SIMG-0-200 Solution Set #6. An optical system consists of thin lenses L (f 00mm)andL 2 (f 2 200mm) separated by d 800mm. Locate and describe the image of an object that is 50 mm high located 50 mm in front of the first lens. At least three ways to do this. Easiest is probably by brute force, i.e., find the image created by the first lens and use it as the object for the second lens. The magnification of the system is the product of the magnifications of the two lenses. + s s 0 f s l f s 00 mm 50 mm 00 mm 2 00 mm 00 mm s 0 00mm (M T ) s0 00 mm s 50 mm 2 s 2 + s 0 2 t s 0 + s 2 800 mm 00 mm + s 2 s 2 500mm f 2 s l 2 f 2 s 2 200 mm 500 mm 5 000 mm 2 000 mm 000 mm s 0 2 000 (M T ) 2 s0 2 s 2 mm mm 000 mm 500 mm 2 50mm 66.7mm 50mm f 00mm 800mm.mm f 200mm M T (M T ) (M T ) 2 2 2 +4 The image is located 000 mm behind the second lens and the magnification is + 4,so the image size is 4 200 50 mm mm 66 2 mm. The image is upright and magnified.
A second way to do this is is to locate the principal points of the lens and measure the object distance to the object-space principal point. Then use the thin-lens imaging equation to find the image distance, which is measured from the image-space principal point. The equivalent focal length of the system is: f eff f + f 2 d f f 2 f f 2 00 mm 200 mm 40 mm (f + f 2 ) d 00 mm 800 mm The BFD is the distance from the rear vertex to the image-space focal point. formula for the BFD was derived in the notes: The BFD V 0 F 0 f 2 (f d) 200 mm (00 mm 800 mm) (f + f 2 ) d 500 mm +280mm The distance from the image-space principal point H 0 to the image-space focal point F 0 is the effective focal length, so the distance from the image-space vertex to the imagespace principal point can be found: V 0 F 0 V 0 F 0 H 0 F 0 280mm ( 40 mm) 20 mm The object-space focal distance (FFD) may be found from the formula that was derived in the notes: FFD FV f (f 2 d) 00 mm (200 mm 800 mm) +20mm (f + f 2 ) d 500 mm The distance from the object-space vertex V to the object-space principal point H can now be found: VH FH FV 40 mm 20 mm 60 mm HV +60 mm H F V 800mm V' F' H' f 00mm f 200mm FH -40mm FV FFD +20mm HV +60mm H'F' -40mm V'F' BFD +280mm V'H' +20mm The distance from the object to the object-space focal point is: s OHOV+VH 50mm+( 60 mm) 0 mm 2
Since the object distance is negative, the object in the equivalent thin-lens equation is virtual. The image distance is now found directly: 0 mm + s 0 40 mm s 0 H 0 O 0 40 mm 0 mm + 40 mm mm The distance from the vertex to the image is obtained by substitution into: s 0 H 0 O 0 H 0 V 0 +V 0 O 0 V 0 O 0 H 0 O 0 H 0 V 0 + 40 000 mm ( 20 mm) mm mm which is the same answer as that obtained above for s 0 2. The image magnification is the ratio of the distances: M T s0 s + 40 mm 0 mm +4 which again is the same answer as that obtained using the brute-force method.
2. A system consists of thin lenses L (f 60 mm) and L 2 (f 2?) separated by d 20 mm. Lens L 2 is made of glass with n.5 and is plano-convex with radius r 60mmfor the curved side. Locate and describe the image of an object that is 5mm high located 80 mm in front of the first lens. We need to first find the focal length of L 2. The focal length is determined from the curvatures of the surfaces and the index of refraction: (n 2 n air ) 0.5 f 2 R R 2 f 2 20 mm 60 mm 0.5 60 mm 20 mm + s s 0 f s 0 f s 60 mm 45 mm 80 mm s 2 t s 0 20mm ( 45 mm) 65 mm s 0 2 f 2 s 2 20 mm +440mm 65 mm M T (M T ) (M T ) 2 s0 s0 2 s s 2 45 mm 80 mm 440 mm 2 65 mm The image is located 440 mm behind the second lens with a magnification of 2,so the image size is 0 mm mm.athinlensismadeofglasswithindexn.5. In air, the lens has a focal length f 254mm. What is its focal length when it is totally immersed in water (n.)? Lensmaker s equation : f (n 2 n ) R R 2 If in air : (.5.0) 254 mm R R 2 0.5 R R 2 R R 2 0.5 254 mm 4.62 mm If in water : (.5.) f R R 2 0.20 f 67.08 mm f 0.5 254 mm The focal length of the lens in water is considerably longer than its focal length in air because the relative refractivity of the glass is much reduced in water. To see that this result makes sense, consider what the focal length of the lens would be if immersed in glass. The relative refractivity of the lens vanishes, so the focal length becomes infinite. 4
4. The surfaces of a thin equiconvex lens have equal radius of curvature: R 50mm. The second surface is aluminized so that it is a mirror. Find the location of the image of an object located 400 mm to the left of the first surface. We can do this in three steps: refraction at the first surface, reflection at the second surface, and refraction back at the first surface. Assume that the refractive index of the glass is n.5 : n + n 2 n 2 n.5 s s 0 R s 0 n 2 00 mm 200 mm 0.5 R 50 mm 00 mm 4 200 mm 200 mm 400 mm s 0 n 2 200 mm 800 mm O' O 400mm V V' 200mm Because the lens is thin, the distance from the first to the second surface is 0mm.For a mirrored surface, the image-space index n 2 n : n 2 + n.5 s 2 s 0 2 800 mm +.5 (.5.5) s 0 2 R 2 50 mm + 50 mm s 0 2.5 50 mm 24 200 mm 200 + 200 mm 200 25 s 0 2.5 +48 mm 72 mm +48mm O' O 72mm V V' 800mm 5
The image is 48 mm behind the second surface, so s 48 mm :.5 + s s 0 s 0 (.5) 50 mm 00 mm.5 400 72 mm 7 mm ' 57.4 mm This is behind the second surface, but since we are going in the negative direction, the distance is in front of the lens. In other words, the distance is positive, so that s 0 + 400 mm if measured in the same space as the object. 7 O' O V V' The image is real and the magnification is 400 7 mm M T s0 s 400 mm 7 6
5. Two thin lenses of focal lengths f 50 mm and f 2 +00mmare separated by a distance d 50mm. Find the focal length of the system of thin lenses and locate the principal points. The focal length of the system may be found from: f eff + t f f 2 f f 2 50 mm + 00 mm 50 mm ( 50 mm 00 mm) 2 00 mm + 00 mm + 0 00 mm The system is AFOCAL; it is a telescope. The focal length is and all cardinal points (focal and principal points) are located at. 6. The solar telescope at the Kitt Peak National Observatory has a primary mirror with adiameterof60 in. The image formed by this mirror is 00 ft away. If the diameter of the sun is 864, 000 mi and its distance is 9, 000, 000 mi. Find the diameter of the image of the sun in mm. The image size is proportional to the ratio of the object distance to the focal length, as shown: 864, 000 mi 9, 000, 000 mi x ' x 00 ft 864, 000 mi 00 ft 9, 000, 000 mi 2.787 ft ' 850 mm 7
7. Consider an optical system composed of two thin lenses, each with focal length of 00 mm. Thefirst lens is located at the front focal point of the second. Find the focal length, focal points, and the principal points of the system. If the focal length of the first lens is changed, determine the effect on these parameters of the system. The focal length of the system is: f eff f eff f + f 2 f f 2 00 mm + 00 mm f eff 00mm t 00 mm 00 mm 00 mm 00 mm The focal length of the system is the same as the focal length of each lens. The imagespace focal point is located at the second lens. The object-space focal point is located at the first lens. Since the lenses are located f eff apart, then the principal points coincide with the lenses, as shown in the sketch. 8
If the focal length of the firstlensischanged,sayf 50mm, then the focal length of the system is: f eff f eff f + f 2 f f 2 50 mm + 00 mm f eff 00mmf 2 t 00 mm 00 mm 50 mm 00 mm Thus the focal length of the system is equal to that of the second lens, regardless of the focal length of the first lens. (a) (OPTIONAL BONUS) Describe any applications for this optical system. Consider the image formed by the second lens in the second system with f 2 00 mm of a nearby object, say at s 500mm. The image formed by the single lens is located at the distance: s 0 f 2 s 00 mm 500 mm 4 25mm 500 mm The magnification is M T s0 mm 25 s 500 mm 4 Now add the first lens. The focal length of the system is unchanged at f eff f 2 00 mm. Now find the image of an object located 500 mm in front of the SECOND lens (i.e., in the same position as before) and find its magnification. Do it brute 9
force: s 500mm t 500mm 00 mm 400 mm + s 0 s s 0 f 50 mm 240mm 400 mm (M T ) s0 240 mm s 400 mm 0.6 The distance to the second lens is: s 2 t s 0 00mm 240 mm 40 mm s 0 2 f 2 s 2 75 mm (M T ) 2 s0 2 s 2 40 mm +5 2 00 mm 40 mm 75 mm 58 mm The transverse magnification of the image formed by the system is the product: M T (M T ) (M T ) 2 5 + 5 2 4 Theimageoftheobjectformedbythefirst lens alone is located 25 mm behind the lens with a magnification of, whereas the image formed by both lenses is 4 located 58 mm behind the second lens with the same magnification. The addition of the first lens located at the front focal point of the second lens moved the image without changing its magnification. This is the action of eyeglasses the optometrist locates a corrective lens at the front focal point of the eyelens to move the image onto the retina without changing the system magnification. In the case just considered, the corrective lens L has positive power and moved the image forward, as required for a farsighted person. Had the power of L been negative, the image would have moved backward onto the retina of a nearsighted person. 0