This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Physics 341: Problem Set #10 solutions 1. In this problem you will calculate a microlensing light curve. In the figure below, the dashed straight line represents the trajectory of a point source passing behind a point mass lens (the solid dot in the center). The circle indicates the Einstein radius. The x and y axes show the angular distance from the lens, measured in units of the Einstein radius: x/ E and y/ E . We will assume that the motion is in the x direction, so that y is constant and equal to the impact parameter . The figure shows the case of an impact parameter y = 0 . 5 E , but we will consider a range of values. x/ E y/ E / E + / E ! / E (a) Use the Pythagorean theorem to write down an expression for / E (the angular separation between the source and lens), in terms of x/ E and y/ E . As shown in the figure above, the angular distance from the lens to the source, , is just given by the Pythagorean theorem: E = s x E 2 + y E 2 1 (b) In class we showed that this configuration produces two images, on opposite sides of the lens. The images, source, and lens all lie along the same line. The angular separation between the lens and each image ( + and - ) is given by: = 1 2 q 2 + 4 2 E = ) E = 1 2 @ E s E 2 + 4 1 A The magnification of each image and the total magnification total are given by = 4 4 - 4 E = ( / E ) 4 ( / E ) 4- 1 total = | + | + | - | (Negative magnifications correspond to flipped images, so the total magnification is the sum of the absolute values.) With these equations you are ready to proceed.is the sum of the absolute values....
View Full Document