{"id":4302,"date":"2018-09-06T13:07:41","date_gmt":"2018-09-06T20:07:41","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=4302"},"modified":"2023-08-05T11:08:25","modified_gmt":"2023-08-05T11:08:25","slug":"image-formation-by-converging-convex-lens","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/image-formation-by-converging-convex-lens.htm","title":{"rendered":"Image formation by converging (convex) lens","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Article about Image formation by converging (convex) lens<\/p>\n

To understand the image formation by the convex lens, learn example problems and solution below. In this case, the object is assumed to be at a certain distance from the convex lens, <\/span><\/span><\/p>\n

then draw the image formation by the convex lens, the image distance from the convex lens and the magnification of the image by the convex lens.<\/span><\/span><\/p>\n

The convex lens has a focal length of 20 cm. Objects with a height of 10 cm are located to the left of the lens. Determine the image distance, the magnification of the image and the image height, if:<\/span><\/span><\/p>\n

a) the object distance from the convex lens is smaller than the focal length of the convex lens (do < f),<\/span><\/span><\/p>\n

b) the object distance from the convex lens is the same as the focal length of the convex lens (do = f),<\/span><\/span><\/p>\n

c) the object distance from the convex lens is greater than the focal length of the convex lens (do > f).<\/span><\/span><\/p>\n

Known:<\/u><\/span><\/span><\/p>\n

The focal length (f) of the convex lens = 20 cm <\/span><\/span><\/p>\n

The focal length of the convex lens signed positive because the focal point of the convex lens is real, where the light passes through the focal point.<\/i><\/span><\/span><\/p>\n

The object height (h) = 10 cm <\/span><\/span><\/p>\n

Solution:<\/u><\/span><\/span><\/p>\n

The object distance is smaller than the focal length of the convex lens (do > f)<\/b><\/u><\/span><\/span><\/p>\n

Suppose the object distance is 5 cm, 10 cm and 15 cm.<\/span><\/span><\/p>\n

a) If the object distance (do) = 5 cm<\/b><\/span><\/span><\/p>\n

\"Image<\/p>\n

The image distance (di)<\/u><\/span><\/span><\/p>\n

1\/di = 1\/f \u2013 1\/do = 1\/20 \u2013 1\/5 = 1\/20 – 4\/20 = -3\/20<\/span><\/span><\/p>\n

di = -20\/3 = -6.7 cm <\/span><\/span><\/p>\n

The image distance is negative, means the image is virtual, where the beam of light does not pass through the image.<\/i><\/span><\/span><\/p>\n

The image distance 6.7 cm, higher than the object distance 5 cm.<\/i><\/span><\/span><\/p>\n

The magnification of image (M)<\/u><\/span><\/span><\/p>\n

M = -di \/ do = -(-6.7) \/ 5 = 6.7 \/ 5 = 1.3<\/span><\/span><\/p>\n

The image magnification exceeds 1 means that the image size is greater than the object size.<\/i><\/span><\/span><\/p>\n

The image height (hi) <\/u><\/span><\/span><\/p>\n

The equation of the image magnification:<\/span><\/span><\/p>\n

M = hi \/ ho<\/span><\/span><\/p>\n

hi = M ho = (1.3)(10 cm) = 13 cm <\/span><\/span><\/p>\n

The image height, signed positive, means the image is upright.<\/i><\/span><\/span><\/p>\n

b) If the object distance (do) = 10 cm<\/b><\/span><\/span><\/p>\n

\"Image<\/p>\n

The image distance (di)<\/u><\/span><\/span><\/p>\n

1\/di = 1\/f \u2013 1\/do = 1\/20 \u2013 1\/10 = 1\/20 – 2\/20 = -1\/20<\/span><\/span><\/p>\n

di = -20\/1 = -20 cm <\/span><\/span><\/p>\n

The image distance is negative means the image is virtual, where the beam of light does not pass through the image.<\/i><\/span><\/span><\/p>\n

The image distance 20 cm is greater than the object distance 10 cm.<\/i><\/span><\/span><\/p>\n

The magnification of image (M)<\/u><\/span><\/span><\/p>\n

M = -di \/ do = -(-20)\/10 = 20\/10 = 2<\/span><\/span><\/p>\n

The magnification of image exceeds 1 means that the image size is greater than the object size.<\/i><\/span><\/span><\/p>\n

The image height (hi) <\/u><\/span><\/span><\/p>\n

The equation of the magnification of image:<\/span><\/span><\/p>\n

M = hi \/ ho<\/span><\/span><\/p>\n

hi = M h = (2)(10 cm) = 20 cm <\/span><\/span><\/p>\n

The image height is positive means that the image is upright.<\/i><\/span><\/span><\/p>\n

c) If the object distance (do) = 15 cm<\/b><\/span><\/span><\/p>\n

The image distance (di)<\/u><\/span><\/span><\/p>\n

1\/di = 1\/f \u2013 1\/do = 1\/20 \u2013 1\/15 = 3\/60 – 4\/60 = -1\/60<\/span><\/span><\/p>\n

di = -60\/1 = -60 cm <\/span><\/span><\/p>\n

The image distance is negative means the image is virtual, where the beam of light does not pass through the image.<\/i><\/span><\/span><\/p>\n

The image distance 60 cm is greater than the object distance 15 cm.<\/i><\/span><\/span><\/p>\n

The magnification of image (M)<\/u><\/span><\/span><\/p>\n

M = -di \/ do = -(-60)\/15 = 60\/15 = 4<\/span><\/span><\/p>\n

The magnification of image exceeds 1 means that the image size is greater than the object size.<\/i><\/span><\/span><\/p>\n

The image height (hi) <\/u><\/span><\/span><\/p>\n

The equation of the magnification of image:<\/span><\/span><\/p>\n

M = hi \/ h<\/span><\/span><\/p>\n

hi = M h = (4)(10 cm) = 40 cm <\/span><\/span><\/p>\n

The image height is positive means the image is upright.<\/i><\/span><\/span><\/p>\n

The object distance is equal to the focal length of the convex lens (do = f)<\/b><\/u><\/span><\/span><\/p>\n

\"Image<\/p>\n

The image is at infinite distances.<\/span><\/span><\/p>\n

The object distance is greater than the focal length of the convex lens (do > f)<\/b><\/u><\/span><\/span><\/p>\n

Suppose the object distance is 30 cm, 40 cm and 50 cm.<\/span><\/span><\/p>\n

a) If the object distance (do) = 30 cm<\/b><\/span><\/span><\/p>\n

\"Image<\/p>\n

The image distance (di)<\/u><\/span><\/span><\/p>\n

1\/di = 1\/f \u2013 1\/do = 1\/20 \u2013 1\/30 = 3\/60 – 2\/60 = 1\/60<\/span><\/span><\/p>\n

di = 60\/1 = 60 cm <\/span><\/span><\/p>\n

The image distance is positive, means the image is real, where the beam of light passes through the image.<\/i><\/span><\/span><\/p>\n

The image distance 60 cm is greater than the object distance 30 cm.<\/i><\/span><\/span><\/p>\n

The magnification of image (M)<\/u><\/span><\/span><\/p>\n

M = -di \/ do = -60\/30 = -2<\/span><\/span><\/p>\n

The magnification of image exceeds 1, means the image size is greater to the object size.<\/i><\/span><\/span><\/p>\n

The image height (hi) <\/u><\/span><\/span><\/p>\n

The equation of the magnification of image:<\/span><\/span><\/p>\n

M = hi \/ h<\/span><\/span><\/p>\n

hi = M h = (-2)(10 cm) = -20 cm <\/span><\/span><\/p>\n

The image height is negative sign means the image is inverted.<\/i><\/span><\/span><\/p>\n

b) If the object distance (do) = 40 cm<\/b><\/span><\/span><\/p>\n

\"Image<\/p>\n

The image distance (di)<\/u><\/span><\/span><\/p>\n

1\/di = 1\/f \u2013 1\/do = 1\/20 \u2013 1\/40 = 2\/40 – 1\/40 = 1\/40<\/span><\/span><\/p>\n

di = 40\/1 = 40 cm <\/span><\/span><\/p>\n

The image distance is positive, means the image is real, where the beam of light passes through the image.<\/i><\/span><\/span><\/p>\n

The image distance 40 cm, equals to the object distance 40 cm.<\/i><\/span><\/span><\/p>\n

The magnification of image (M)<\/u><\/span><\/span><\/p>\n

M = -di \/ do = -40\/40 = -1<\/span><\/span><\/p>\n

The magnification of image is 1 means the image size is equal to the object size. <\/i><\/span><\/span><\/p>\n

The image height (hi) <\/u><\/span><\/span><\/p>\n

The equation of the magnification of image:<\/span><\/span><\/p>\n

M = hi \/ ho<\/span><\/span><\/p>\n

hi = M h = (-1)(10 cm) = -10 cm <\/span><\/span><\/p>\n

The image height is negative means the image is inverted.<\/i><\/span><\/span><\/p>\n

b) <\/b>If the object distance <\/b>(do) = 50 cm<\/b><\/span><\/span><\/p>\n

\"Image<\/p>\n

The image distance <\/u>(di)<\/u><\/span><\/span><\/p>\n

1\/di = 1\/f \u2013 1\/do = 1\/20 \u2013 1\/50 = 5\/100 – 2\/100 = 3\/100<\/span><\/span><\/p>\n

di = 100\/3 = 33.3 cm <\/span><\/span><\/p>\n

The image distance is positive, means the image is real, where the beam of light passes through the image.<\/i><\/span><\/span><\/p>\n

The image distance 33.3 cm is smaller than the object distance 50 cm.<\/i><\/span><\/span><\/p>\n

The magnification of image <\/u>(M)<\/u><\/span><\/span><\/p>\n

M = -di \/ do = -33.3\/40 = -0.8<\/span><\/span><\/p>\n

The magnification of image is smaller than 1, means the image size is smaller than the object size.<\/i><\/span><\/span><\/p>\n

The image height <\/u>(hi) <\/u><\/span><\/span><\/p>\n

The equation of the magnification of image:<\/span><\/span><\/p>\n

M = hi \/ h<\/span><\/span><\/p>\n

hi = M h = (-0.8)(10 cm) = -8 cm <\/span><\/span><\/p>\n

The image height is negative, means the image is inverted.<\/i><\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

Article about Image formation by converging (convex) lens To understand the image formation by the convex lens, learn example problems and solution below. In this case, the object is assumed to be at a certain distance from the convex lens, then draw the image formation by the convex lens, the image distance from the convex … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Image formation by converging (convex) lens","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[2],"tags":[],"class_list":["post-4302","post","type-post","status-publish","format-standard","hentry","category-basic-physics-tutorials"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/4302","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=4302"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/4302\/revisions"}],"predecessor-version":[{"id":8440,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/4302\/revisions\/8440"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=4302"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=4302"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=4302"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}