta_2 = \sin^{-1}(\sin 30^\circ/1.923) = 15^\circ. $
The angle of refraction is higher in zircon because it has lower index of refraction.
===== - Determining critical angle =====
What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air? The index of refraction for polystyrene is 1.49.
The critical angle $\theta_c = \sin^{-1}(n_2/n_1)$.
The pipe is surrounded by air. Light needs to go from the pipe to the air. So the pipe is the first medium and air the second. Hence $n_1\equiv n = 1.49$ and $n_2=1$.
Therefore $\theta_c = \sin^{-1}(1/n) = 42^\circ. $
===== - Dispersion by flint glass =====
A beam of white light goes from air into flint glass at an incidence angle of 43.2°. What is the angle between the red (660 nm) and violet (410 nm) parts of the refracted light?
{{ :courses:phy101:11:11a.jpg?nolink&320 |}}
The index of refraction of different media at different wavelengths are given in the following table.
^ Medium ^ Red \\ 660 nm ^ Orange \\ 610 nm ^ Yellow \\ 580 nm ^ Green \\ 550 nm ^ Blue \\ 470 nm ^ Violet \\ 410 nm ^
| Water | 1.331 | 1.332 | 1.333 | 1.335 | 1.338 | 1.342 |
| Diamond | 2.410 | 2.415 | 2.417 | 2.426 | 2.444 | 2.458 |
| Glass, crown | 1.512 | 1.514 | 1.518 | 1.519 | 1.524 | 1.530 |
| Glass, flint | 1.662 | 1.665 | 1.667 | 1.674 | 1.684 | 1.698 |
| Polystyrene | 1.488 | 1.490 | 1.492 | 1.493 | 1.499 | 1.506 |
| Quartz, fused | 1.455 | 1.456 | 1.458 | 1.459 | 1.462 | 1.468 |
----
For red light Snell's law becomes $n_1 \sin\theta_1 = n_r \sin\theta_r $ where $n_1=1$ for air and $n_r$ is the index of refraction of flint glass at red wavelengths.
For red light, $n_r = 1.662$ and, hence, $\theta_r = \sin^{-1}(\sin\theta_1 / n_r) = \sin^{-1}(\sin 43.2^\circ / 1.662) = 24.3^\circ$.
For violet light, $n_v = 1.698$ and, hence, $\theta_v = \sin^{-1}(\sin\theta_1 / n_v) = \sin^{-1}(\sin 43.2^\circ / 1.698) = 23.7^\circ$.
Violet light bends more than the red light because it has smaller wavelength, but the difference between the two angles $\theta_r-\theta_v = 0.6^\circ = 36$ arcmin, which is very small.
===== - Solar power =====
One of the solar technologies used today for generating electricity involves a device (called a parabolic trough or concentrating collector) that concentrates sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where the thermal energy is transferred to another system that is used to generate steam and eventually generates electricity through a conventional steam cycle. The figure shows such a working system in southern California. The real mirror is a parabolic cylinder with its focus located at the pipe; however, we can approximate the mirror as exactly one-quarter of a circular cylinder.
{{ :courses:phy101:11:11b.jpg?nolink&600 |}}
a) If we want the rays from the sun to focus at 40.0 cm from the mirror, what is the radius of the mirror? \\ \\
b) What is the amount of sunlight concentrated onto the pipe, per meter of pipe length, assuming the insolation (incident solar radiation) is 900 W/m$^2$? \\ \\
c) If the fluid-carrying pipe has a 2.00-cm diameter, what is the temperature increase of the fluid per meter of pipe over a period of 1 minute? Assume that all solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil (density $\rho=800$ kg/m$^3$; heat capacity $c=1670$ J kg$^{-1}$ K$^{-1}$).
a) The radius of the mirror can be found from the mirror equation remembering that the sun is the object here and, hence, distance of the object $d_o \approx \infty$ compared to image distance $d_i$. So
$$ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \Rightarrow f = \left(\frac{1}{d_o} + \frac{1}{d_i}\right)^{-1} = \left(\frac{1}{\infty} + \frac{1}{40}\right)^{-1} = 40 \text{ cm}. $$
Therefore, the radius of curvature of the mirror $R = 2f = 80$ cm.
b) The insolation is given; 900 W of power falls within 1 m$^2$ area of the mirror surface. We have to find the total power incident on 1 m length of the mirror. The mirror is one-fourth part or quarter-section of a cylinder. Area of a cylindrical surface $A = 2\pi R L$ where $R$ is the radius and $L$ the length.
The area of 1-m length of our mirror is, then, $A = 2\pi R / 4 = \pi R / 2 = 1.26$ m$^2$ because $L=1$ m.
So the amount of sunlight concentrated on a 1-m length of the pipe is $900\times 1.26 = 1134$ W.
c) The increase in temperature can be calculated using the heat capacity equation: $Q=mc\Delta T$. The radius of the pipe $r=d/2=1$ cm = 0.01 m.
The mass of a 1-m section of the cylindrical pipe $m=\rho V = \rho \pi r^2 L = \rho \pi r^2 $ because length $L=1$ m. So $m=800\times 3.1416 \times 0.01^2 = 0.251$ kg.
The energy transferred in 1 min $Q=Pt=1134\times 60=68040$ J. So the increase in temperature of the fluid (mineral oil) inside the pipe
$$ \Delta T = \frac{Q}{mc} = \frac{68040}{0.251\times 1670} = 162 \text{ K} = 162^\circ \text{ C}.$$
So the temperature of the fluid increases by $162^\circ$ C in just 1 minute.
===== - Using the lensmaker's equation =====
Find the radius of curvature of a biconcave lens symmetrically ground from a glass with index of refraction 1.55 so that its focal length in air is 20 cm (for a biconcave lens, both surfaces have the same radius of curvature).
We have to use the lensmaker's equation to find the focal length and then the radius.
$$ \frac{1}{f} = \left(\frac{n_2}{n_1}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right) \Rightarrow R = - 2f \left(\frac{n_2}{n_1}-1\right) $$
where $f=-20$ cm (negative because concave lenses are diverging), $n_1=1$ and $n_2=1.55$. So $R=22$ cm.
===== - Converging lenses =====
Find the location, orientation, and magnification of the image for a 3.0 cm high object at each of the following positions in front of a convex lens of focal length 10.0 cm.\\
a) $d_o=50$ cm \\
b) $d_o=5$ cm \\
c) $d_o=20$ cm.
Use the thin lens equation: $f^{-1} = d_o^{-1} + d_i^{-1}$.
a) Location of the image $d_i = (f^{-1}-d_o^{-1})^{-1} = 1/(1/10-1/50) = 12.5$ cm. Focal length is positive because convex lenses are converging.
Magnification of the image $m=-d_i/d_o = -12.5/50 = -0.25$ which is negative meaning the image is inverted. The orientation of the image is opposite to the object.
Height of the image $h_i = |m| h_o = 0.25\times 3 = 0.75$ cm.
b) Location of the image $d_i = 1/(1/10-1/5) = -10$ cm. Magnification $m=-(-10)/5=+2$. The image is upright. The height of the image $h_i=2\times 3 = 6$ cm.
c) Location of the image $d_i = 1/(1/10-1/20) = 20$ cm. Magnification $m=-20/20=-1$. The image is inverted. The height of the image $h_i=1\times 3 = 3$ cm which is equal to the object.
===== - Ray tracing and the lens equations =====
To project an image of a light bulb on a screen 1.50 m away, you need to choose what type of lens to use (converging or diverging) and its focal length (Figure below). The distance between the lens and the light bulb is fixed at 0.75 m. Also, what is the magnification and orientation of the image?
{{ :courses:phy101:11:11.b.png?nolink&530 |}}
The image must be real. So we choose a converging lens. Ray tracing can be used to prove that converging lenses create real image.
Use thin-lens equation for finding focal length. $f = 1/(1/d_o + 1/d_i) = 1/(1/0.75 + 1/1.5) = 0.5$ m.
Magnification of the image $m = -d_i/d_o = 1.5/0.75 = -2$. So the image will be inverted.
The image is inverted, real and larger than the object.