Trigonometry and TelescopesDate: 12/15/2001 at 15:27:02 From: Maitri Subject: Trigonometry and Telescopes How is trigonometry related to how reflector and refractor telescopes work? I think that it might be that the farther you move the lens or mirror from the eyepiece, the smaller the image is magnified. But I don't fully understand that. Thank you. Date: 12/16/2001 at 07:49:58 From: Doctor Jerry Subject: Re: Trigonometry and Telescopes Hi Maitri, There are many kinds of telescopes. Some telescopes involve a reflecting mirror with a parabolic cross-section. Using trigonometry, one can prove that light coming in along the axis of symmetry of the mirror will reflect from the mirror to a common point, called the focus. Other telescopes involve hyperbolic mirrors, which have similar reflecting properties. The necessary calculations could not be done without using trigonometry. There's also Snell's Law, governing how light reflects/refracts as it travels from air to glass or glass to air. Snell's law involves the sine of a certain angle together with the velocity of light in air and in glass. You can look Snell's law up on the net, or try Index of Refraction - Dr. Math archives http://mathforum.org/dr.math/problems/shelby10.10.01.html - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 12/16/2001 at 13:05:00 From: Maitri Subject: Trigonometry and Telescopes Dear Dr. Math, How can trigonometry help you prove that light coming in along the axis of symmetry of the mirror will reflect from the mirror to a common point? Please explain this theory to me again. Thank you. Date: 12/17/2001 at 08:12:06 From: Doctor Jerry Subject: Re: Trigonometry and Telescopes Hi Maitri, I'll try to show how one might go about proving the reflecting property of a parabola. I'll use the simple parabola y = x^2 and I'll need to use the fact that at the point (x,x^2) of the parabola, the slope of the tangent line is 2x. I'll also need the fact that the focus of this parabola is (0,1/4). Imagine a beam of light coming down from high in the first quadrant, traveling in a line parallel to the y-axis. I'm drawing a figure with this line coming down and hitting the parabola at something like (2,4). Draw a line tangent to the parabola at (2,4). The equation of this tangent line is y - 4 = (2*2)(x - 2) I just used the fact that the slope at (2,4) is 2*x = 2*2 = 4. We know that light bounces off the parabola (imagine that it is silvered) so that the angle of incidence is equal to the angle of reflection. Because the slope 4 is equal to the tangent of the angle of inclination alpha, we know that tan(alpha) = 4. We see, then, that the angle between the incoming vertical ray and the parabola is 90-alpha. We want to show that the angle gamma between the tangent line and the line from (2,4) to the focus (0,1/4) is the same as 90-alpha. By extending the line from (2,4) through (0,1/4) to the x-axis and letting the angle of inclination of this line be beta, we see that gamma = alpha-beta. So, all we must do is to show that tan(alpha-beta) = tan(90-alpha). Well, tan(alpha-beta) = [tan(alpha)-tan(beta)]/[1+tan(alpha)*tan(beta)]. = [4-tan(beta)]/[1+4*tan(beta)]. Because beta is the inclination of the line with equation y - 4 = (15/8)(x-2), we see that tan(beta) = 15/8. So, tan(alpha-beta) = [4-15/8]/[1+15/2] = 17/68=1/4. But, tan(90-alpha) = cot(alpha) = 1/tan(alpha) = 1/4. So, we've proved that the angle of incidence is equal to the angle of reflection (well, we actually proved that the tangents of these angles are equal; because they're both first quadrant angles, the other follows). This argument can be generalized to any parabola. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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