1. Optical Instruments, Camera A single lens camera consists basically of an opaque box, converging lens and film. Focusing depends on the object distance and the focal length of the lens and is accomplished by varying the distance of the lens from the film. The shutter, behind the lens, opens for short time interval. Typical settings are 1/30, 1/60, 1.125. An aperture adjustment control light intensity and when small reduces spherical aberrations.

2. Cont. The intensity of light (energy per unit area of film), reaching the film is proportional to the area of the lens, thus proportional to D 2. The area of the image is proportional to q 2 and as q  f then The intensity is proportional to f/D. f/D is called the focal number of the lens and often refers to lens speed. A low f-number (1.2) refers to a fast speed. Increasing the f-number per setting decreases aperture by a factor of 2.

3. The Eye Light enters the eye and is first refracted by the cornea, and then passes through the pupil (crystalline lens). The iris, a muscular diaphragm, controls the pupils aperture, regulating light intensity. The cornea/lens system focuses the light to the back of the eye, the retina. The retina contain millions of receptors (rod and cones) which send impulses to the brain via the optic nerve.

4. Cont. The rod and cones chemically adjust according to the light intensity. Accommodation, shaping of the pupils, is accomplished the Ciliary muscle and thin filaments, Zonules, located in the iris. Relaxing the ciliary muscle tightens the zonules, which in turn pull on the edge of the lens flattening it. A flattened lens increases its focal length increasing distance of sight. A bulged lens, tight zonulus, allow close sight.

5. Cont. The near point is about 20 cm and increases with age. The far point is infinity assuming sight is normal. Farsightedness (Hyperopia) occurs when the image is focused beyond the retina. Far away images can be seen clearly but not near objects Nearsightedness, (myopia) occurs when the image is focused before the retina.

6. Cont. Old age vision, (presbyopia) occurs with age as the ciliary muscle weaken and is unable to focus the lens well. Astigmatism, causes the production of a line image on the retina as a result of either the cornea or pupil or both not being symmetrical. For corrective lenses the power P = 1/f in diopters. If f = 20cm, then P = 1/.20 = 5.0

7. The Simple Magnifier Using a converging lens, if the object is place just inside its focal point, a virtual, upright, enlarged image is formed allowing the eye to focus closer than its near point. Angular magnification, (m) is the ratio of the angle caused by an object with a lens in use to the angle caused by no lens. M =  /  o eye p 

8. Cont. Magnification is max. when the image formed is at the near point about 25 cm. Object distance is thus p = (25f)/(25 + f ) and m = 1 + 25/f With a single lens, magnification of about 4 is possible without aberrations. Magnification can be increased by using multiple lenses.

9. Compound Microscope Consisting of two lenses, the objective lens has a focal point (f o ), less than 1cm and the ocular lens with (f e ) a couple of cm. The distance between the lenses, L is >> than either f o and f e. The object viewed is placed just outside the focal point of the objective lens, forming a real reduced inverted image at or inside the focal point of the ocular lens which enlarges this inverted image.

10. Cont. Lateral magnification of the first image M 1 = -q/p. Note that q almost equals L and that p 1 almost equals f o. Therefore M 1 = -q/p = -L/f o As m e = 25/f e the overall magnification is the product of the lateral and angular magnifications. m = M 1 m e = (-L/f o )(25/f e ) The negative sign indicates that the image is inverted with respect to the object.

11. Telescope The refracting telescope uses lens combination to form an image. The objective lens form a real inverted image I 1, at the focal point of the ocular lens, ( object at infinity). The two lenses are separated by the distance = f o + f e, the length of the telescope. The ocular lens forms an enlarge inverted image I 2. Angular magnification m =(h’/f e )/(h’/f o ) = f o /f e

12. Cont. The reflecting telescope uses curved mirrors and lenses to form an image. Aberrations can be overcome by using a parabolic mirror. Light passes down the barrel of the telescope and would be focused past the point of a flat mirror that redirects the image ( usually to the side) where it passes through an ocular lens for viewing. Chromatic aberrations are minimal as th objective rays don’t pass through glass

13. Rayleigh’s Criterion If two distant light sources are close together, due to diffraction, they may appear as two blurred spots circled by dark and bright bands. Rayleigh’s criterion states that if the central maxima of one falls on the first minima of the second, they are said to be resolved. The minimum angular separation to be just resolved is sin  = /a where a is slit width.

14. Cont. The limiting angle of resolution for a slit  min = /a where  min is the radius of the slit For circular apertures  min = 1.22 /D where D is the diameter of the aperture.

15. Resolving Power of the Diffraction Grating The diffraction of grating is better than that of a prism, however there is a limit to the resolving power of a spectroscope. Consider two wavelengths of close value the resolving power is given by R = λ/(λ 2 -λ 1 ) = λ/Δλ If N lines on the grating is illuminated the resolving power of the mth order is R=Nm

16. Cont. If m=0 then R = 0 which indicates that all wavelengths are indistinguishable at the zeroth order. Consider a λ = 600nm and 5000 lines illuminated for a second order diffraction pattern. Then R = 5000x2 = 10000 Therefore the minimum wavelength separation Δλ, = λ/R = 6x10 -2 nm

17. The Michelson Interferometer An optical instrument that splits light into two part and then recombines it, for accurate length measurements. Read 25.7 and make a brief summary of the passage, to be handed in after the vacation.