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Photograph of several parabolic trough collectors set in a row in an open area.
Parabolic trough collectors are used to generate electricity in southern California. (credit: “kjkolb”/Wikimedia Commons)
  1. If we want the rays from the sun to focus at 40.0 cm from the mirror, what is the radius of the mirror?
  2. 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 ?
  3. 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.

Strategy

First identify the physical principles involved. Part (a) is related to the optics of spherical mirrors. Part (b) involves a little math, primarily geometry. Part (c) requires an understanding of heat and density.

Solution

  1. The sun is the object, so the object distance is essentially infinity: d o = . The desired image distance is d i = 40.0 cm . We use the mirror equation to find the focal length of the mirror:
    1 d o + 1 d i = 1 f f = ( 1 d o + 1 d i ) −1 = ( 1 + 1 40.0 cm ) −1 = 40.0 cm

    Thus, the radius of the mirror is R = 2 f = 80.0 cm .
  2. The insolation is 900 W/m 2 . You must find the cross-sectional area A of the concave mirror, since the power delivered is 900 W/m 2 × A . The mirror in this case is a quarter-section of a cylinder, so the area for a length L of the mirror is A = 1 4 ( 2 π R ) L . The area for a length of 1.00 m is then
    A = π 2 R ( 1.00 m ) = ( 3.14 ) 2 ( 0.800 m ) ( 1.00 m ) = 1.26 m 2 .

    The insolation on the 1.00-m length of pipe is then
    ( 9.00 × 10 2 W m 2 ) ( 1.26 m 2 ) = 1130 W .
  3. The increase in temperature is given by Q = m c Δ T . The mass m of the mineral oil in the one-meter section of pipe is
    m = ρ V = ρ π ( d 2 ) 2 ( 1.00 m ) = ( 8.00 × 10 2 kg/m 3 ) ( 3.14 ) ( 0.0100 m ) 2 ( 1.00 m ) = 0.251 kg

    Therefore, the increase in temperature in one minute is
    Δ T = Q / m c = ( 1130 W ) ( 60.0 s ) ( 0.251 kg ) ( 1670 J · kg/°C ) = 162 °C

Significance

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 400 °C . We are considering only one meter of pipe here and ignoring heat losses along the pipe.

Image in a convex mirror

A keratometer is a device used to measure the curvature of the cornea of the eye, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12 cm from the cornea and the image magnification is 0.032, what is the radius of curvature of the cornea?

Strategy

If you find the focal length of the convex mirror formed by the cornea, then you know its radius of curvature (it’s twice the focal length). The object distance is d o = 12 cm and the magnification is m = 0.032 . First find the image distance d i and then solve for the focal length f .

Solution

Start with the equation for magnification, m = d i / d o . Solving for d i and inserting the given values yields

d i = m d o = ( 0.032 ) ( 12 cm ) = −0.384 cm

where we retained an extra significant figure because this is an intermediate step in the calculation. Solve the mirror equation for the focal length f and insert the known values for the object and image distances. The result is

1 d o + 1 d i = 1 f f = ( 1 d o + 1 d i ) −1 = ( 1 12 cm + 1 −0.384 cm ) −1 = −0.40 cm

The radius of curvature is twice the focal length, so

R = 2 f = −0.80 cm

Significance

The focal length is negative, so the focus is virtual, as expected for a concave mirror and a real object. The radius of curvature found here is reasonable for a cornea. The distance from cornea to retina in an adult eye is about 2.0 cm. In practice, corneas may not be spherical, which complicates the job of fitting contact lenses. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror. Thus, the image is virtual because no rays actually pass through it. In the problems and exercises, you will show that, for a fixed object distance, a smaller radius of curvature corresponds to a smaller the magnification.

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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