[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-657052-105":3,"detail-sidebar-cat-0-en-105":81,"doc-detail-657052-en":130},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":74,"head_meta":76,"extra_data":78,"updated_unix":80},105,"en","lecture-9-notes-multiple-lens-systems","Lecture 9 Notes - Multiple-lens systems","","Lecture 9 notes explain how to analyze multiple-lens optical systems using the lens equation and ray diagrams with sign conventions. It shows that the image formed by the first lens can act as the object for the second lens, enabling calculation of q values and final image location, including cases where the final image becomes virtual. The notes derive total magnification from sequential magnifications, relate magnification to lens separation L, and treat parallel input from an object at infinity using effective focal-point reasoning. They also define lens power P=1/f, discuss additivity for closely spaced lenses, and address special ray-diagram problems such as objects at the first focal point and other troublesome cases.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/technology/","Technology",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/lecture-9-notes-multiple-lens-systems/657052/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/lecture-9-notes-multiple-lens-systems/657052.png","ImageObject",300,407,{"name":42,"@type":43},"Jasmine","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-11-01","2026-10-26",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",6,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"How can the image from the first lens be used for the second lens?","Question",{"text":63,"@type":64},"Treat the image formed by the first lens as the object for the second lens, then apply the lens equation to determine the second lens’ image and location.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"What determines whether the final image is virtual or real in a two-lens system?",{"text":68,"@type":64},"The sign of the relevant image distances (e.g., q values) under the stated sign convention determines whether the final image lies on the same side as the object (virtual) or opposite side (real).",{"name":70,"@type":61,"acceptedAnswer":71},"How does total magnification relate to the two lenses?",{"text":72,"@type":64},"Total magnification is the product of the individual magnifications, M1M2, where M1 describes the first lens and M2 describes the second lens using the first image as the second lens’ object.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},657052,1793030091,{"code":4,"msg":82,"data":83},"success",[84,88,92,96,101,104,109,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":89,"show_sort_weight":90,"slug":91},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":93,"show_sort_weight":94,"slug":95},"Exam",70,"exam",{"id":97,"doc_module":4,"doc_module_name":25,"category_name":98,"show_sort_weight":99,"slug":100},5,"Comic",60,"comic",{"id":55,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":102,"slug":103},50,"technology",{"id":105,"doc_module":4,"doc_module_name":25,"category_name":106,"show_sort_weight":107,"slug":108},7,"Healthcare",40,"healthcare",{"id":110,"doc_module":4,"doc_module_name":25,"category_name":111,"show_sort_weight":112,"slug":113},8,"Research & Report",30,"research-report",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},9,"Religion & Spirituality",20,"religion-spirituality",{"id":117,"doc_module":4,"doc_module_name":25,"category_name":120,"show_sort_weight":117,"slug":121},"World Cup","world-cup",{"id":123,"doc_module":4,"doc_module_name":25,"category_name":124,"show_sort_weight":123,"slug":125},10,"Lifestyle","lifestyle",{"id":127,"doc_module":4,"doc_module_name":25,"category_name":128,"show_sort_weight":97,"slug":129},19,"General","general",{"code":4,"msg":82,"data":131},{"doc_id":79,"user_id":132,"nickname":42,"user_avatar":133,"doc_module":4,"category_id":55,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":110,"language":139,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":12,"update_tm":143,"read_time":117},2336478487870,"https://ap-avatar.wpscdn.com/davatar_085a072bc5b1113ac321206ff7593b45","Lecture 9 Notes: 07 / 13  \nMultiple-lens systems  \nIf we have one lens behind another, we can simply treat the image formed by the first lens as an object for the second lens. For example, suppose we have two convergent lenses, with focal lengths f1 and f2, separated by a distance L. The object is located at a distance p1 in front of the first lens. We want to locate the image and find the magnification.  \nThis is a typical ray diagram, to show our distances and sign conventions:  \nFor this particular diagram, all the quantities are positive. If one of the lenses was divergent, its focal length would be negative; if one or both of the images was on the same side as its corresponding object, the value of q for that image would be negative.  \nLet us calculate q2 for given p1 , L, f1 and f2. The equations for the two lenses are  \nWe can simply use the first equation to calculate q1 and plug into the second. For example, suppose that the object is 50cm away, the lenses are 120cm apart, the first lens has a focal length of 30cm and the second lens has a focal length of 100cm. Then, our  \nequation will give us  \nThis means that the final image will be virtual, and will be 82 centimeters in front of the second lens (in the direction of the object) . This is unlike the ray diagram above, where the final image is real and behind the second lens.  \nWe can calculate the magnification of this two-lens contraption. Since the magnification of the first lens is M1 = -h1 / h, where h is the height of the object and h1 is the height of the first image, and M2 = -h2 / h1 , since the first image is now the object and h2 is the height of the second image, M1M2 = h2 / h = Mis the total magnification from the object to the final image (the sign cancels because a sequence of two inverted real images give an upright real image, so the magnification for a real image is positive.) Since  \nM1 = -q1 / p1 and M2 = -q2 / p2 = q2 / (L-q1), the magnification is  \nFor our example, this gives  \nThe final image is therefore inverted, and 2.7 times larger than the original.  \nExample: We might ask how the magnification depends on the distance L; perhaps our apparatus is adjustable, and we want find a length that gives a certain magnification.  \nM1 = -q1 / p1 does not depend on the distance L, but M2 does. We have  \nThe overall magnification is thus equal to  \nWhat should we make the length of our apparatus if we want a magnification of, say, -5? Solve this equation for L:  \nSo if we separated the lenses by 145cm and left the object 50cm from the front lens, the magnification would now be-5.  \nSuppose we have an object at infinity, and we look at it using lenses of focal lengths f1 and f2 separated by a distance L. Where does the image form?  \nThe first lens will form an image at the focal point, since parallel rays coming from the far-away object will cross at the focal distance. Thus, q1 = f1. For the second image,  \nThe location of the second image for a far-away source can be treated as an effective focal point of the entire two-lens image. Thus,  \nIf the separation between the lenses is very small compared to the focal length of the first lens (for example, if we put two lenses together, front to back), we can neglect L, and obtain the following expression:  \nThe two lenses thus function together as a single lens, with a combined focal length. Note that this equation is only valid if the lenses are held very close together; otherwise the more general equation above should be used.  \nWe define the power of a lens as the reciprocal of its focal length: P = 1 /f. The units of lens power are just m-1 , but these units are traditionally referred to as diopters when used in this context, just like Hertz are seconds-1 in the context of frequency. As we can see, lens power is additive when two lenses are put close together:  \nExample: Consider a system with a convergent lens, f1 = 40 cm, followed by a divergent lens, f2 = -60 cm, at a distance of 100 cm. An object i","cbCaiqsSEPWo6uZJ","https://ap.wps.com/l/cbCaiqsSEPWo6uZJ","pdf",115612,"English","# Multiple-lens systems\n## Two-lens image formation and sign conventions\n## Calculating image location (q2) and virtual images\n## Total magnification and dependence on separation L\n## Object at infinity and effective focal behavior\n## Combined focal length for closely spaced lenses\n## Lens power and diopters\n## Example with convergent and divergent lenses\n## Potential ray-diagram problems and special cases","[{\"question\":\"How can the image from the first lens be used for the second lens?\",\"answer\":\"Treat the image formed by the first lens as the object for the second lens, then apply the lens equation to determine the second lens’ image and location.\"},{\"question\":\"What determines whether the final image is virtual or real in a two-lens system?\",\"answer\":\"The sign of the relevant image distances (e.g., q values) under the stated sign convention determines whether the final image lies on the same side as the object (virtual) or opposite side (real).\"},{\"question\":\"How does total magnification relate to the two lenses?\",\"answer\":\"Total magnification is the product of the individual magnifications, M1M2, where M1 describes the first lens and M2 describes the second lens using the first image as the second lens’ object.\"}]","Lecture 9 Notes - Multiple-lens systems | PDF",1792972977]