10.15 Image formation by mirrors  (Page 4/9)

1. If we wish to place the fluid-carrying pipe 40.0 cm from the concave mirror at the mirror’s focal point, what will be the radius of curvature of the mirror?
2. Per meter of pipe, what will be the amount of sunlight concentrated onto the pipe, assuming the insolation (incident solar radiation) is $\text{0.900 k}{\text{W/m}}^2$ ?
3. If the fluid-carrying pipe has a 2.00-cm diameter, what will be the temperature increase of the fluid per meter of pipe over a period of one minute? Assume all the solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.

Strategy

To solve an Integrated Concept Problem we must first identify the physical principles involved. Part (a) is related to the current topic. Part (b) involves a little math, primarily geometry. Part (c) requires an understanding of heat and density.

Solution to (a)

To a good approximation for a concave or semi-spherical surface, the point where the parallel rays from the sun converge will be at the focal point, so $R=2f=\text{80.0 cm}$ .

Solution to (b)

The insolation is $\text{900 W}{\text{/m}}^2$ . We must find the cross-sectional area $\mathrm{A}$ of the concave mirror, since the power delivered is $\text{900 W}{\text{/m}}^2\times \text{A}$ . The mirror in this case is a quarter-section of a cylinder, so the area for a length $\mathrm{L}$ of the mirror is $\mathrm{A}=\frac{1}{4}\mathrm{(2}\mathrm{\pi R}\mathrm{)L}$ . The area for a length of 1.00 m is then

The insolation on the 1.00-m length of pipe is then

Solution to (c)

The increase in temperature is given by $Q=\mathrm{mc}\mathrm{\Delta }T$ . The mass $m$ of the mineral oil in the one-meter section of pipe is

Therefore, the increase in temperature in one minute is

Discussion for (c)

An array of such pipes in the California desert can provide a thermal output of 250 MW on a sunny day, with fluids reaching temperatures as high as $\text{400º}\text{C}$ . We are considering only one meter of pipe here, and ignoring heat losses along the pipe.

What happens if an object is closer to a concave mirror than its focal length? This is analogous to a case 2 image for lenses ( $d_{\text{o}}<f$ and $f$ positive), which is a magnifier. In fact, this is how makeup mirrors act as magnifiers. [link] (a) uses ray tracing to locate the image of an object placed close to a concave mirror. Rays from a common point on the object are reflected in such a manner that they appear to be coming from behind the mirror, meaning that the image is virtual and cannot be projected. As with a magnifying glass, the image is upright and larger than the object. This is a case 2 image for mirrors and is exactly analogous to that for lenses.