{"id":785,"date":"2018-02-17T13:47:57","date_gmt":"2018-02-17T05:47:57","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=785"},"modified":"2018-02-17T13:47:57","modified_gmt":"2018-02-17T05:47:57","slug":"converging-lens-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/converging-lens-problems-and-solutions.htm","title":{"rendered":"Converging lens \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A 5-cm high object is placed 5 cm from a <\/span><\/span>15<\/span><\/span>-cm focal length <\/span><\/span>converging <\/span><\/span>lens. Determine the image distance<\/a>, the magnification of the image, the image height and <\/span><\/span>the <\/span><\/span>properties o<\/span><\/span>f <\/span><\/span>the image.<\/span><\/span><\/p>\n

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

The focal length <\/span><\/span>(f) = 15 cm<\/span><\/span><\/p>\n

The plus sign indicates that the focal point is real or the rays pass through the point.<\/span><\/span><\/i><\/p>\n

The object height <\/span><\/span>(h<\/span><\/span>o<\/span><\/sub>) = 5 cm<\/span><\/span><\/p>\n

The object distance <\/span><\/span>(<\/span><\/span>d<\/span><\/span>o<\/span><\/sub>) = 5 cm<\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>:<\/u><\/span><\/span> the image distance<\/span><\/span>, <\/span><\/span>the magnification of image<\/span><\/span>, <\/span><\/span>the image height and the properties of image <\/span><\/span><\/p>\n

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

Formation of image by converging lens<\/a> :<\/span><\/span><\/p>\n

\"Converging<\/p>\n

The image distance<\/b><\/u><\/span><\/span> (<\/b><\/u><\/span><\/span>d<\/b><\/u><\/span><\/span>i<\/b><\/u><\/span><\/sub>) :<\/b><\/u><\/span><\/span><\/p>\n

1\/<\/span><\/span>d<\/span><\/span>i<\/span><\/sub> = 1\/f \u2013 1\/<\/span><\/span>d<\/span><\/span>o <\/span><\/sub>= 1\/15 \u2013 1\/5 = 1\/15 \u2013 3\/15 = -2\/15<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/sub> <\/span><\/span>= -15\/2 = -7.5 cm<\/span><\/span><\/p>\n

The minus sign indicates that the image is virtual or the rays do not pass through the image.<\/span><\/span><\/i><\/p>\n

The magnification of image <\/b><\/u><\/span><\/span>(<\/b><\/u><\/span><\/span>m<\/b><\/u><\/span><\/span>) :<\/b><\/u><\/span><\/span><\/p>\n

m<\/span><\/span> = – <\/span><\/span>d<\/span><\/span>i<\/span><\/sub> \/ <\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = -(-7.5)\/5 = 7.5\/5 = 1.5 <\/span><\/span><\/p>\n

The plus sign indicates that the image is upright.<\/span><\/span><\/p>\n

The image height <\/b><\/u><\/span><\/span>(h<\/b><\/u><\/span><\/span>i<\/b><\/u><\/span><\/sub>) :<\/b><\/u><\/span><\/span><\/p>\n

m<\/span><\/span> = h<\/span><\/span>i<\/span><\/sub> \/ h<\/span><\/span>o<\/span><\/sub><\/p>\n

h<\/span><\/span>i <\/span><\/sub>= <\/span><\/span>m <\/span><\/span>h<\/span><\/span>o<\/span> <\/sub>= (1.5)5 = 10\/3 = 7.5 cm <\/span><\/span><\/p>\n

The plus sign indicates that the image is upright.<\/i><\/span><\/span><\/p>\n

The properties of the image<\/b><\/u><\/span><\/span><\/p>\n

– <\/span><\/span>virtual<\/span><\/span><\/p>\n

– <\/span><\/span>upright<\/span><\/span><\/p>\n

– <\/span><\/span>The image greater than the object<\/span><\/span><\/p>\n

– <\/span><\/span>The image distance is greater than the object distance<\/span><\/span><\/p>\n

<\/p>\n

2. A <\/span><\/span>10<\/span><\/span>-cm high object is placed <\/span><\/span>30<\/span><\/span> cm from a <\/span><\/span>15<\/span><\/span>-cm focal length <\/span><\/span>converging <\/span><\/span>lens. Determine the image distance, the magnification of the image, the image height and <\/span><\/span>the <\/span><\/span>properties o<\/span><\/span>f <\/span><\/span>the image.<\/span><\/span><\/p>\n

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

The focal length <\/span><\/span>(f) = 15 cm<\/span><\/span><\/p>\n

The plus sign indicates that the focal point is real or the rays pass through the point.<\/span><\/span><\/i><\/p>\n

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

The object distance <\/span><\/span>(s) = 30 cm<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The image distance, the magnification of the image, the image height and the properties of the image<\/span><\/span><\/p>\n

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

Formation of the image by the converging lens :<\/span><\/span><\/p>\n

\"Converging<\/p>\n

The image distance <\/b><\/u><\/span><\/span>(<\/b><\/u><\/span><\/span>d<\/b><\/u><\/span><\/span>i<\/b><\/u><\/span><\/sub>) :<\/b><\/u><\/span><\/span><\/p>\n

1\/<\/span><\/span>d<\/span><\/span>i <\/span><\/sub>= 1\/f \u2013 1\/<\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = 1\/15 \u2013 1\/30 = 2\/30 \u2013 1\/30 = 1\/30<\/span><\/span><\/p>\n

d<\/span><\/span>i <\/span><\/sub>= 30\/1 = 30 cm<\/span><\/span><\/p>\n

The plus sign indicates that the image is real or the rays<\/a> pass through the image.<\/span><\/span><\/p>\n

The magnification of image <\/b><\/u><\/span><\/span>(<\/b><\/u><\/span><\/span>m<\/b><\/u><\/span><\/span>) :<\/b><\/u><\/span><\/span><\/p>\n

m<\/span><\/span> = – <\/span><\/span>d<\/span><\/span>i<\/sub> <\/span><\/span>\/ <\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = -(30)\/30 = -30\/30 = -1 <\/span><\/span><\/p>\n

The minus sign indicates that the image is inverted.<\/span><\/span><\/p>\n

The image height <\/b><\/u><\/span><\/span>(h<\/b><\/u><\/span><\/span>i<\/b><\/u><\/span><\/sub>) :<\/b><\/u><\/span><\/span><\/p>\n

m<\/span><\/span> = h<\/span><\/span>i<\/span><\/sub> \/ h<\/span><\/span>o<\/span><\/sub><\/p>\n

h<\/span><\/span>i<\/span><\/sub> = <\/span><\/span>m<\/span><\/span> h<\/span><\/span>o<\/span> <\/sub>= (-1)10 = -10 cm <\/span><\/span><\/p>\n

The minus sign indicates that the image is inverted.<\/span><\/span><\/p>\n

The properties of the image<\/b><\/u><\/span><\/span><\/p>\n

– <\/span><\/span>real or the rays pass through the image<\/span><\/span><\/p>\n

– <\/span><\/span>inverted<\/span><\/span><\/p>\n

– <\/span><\/span>The image height is the same as the object height.<\/span><\/span><\/p>\n

– <\/span><\/span>The image distance is the same as the object distance.<\/span><\/span><\/p>\n

<\/p>\n

3. A <\/span><\/span>10<\/span><\/span>-cm high object is placed <\/span><\/span>30<\/span><\/span> cm from a <\/span><\/span>20<\/span><\/span>-cm focal length <\/span><\/span>converging <\/span><\/span>lens. Determine the image distance, the magnification of the image, the image height and <\/span><\/span>the <\/span><\/span>properties o<\/span><\/span>f <\/span><\/span>the image.<\/span><\/span><\/p>\n

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

The focal length <\/span><\/span>f) = 20 cm<\/span><\/span><\/p>\n

The plus sign indicates that the focal point is real or the rays pass through the point.<\/span><\/span><\/i><\/p>\n

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

The object distance <\/span><\/span>(<\/span><\/span>d<\/span><\/span>o<\/span><\/sub>) = 30 cm<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The image distance<\/span><\/span>, <\/span><\/span>the magnification of image<\/span><\/span>, <\/span><\/span>the image height and the properties of image <\/span><\/span><\/p>\n

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

The image distance<\/b><\/u><\/span><\/span> (<\/b><\/u><\/span><\/span>d<\/b><\/u><\/span><\/span>i<\/b><\/u><\/span><\/sub>) :<\/b><\/u><\/span><\/span><\/p>\n

1\/<\/span><\/span>d<\/span><\/span>i<\/span><\/sub> = 1\/f \u2013 1\/<\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = 1\/20 \u2013 1\/30 = 3\/60 \u2013 2\/60 = 1\/60<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/span><\/sub> = 60\/1 = 60 cm<\/span><\/span><\/p>\n

The plus sign indicates that the image is real or the rays pass through the image.<\/span><\/span> <\/i><\/p>\n

The magnification of image <\/b><\/u><\/span><\/span>(<\/b><\/u><\/span><\/span>m<\/b><\/u><\/span><\/span>) :<\/b><\/u><\/span><\/span><\/p>\n

m<\/span><\/span> = – <\/span><\/span>d<\/span><\/span>i<\/span> <\/sub>\/ <\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = -(60)\/30 = -2 <\/span><\/span><\/p>\n

The minus sign indicates that the image is inverted.<\/span><\/span><\/i><\/p>\n

The image is 2 times greater than the object.<\/i><\/span><\/span><\/p>\n

The image height <\/b><\/u><\/span><\/span>(h<\/b><\/u><\/span><\/span>i<\/b><\/u><\/span><\/sub>) :<\/b><\/u><\/span><\/span><\/p>\n

m<\/span><\/span> = h<\/span><\/span>i<\/span><\/sub> \/ h<\/span><\/span>o<\/span><\/sub><\/p>\n

h<\/span><\/span>i<\/span><\/span> = <\/span><\/span>m<\/span><\/span> h<\/span><\/span>o<\/span><\/sub> = (-2)10 = -20 cm <\/span><\/span><\/p>\n

The minus sign indicates that the image is inverted.<\/span><\/span><\/p>\n

The properties of the image<\/b><\/u><\/span><\/span><\/p>\n

– <\/span><\/span>real<\/span><\/span><\/p>\n

– <\/span><\/span>inverted<\/span><\/span><\/p>\n

– <\/span><\/span>the image height is greater than the object height<\/span><\/span><\/p>\n

– <\/span><\/span>the image distance is greater than the object distance<\/span><\/span><\/p>\n

<\/p>\n

4. Based on the figure below, determine the focal length of the converging lens!<\/span><\/span><\/p>\n

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

Object distance<\/span><\/span> (<\/span><\/span>d<\/span><\/span>o<\/span><\/span><\/sub>) = 20 cm<\/span><\/span><\/p>\n

Image distance <\/span><\/span>(<\/span><\/span>d<\/span><\/span>i<\/span><\/span><\/sub>) = 30 cm<\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>:<\/u><\/span><\/span> The focal length <\/span><\/span>(f)<\/span><\/span><\/p>\n

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

1\/<\/span><\/span>d<\/span><\/span>o<\/span><\/span><\/sub> + 1\/<\/span><\/span>d<\/span><\/span>i<\/span><\/span><\/sub> = 1\/f<\/span><\/span><\/p>\n

d<\/span><\/span>o<\/span><\/span><\/sub> = <\/span><\/span>object distance <\/span><\/span>(<\/span><\/span>plus sign <\/i><\/span><\/span>because <\/i><\/span><\/span>rays pass through <\/i><\/span><\/span>object<\/i><\/span><\/span>)<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/span><\/span><\/sub> = <\/span><\/span>image distance<\/span><\/span> (<\/span><\/span>plus sign <\/i><\/span><\/span>because <\/i><\/span><\/span>rays pass through <\/i><\/span><\/span>image<\/i><\/span><\/span>)<\/span><\/span><\/p>\n

f = <\/span><\/span>focal length <\/span><\/span>(<\/span><\/span>plus sign <\/i><\/span><\/span>because rays pass through the focal point or the focal point is real<\/i><\/span><\/span>)<\/span><\/span><\/p>\n

The focal length :<\/span><\/span><\/p>\n

1\/f = 1\/20 + 1\/30 = 3\/60 + 2\/60 = 5\/60<\/span><\/span><\/p>\n

f = 60\/5 = 12 cm<\/span><\/span><\/p>\n

5. <\/span><\/span>Object’s distance is 30 cm and the focal length is 20 cm. Determine the magnification of the image.<\/span><\/span><\/p>\n

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

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

The plus sign indicates that the focal point is real or rays <\/i><\/span><\/span>pass through <\/i><\/span><\/span>the image.<\/i><\/span><\/span><\/p>\n

Object distance <\/span><\/span>(<\/span><\/span>d<\/span><\/span>o<\/span><\/span><\/sub>) = 30 cm<\/span><\/span><\/p>\n

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

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

Image distance :<\/span><\/span><\/p>\n

1\/<\/span><\/span>d<\/span><\/span>i<\/span><\/span><\/sub> = 1\/f \u2013 1\/<\/span><\/span>do<\/span><\/span> = 1\/20 \u2013 1\/30 = 3\/60 \u2013 2\/60 = 1\/60<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/span><\/span><\/sub> = 60\/1 = 60 cm<\/span><\/span><\/p>\n

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

M = <\/span><\/span>d<\/span><\/span>i<\/span><\/span><\/sub> \/ <\/span><\/span>d<\/span><\/span>o<\/span><\/span><\/sub> = 60 cm \/ 30 cm = 60\/30 = 2 <\/span><\/span>times<\/span><\/span><\/p>\n

[wpdm_package id=’864′]<\/p>\n

    \n
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  6. Optical instrument contact lenses problems and solutions<\/a><\/li>\n
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  10. Optical instrument telescopes problems and solutions<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    1. A 5-cm high object is placed 5 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image. Known : The focal length (f) = 15 cm The plus sign indicates that the focal point is real or the … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","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":"Converging lens \u2013 problems and solutions","_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":[3],"tags":[],"class_list":["post-785","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/785","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=785"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/785\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=785"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=785"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=785"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}