MIT2_71S09_practice3_sol

MIT2_71S09_practice3_sol - Solution: The first system is...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution: The first system is the usual single-lens imaging system, so it satisfies: 1 1 1 imaging condition: + = S 1 S 2 f S 2 lateral magnification: m = S 1 where, from the way the lens is described, 1 1 1 2 = ( n f g 1) R = ( n R g 1) R The second system is best modeled anew using the matrix formulation for ray propa- gation: 1 2.71/2.710 Optics Practice Exam 3- Solutions Spring 09 1. A thin bi-convex lens with the same absolute curvature on both faces is used in the two imaging systems shown below. In the first, both object and image are in air, whereas in the second the object is immersed in a material of index n < n g , where n g is the index of the glass used to make the lens. Compare the two imaging systems in terms of imaging condition and magnification. 1 1 n g img 1 1 = R n g n 1 n obj R x img S 2 1 1 1 S 1 1 n x obj 1 1 1 1 f n obj = S S 1 2 1 1 1 n x obj S 1 1 1 = n f f n obj S 1 S 1 S 2 S 2 S + 1 x 2 n n f f obj 1 n + 1 2 where = n (note f > f ) f g 2 R x img S 1 S 1 S n Imaging condition: = 0 1 1 S + 2 2 = 0 + = obj n n f S 1 S 2 f Assuming the imaging condition is satisfied, the system equation becomes: S 1 1 = img 1 n f f n obj x img 1 S 2 f x obj x img S 2 n S 2 m = = 1 = (from the imaging condition) x obj f S 1 2. In the configuration below, lenses L1 and L2 are identical with focal length f , and we consider them to be infinite aperture. The system is illuminated coherently by an on-axis plane wave. 2 (a) Write an expression for the field at x in terms of the thin complex transparencies g 1 ,g 2 . Solution: The first part (to the left of g 2 ( x )) is a Fourier-transforming system: The second part (to the right of g 2 ) is a single-lens imaging system with unit magnification: x The field at 2 2 x = exp i ( x + y ) G 1 g 2 ( x ) 2 f f (Note that the first exponential term could have been omitted...) (b) Consider the specific case with f = 10 cm and g 1 , g 2 defined as: 3 If = 1 m, derive and sketch the intensity at the output plane x . Solution: The binary grating has fundamental period = 20 m and duty cycle 50%, i.e. it is of the form: 1 2 x 1 n = x g 1 ( x ) = 1 + sgn cos sinc e i 2 2 2 2 n = x 1 G 1 = sinc f 2 n = n x 1 2 f Mind the coordinates! f 1 m 20cm Since = = 1 cm, the g 2 transparency only lets orders 1 , , +1 20 m to pass through, therefore the field (or intensity) at the output plane is three bright spots. 4 Example: OTF of the Zernicke phase mask The thin phase transparency whose schematic is given below is placed at the Fourier plane of a unit magnification telescope with focal length f = 10cm. What is the optical transfer function for quasimonochromatic illumination at wavelength = 1 m?...
View Full Document

This note was uploaded on 02/24/2012 for the course MECHANICAL 2.710 taught by Professor Sebaekoh during the Spring '09 term at MIT.

Page1 / 21

MIT2_71S09_practice3_sol - Solution: The first system is...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online