1. Chapter 25:Optical Instruments Cameras Homework assignment : Read Chap.25, Sample exercises : 4,21,24,41,43  Principle of a camera ss’ D Intensity of light at film: Area of image : Focal length vs. s’ : Area of image : Define f-number as : Intensity of light at film:

2. Eyes  Structure of the eye near point (~25 cm, changes): the closest distance for which the lens can accommodate to focus light on retina. far point : the farthest distance for which the lens of the relaxed eye can accommodate to focus light on retina.

3. Eyes  Conditions of the eye farsightedness (hyperopia)nearsightedness (myopia) Presbyopia (old age-vision) : due to a reduction in accommodation ability Antigmatism : due to asymmetry in the cornea or lens Power of a lens (unit diopter): P=1/f P in diopter, f in m, f=+20 m -> P=+5.0 diopter

4. Angular size angular size : d (usually d min =25 cm) h

5. Magnifying glass whenbut for human eye. the minimum distance at which an eye can see image of an object comfortably and clearly. virtual image s’ the eye is most relatex

6. Microscope small Object is placed near F 1 (s 1 ~f 1 ). Image by lens1 is close to the focal point of lens2 at F 2. magnifier

7. Refracting Telescope angular size of image by lens2; eye is close to eyepiece image height by lens1 at its focal point Image by lens1 is at its focal point which is the focal point of lens 2 image distance after lens1 magnifier

8. Reflecting Telescope

9. Resolution of Single-Slit and Circular Apertures  Resolution of single-slit aperture The ability of an optical system such as the eye, a microscope, or a telescope to distinguish between closely spaced objects is limited because of wave nature of light. - Light from two independent sources which are not coherent. - If the angle subtended by the sources at the aperture is large enough, then the diffraction patterns are distinguishable (resolvable). - If the angle is too small, then the two sources are not distinguishable (unresolvable).

10. Resolution of Single-Slit and Circular Apertures  Rayleigh’s criterion When the central maximum of one image falls on the first minimum of another image, the images are said to be just resolved. This limiting condition of resolution is know as Rayleigh’s condition. a From what we learned in Chapter 24 about the diffraction due to a single-slit the first minimum occurs at: According to Rayleigh’s criterion, this gives the smallest angular separation for which two images are resolvable.

11. Resolution of Single-Slit and Circular Apertures  Resolution of circular aperture D Central maximum has an angular width given by: Comparable to the slit geometry, where the central maximum is defined by sin  = /a. Following the same argument as for the single- slit case, the limiting angle of resolution of the circular aperture is:

12. Resolution of Single-Slit and Circular Apertures  Resolution of circular aperture (cont’d)

13. Resolution of Single-Slit and Circular Apertures  Example 25.6 : Resolution of a microscope Sodium light of wavelength 589 nm is used to view an object under a microscope. The aperture has a diameter of 0.90 cm. (a) Find the limiting angle of resolution. (b) Using visible light of any wavelength you desire, find the maximum limit of resolution (the shortest visible wavelength - violet ). (c) What effect does water between the object and the objective lens have on the resolution with 589-nm light?

14. Resolution of Single-Slit and Circular Apertures  Example 25.7 : Resolving craters on the Moon The Hubble Space Telescope has an aperture of diameter 2.40 m. (a) What is its limiting angle resolution at a wavelength of 600 nm? (b) What is the smallest lunar crater the Hubble Space telescope can resolve (The Moon is 3.84x10 8 m from Earth)?

15. Resolution of Single-Slit and Circular Apertures  Resolving power of the diffraction grating Devices based on both the prism and the diffraction grating can be used to make accurate wavelength measurements. However, the diffraction grating device has a higher resolution. Resolving power of the diffraction grating: N : the number of grating slits illuminated m : the order number

16. Resolution of Single-Slit and Circular Apertures  Example : Light from sodium atoms Two bright lines in the spectrum of sodium have wavelengths of 589.00 nm and 589.59 nm, respectively. (a)What must the resolving power of a grating be to distinguish these wavelengths? (b) To resolve these lines in the 2 nd -order spectrum, how many lines of the grating must be illuminated?

17. The Michelson Interferometer d1d1 d2d2 Light from a source is split into 2 beams, reflected from 2 mirrors, and recombined. Path difference  r = 2(d 2 – d 1 ) Recombined light shows an interference pattern at the detector If one mirror is moved a distance /2 then the path difference changes by and exactly one fringe moves across the detector. Precise distance measurement by counting fringes! Alternatively, if one mirror is moved a distance of /4, a bright fringe becomes a dark fringe. This shift of the fringe makes it possible to measure the wavelength accurately.