Physics Help Forum multi-surface problem in refraction at spherical surface.

 Oct 3rd 2010, 08:41 PM #1 Junior Member   Join Date: Oct 2010 Posts: 1 multi-surface problem in refraction at spherical surface. i've tried this about 9 times and none of my answers seem right and would LOVE some help. any suggestions? A thick-walled spherical shell made of transparent glass (n=1.50) has internal radius R and external radius 2R. The central cavity and region surrounding the shell are filled with air (n=1.00). The center of symmetry of the object is located at the origin of an xy coordinate system. Paraxial light rays travelling parallel to the x-axis enter the sphere from the left as shown. These rays will refract four times before they passing entirely through the spherical shell and emerge into the air beyond x = 2R. The final image is located at one of the two focal points of this complex optical system. (a) Calculate the x-coordinates of the three intermediate image points and the final focal point, and identify each image as real or virtual. Note: the image due to each successive refraction acts as the object for the next refraction. You will find that the real image due to the first surface acts as a virtual object at the second surface, with negative object distance. (b)b) Calculate the position of the focal point for parallel light entering from the left as before, but now with the interior region filled with glass (n=1.50) to make a homogenous solid glass sphere.
 Feb 18th 2011, 02:41 AM #2 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 Consider the pole of the outer sphere at origin of the coordinate axis. ( I am taking real positive, virtual negative sign convention.) Position of the first image is given by 1/vo + μ/vi = (μ -1)/2R In the first case object is at infinity. So 1.5/vi = 0.5/2R = 1/4R Or vi = 6R. It is real and inside the glass. Position of the second image: Distance of the first image from the inner sphere is 6R - R = 5R. It acts like a virtual object for the inner sphere. R of the inner sphere is negative. So -1.5/5R + 1/v'i = - 0.5/R or v'i is -5R from the first pole of the inner sphere. Position of the third image: The distance of the second image from the second pole of the inner sphere is 5R + 2R = 7R. It acts like a real object, because the rays are diverging from this point. So 1/7R + 1.5/v''i = - 0.5/R or v"i = -14R/9 from the second pole of the inner sphere. The position of the final image: The distance of the third image from the second pole of the outer sphere is 14R/9 + R = 23/9R. It acts like a real object inside the glass. So 1.5/(23R/9) + 1/v'''i = 0.5/2R or 1/v'''i = 1/4R - 27/46R or 92R/31 from the second pole of the outer sphere. Mark these images along the x-axis and find their co-ordinates.