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Lecture 3.3. Image formation Ray tracing Calculation Lenses Convex Concave Mirrors Convex Concave Optical instruments Image formation Laws of refraction and reflection can be used to explain how lenses and mirrors operate Spherical Lenses Convex (converging) lens Parallel rays from the Sun passing through a convex lens F F: focal point f: focal length f Sunlight concentrated by magnifying glass may burn hole in paper placed at focal point F Image formation F f F is the focal point and f is the focal length Refractive power (or Strength) of the lens in diopters (D) = 1 f ( f is in metres) Example A farsighted person requires an eyeglass of strength 2.5 diopters. What is the focal length of the eyeglass lens? (a) 4 cm, (b) 40 cm or (c) 400cm D =2.5m-1 f = 1 =1/2.5m = 0.4 m = 40cm. D Image Formation (ray tracing) Object at a distance greater than the focal length from the lens 1 h 3 2 F h′ F s s′ Ray 1 entering lens parallel to optic axis will exit and pass through the focal point Ray 2 passing through the focal point will exit the lens and travel parallel to optic axis Ray 3 will undergo small deviation (displacement (not shown) (thin lens) Real, inverted image formed Real image (may be projected and displayed on a screen) Rays are reversible Image Formation (ray tracing) Object at a distance less than the focal length from the lens h′ h F s s′ Image • virtual • upright •magnified. Simple magnifier F Concave (diverging) lens Optic axis F f Rays entering lens parallel to axis appear to originate at focal point Dashed lines indicate the direction from which the rays appear to come Concave (diverging) lens Object outside the focal length h h′ s′ F Optic axis s Virtual image always produced by a concave lens Cannot be viewed on screen since rays are diverging on the right of the lens However can be viewed with the eye since the eye converges the rays onto the retina. Image Formation---calculation Ray diagrams are useful in sketching the relationship between object and image Relationship may also be calculated A E h F B C O F h′ D s s′ Triangles AOB and DOC are similar Triangles EFO and DFC are similar h h' = − s s' h h' = − f s '− f h' s' = − h s s ' s '− f = s f Equating the expressions h' s '− f = − h f 1 1 1 + = s' s f Image ---calculation 1 1 1 Thin lens formula + = s' s f h F B h′ F s s′ Object distance s positive if object is in front of lens negative if object is behind lens Image distance s′ positive if image is formed behind the lens (real) negative if is formed in front of the lens (virtual) Focal length f •positive -- convex lens •negative --concave lens Magnification h F B h′ F s s′ Magnification is defined as s' M= − s or M Negative : inverted image M Positive : upright image h' M= h Simple magnifier h′ h F s F s′ Object placed inside focal length of converging lens; Image viewed. •Virtual, •Magnified •Upright Example An object 0.5 cm in height is placed 8 cm from a convex lens of focal length 10 cm. Determine the position, magnification, orientation and height of the image. 1 1 1 1 1 1 + = s' s f 1 1 1 = − s ' 10 8 f − s 10 x 8 s′ = = - 40cm. 8-10 s' −40cm M= − = − = 5 s 8cm h' M= h = s' M +ve image upright h′ = M x h = 5 x 0.5cm = 2.5cm h′ h F s s′ F Combining Lenses Effective focal length (feff) of combination of a number of thin lenses close together 1 1 1 = + + ...... f eff f1 f 2 Effective refractive power (or strength) (Seff) of combination of thin lenses close together Seff = S1 + S 2 + ..... Determine the combined strength of a thin convex lens and a thin concave lens placed close together if their respective focal lengths are 10cm and -20cm. Strength S, in diopters (D) = 1 1 1 = Seff= + f eff f1 f 2 1 f ( f is in metres) 1 1 S= − eff .1m .2m Seff = 10 − 5 = 5diopters Example An object is placed 45 cm from a lens of focal length -25 cm. Determine the position, magnification, and orientation of the image. 1 1 1 + = s' s f 1 S’ 1 1 1 = − s' f s 1 1 = -25 cm- 45 cm M=- S’ S S’ = -16.1 cm 16.1 cm M=45 cm = 0.36 M +ve image upright h h′ s′ F s Optic axis Mirrors Flat Mirror Concave Mirror Convex Mirror Curved mirrors analogous to lenses Ray tracing and thin lens equation also valid. real and virtual images are also formed Flat Mirror image object d d Object and image distance equal Upright, virtual image Spherical Mirrors Hollow sphere Spherical mirror C R Principal or optic axis Spherical mirror is a section hollow sphere Radius of curvature R = 2f Curved (Spherical) Mirrors Concave mirror (converging) F f Positive focal length Convex mirror (diverging) Negative focal length f F Thin lens formula may be used to determine object and image distances and focal lengths etc For all single lenses and mirrors: Real image : inverted (h′ negative), positive image distance s′ (RHS of lens, LHS of mirror) Virtual image: upright (h′ positive), negative image distance s′ (LHS of lens, RHS of mirror) Mirrors Concave shaving/makeup mirrors C F Object placed at distance < f from mirror Image is virtual, upright and enlarged. Question: if object is placed at the focal point, where is image located? Application: searchlight Mirrors Example An object is positioned 5 cm in front of a concave mirror of focal length 10 cm. Determine the location of its image and its characteristics. s = 5 cm 1 1 1 + = s' s f 1 1 1 = − s' f s f = 10 cm 1 1 1 = − s ' 10cm 5cm 1 1 2 1 =− = − s ' 10cm 10cm 10cm Characteristics. •Image virtual •Located behind mirror S’ = -10cm s' M= − s −10cm M= − 5cm M =2 Optical instruments Microscopes, telescopes, cameras etc System may have many optical elements (example, lenses and mirrors) Thin lens formula or ray tracing may be used to analyse behaviour of such systems Simple compound microscope two convex lenses Objective lens eyepiece Fo Object (height h) Fe h′ Fe Fo Final image h′′ image formed by objective lens is inside focal length of eyepiece lens. Optical instruments Simple refracting telescope Objective lens f0 fe eyepiece Fo,Fe Virtual image at infinity, Magnified and inverted f0 M= − fe Objective lens forms image (real, inverted) at focal point F0 which is also the focal point Fe of the eyepiece; virtual image is then formed at infinity. Optical instruments Simple reflecting telescope Flat mirror Objective, concave mirror Eyepiece lens Example Effective focal length of the objective in the Hubble telescope is 57.8 m. What focal length eyepiece is required to give a magnification of -8.0 x 103. f M= − 0 fe fe = − f0 M 57.8m −3 fe = 7.23 10 m − = × 3 8.0 ×10 Optical instruments Medical loupes Galilean Design objective Operating site eyepiece Typical working distance 28-38 cm Typically Magnification Optical design allows m ≈ 2.5 → 4.5 observer focus at infinity thereby relieving eyestrain Optical instruments Optical fibre: Total Internal Reflection Endoscope Endoscope for medical investigations— inserted through small incision or orifice to inspect and facilitate operation on interior parts of the body flexible shaft includes: light source to illuminate area, image channel to view area under investigation, air or water conduit to clear debris, instrument conduit eyepiece Typical endoscope Instrument entry Flexible shaft Transmits light, air , water Optical instruments Camera aperture CCD array Lens translated to change image distance to adjust for different object distances. Real, inverted image formed on CCD array Example Microscope as shown in the previous figure has the following characteristics objective lens focal length = 0.7 cm eyepiece lens focal length = 4.5 cm separation between lenses = 21 cm. Object size = 0.01 cm. object to objective lens distance = 0.73 cm. Calculate total magnification produced and size of the final image. image distance s′ due to objective lens 1 1 1 + = s' s f 1 1 1 = − s' f s 1 1 1 = − s ' 0.7 0.73 s′ = 17 cm Magnification M0 = -s′/s = -17/0.73 = -23.3 Magnification negative because image is inverted M0 = h′/h size of the image = h′ =M0h = (-23.3)(0.01)cm = -0.233 cm. Image produced by objective is situated at a distance (21-17) cm = 4 cm from the eyepiece. 1 1 1 + = s' s f 1 1 1 = − s' f s 1 1 1 = − s ' 4.5 4 s′ = -36 cm s′ negative: image is on same side of eyepiece as object Magnification Me = -s′/s = -(-36/4) = 9 Size of the final image = h′ ′ =Meh′ = (9)( -0.233) = -2.1cm. Total magnification MT = (-2.1cm)/(0.01 cm) = -210 Total magnification MT = M0 x Me = (-23.3) x 9 = - 210