LL2 Section 56 Problem 1 Find the focal distance for image formation with the aid of two axially- symmetric optical systems whose axes coincide.

Let f 1 and f 2 be the focal lengths of the two systems. For each system separately, Rear principal focus of the first system Front principal focus of the second system. O1’O1’O2O2

The image produced by the first system serves as the object for the second. the distance from the rear principal focus of the first system to the front principal focus of the second system.

Now complete the square

Position of the object relative to the first principal focus of the composite system Position of the image relative to the second principal focus of the composite system Square of the principal focal length of the composite system

The front principal focus of the composite system occurs at X c = 0, or X 1 = -f 1 2 /l. Light emitted from a point source at this position will emerge as a parallel beam after the composite system. The rear principal focus of the composite system occurs at X c ’ = 0, or X 2 ’ = +f 2 2 /l. Parallel rays incident on the composite system will be focused to a point at this position. X c = X’ c = Principal foci are not symmetric with respect to front and back surfaces of composite system.

Principal focal length of composite system Sign is determined by the equation for the transverse magnification When the object is far away, X is a large negative number. According to the example diagram, the final image is erect. That requires f to be negative.

If l = 0, O 1 ’ coincides with O 2 and This is the condition for telescopic image formation. See slide 5 X 1 = 0 X 2 ‘ = 0 X 1 = -  X 2 ‘ = 