{"id":777,"date":"2018-02-17T10:53:42","date_gmt":"2018-02-17T02:53:42","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=777"},"modified":"2018-02-17T10:53:42","modified_gmt":"2018-02-17T02:53:42","slug":"diverging-lens-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/diverging-lens-problems-and-solutions.htm","title":{"rendered":"Diverging lens \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A 5-cm high object is placed 15 cm from a 30-cm focal length diverging lens<\/a>. Determine the image distance<\/a>, the magnification of the image, the image height, and properties of the image.<\/span><\/span><\/p>\n

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

The focal length (f) = -30 cm<\/span><\/span><\/p>\n

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

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

The object distance (d<\/span><\/span>o<\/span><\/sub>) = 15 cm<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The image distance (d<\/span><\/span>i<\/span><\/span><\/sub>), the magnification of image (m), the image height (h<\/span><\/span>i<\/span><\/span><\/sub>) and the properties of the image.<\/span><\/span><\/p>\n

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

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

\"Diverging<\/p>\n

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

1\/d<\/span><\/span>i <\/span><\/sub>= 1\/f \u2013 1\/d<\/span><\/span>o <\/span><\/sub>= -1\/30 \u2013 1\/15 = -1\/30 \u2013 2\/30 = -3\/30<\/span><\/span><\/p>\n

d<\/span><\/span>i <\/span><\/sub>= -30\/3 = -10 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><\/p>\n

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

m = – d<\/span><\/span>i <\/span><\/sub>\/ d<\/span><\/span>o<\/span><\/sub> = -(-10)\/15 = 10\/15 = 2\/3 <\/span><\/span><\/p>\n

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

The image 2\/3 smaller than the object.<\/span><\/span><\/p>\n

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

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

h<\/span><\/span>i <\/span><\/sub>= m h<\/span><\/span>o <\/span><\/sub> = (2\/3)5 = 10\/3 = 3.3 cm <\/span><\/span><\/p>\n

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

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

The properties of the image formed by a diverging mirror :<\/span><\/span><\/p>\n

– virtual<\/span><\/span><\/p>\n

– upright <\/span><\/span><\/p>\n

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

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

<\/p>\n

2. A 10-cm high object is placed 60 cm from a 30-cm focal length diverging lens. Determine the image distance, the magnification of the image, the image height, the properties of the image. <\/span><\/span><\/p>\n

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

The focal length (f) = -30 cm<\/span><\/span><\/p>\n

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

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

The object distance (d<\/span><\/span>o<\/span><\/sub>) = 60 cm<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The image distance (d<\/span><\/span>i<\/span><\/sub>), the magnification of image (m), the image height (h<\/span><\/span>i<\/span><\/span><\/sub>) and the properties of image <\/span><\/span><\/p>\n

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

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

\"Diverging<\/p>\n

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

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

d<\/span><\/span>i<\/span><\/sub> = -60\/3 = -20 cm<\/span><\/span><\/p>\n

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

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

m = -d<\/span><\/span>i<\/span><\/sub>\/d<\/span><\/span>o<\/span><\/sub> = -(-20)\/60 = 20\/60 = 1\/3 <\/span><\/span><\/p>\n

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

The image 1\/3 smaller than the object.<\/i><\/span><\/span><\/p>\n

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

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

h<\/span><\/span>i <\/span><\/sub>= m h<\/span><\/span>o<\/span><\/sub> = (1\/3)10 = 10\/3 = 3.3 cm<\/span><\/span><\/p>\n

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

Properties of image<\/b><\/u><\/span><\/span><\/p>\n

– virtual because the rays do not pass through the image<\/span><\/span><\/p>\n

– upright<\/span><\/span><\/p>\n

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

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

[wpdm_package id=’862′]<\/p>\n

    \n
  1. Concave mirror problems and solutions<\/a><\/li>\n
  2. Convex mirror problems and solutions<\/a><\/li>\n
  3. Diverging lens problems and solutions<\/a><\/li>\n
  4. Converging lens problems and solutions<\/a><\/li>\n
  5. Optical instrument human eye problems and solutions<\/a><\/li>\n
  6. Optical instrument contact lenses problems and solutions<\/a><\/li>\n
  7. Optical instrument eyeglasses<\/a><\/li>\n
  8. Optical instrument magnifying glass problems and solutions<\/a><\/li>\n
  9. Optical instrument microscope – problems and solutions<\/a><\/li>\n
  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 15 cm from a 30-cm focal length diverging lens. Determine the image distance, the magnification of the image, the image height, and properties of the image. Known : The focal length (f) = -30 cm The minus sign indicates that the focal point is virtual or the rays … 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":"Diverging 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-777","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\/777","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=777"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/777\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=777"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=777"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=777"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}