{"id":728,"date":"2018-02-16T08:18:27","date_gmt":"2018-02-16T00:18:27","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=728"},"modified":"2018-02-16T08:18:27","modified_gmt":"2018-02-16T00:18:27","slug":"convex-mirror-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/convex-mirror-problems-and-solutions.htm","title":{"rendered":"Convex mirror \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. <\/span><\/span>The focal length of a convex mirror<\/a> is 10 cm and the object distance<\/a> is 20 cm. Determine (a) the image distance (b) the magnification of image <\/span><\/span><\/p>\n

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

The focal length <\/span><\/span>(f) = -10 cm<\/span><\/span><\/p>\n

The minus sign indicates that the focal point of convex mirror is virtual<\/span><\/span><\/i><\/p>\n

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

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

Formation of image by concave mirror :<\/span><\/span><\/p>\n

\"Convex<\/p>\n

The image distance (di<\/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\/10 \u2013 1\/20 = -2\/20 \u2013 1\/20 = -3\/20 <\/span><\/span><\/p>\n

d<\/span><\/span>i<\/span> <\/sub>= -20\/3 = -6.7 cm<\/span><\/span><\/p>\n

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

The magnification of image :<\/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> = -(-6.7)\/20 = 6.7\/20 = 0.3 <\/span><\/span><\/p>\n

m = <\/i><\/span><\/span>0,3 <\/i><\/span><\/span>time smaller than the object.<\/i><\/span><\/span><\/p>\n

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

<\/p>\n

2. <\/span><\/span>A 10-cm high object is placed in front of a convex mirror with focal length 20 cm. Determine the image height if the object distance is (a) 10 cm (b) 30 cm (c) 40 cm (d) 50 cm<\/span><\/span><\/p>\n

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

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

The minus sign indicates that the focal point is virtual<\/span><\/span><\/i><\/p>\n

The radius of curvature <\/span><\/span>(<\/span><\/span>r<\/span><\/span>) = 2 f = 2(20) = 40 cm<\/span><\/span><\/p>\n

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

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

a) <\/b><\/u><\/span><\/span>the focal length <\/b><\/u><\/span><\/span>(f) = -20 cm <\/b><\/u><\/span><\/span>and the object distance <\/b><\/u><\/span><\/span>(do<\/sub>) = 10 cm <\/b><\/u><\/span><\/span><\/p>\n

\"Convex<\/p>\n

The image distance (d<\/u><\/span><\/span>i<\/u><\/span><\/span><\/sub>) <\/u><\/span><\/span>:<\/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\/10 = -1\/20 \u2013 2\/20 = -3\/20<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/sub> <\/span><\/span>= -20\/3 = -6.7<\/span><\/span><\/p>\n

The minus sign indicates that the image is virtual or the image is behind the mirror.<\/span><\/span><\/p>\n

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

m<\/span><\/span> = –<\/span><\/span>d<\/span><\/span>i <\/span><\/sub>\/ <\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = -(-6.7)\/10 = 6.7\/10 = 0.67 <\/span><\/span><\/p>\n

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

The image is 0.67 smaller than the object.<\/i><\/span><\/span><\/p>\n

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

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

h<\/span><\/span>i <\/span><\/sub>= h<\/span><\/span>o<\/span><\/sub> m<\/span><\/span> = (10 cm)(0.67) = 6.7 cm <\/span><\/span><\/p>\n

b) <\/b><\/u><\/span><\/span>the focal length <\/b><\/u><\/span><\/span>(f) = -20 cm <\/b><\/u><\/span><\/span>and the object distance <\/b><\/u><\/span><\/span>(<\/b><\/u><\/span><\/span>d<\/b><\/u><\/span><\/span>o<\/b><\/u><\/span><\/sub>) = 30 cm <\/b><\/u><\/span><\/span><\/p>\n

\"Convex<\/p>\n

The image distance <\/u><\/span><\/span>(<\/u><\/span><\/span>d<\/u><\/span><\/span>i<\/u><\/span><\/sub>) :<\/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 = -5\/60<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/span><\/sub> =<\/span><\/span> -60\/5 = -12<\/span><\/span><\/p>\n

The minus sign indicates that the image is virtual or the image is behind the mirror.<\/i><\/span><\/span><\/p>\n

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

m<\/span><\/span> = –<\/span><\/span>d<\/span><\/span>i<\/span> <\/sub>\/ <\/span><\/span>d<\/span><\/span>o<\/span> <\/sub>= -(-12)\/30 = 12\/30 = 0.4 <\/span><\/span><\/p>\n

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

The image is 0,4 times smaller the object.<\/i><\/span><\/span><\/p>\n

The height of image <\/u><\/span><\/span>(h<\/u><\/span><\/span>i<\/u><\/span><\/span><\/sub>) :<\/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>= h<\/span><\/span>o<\/span><\/sub> m<\/span><\/span> = (10 cm)(0.4) = 4 cm <\/span><\/span><\/p>\n

c) <\/b><\/u><\/span><\/span>The focal length <\/b><\/u><\/span><\/span>(f) = -20 cm <\/b><\/u><\/span><\/span>and the object distance <\/b><\/u><\/span><\/span>(<\/b><\/u><\/span><\/span>d<\/b><\/u><\/span><\/span>o<\/b><\/u><\/span><\/sub>) = 40 cm <\/b><\/u><\/span><\/span><\/p>\n

\"Convex<\/p>\n

The image distance <\/u><\/span><\/span>(<\/u><\/span><\/span>d<\/u><\/span><\/span>i<\/u><\/span><\/span><\/sub>) :<\/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\/40 = -2\/40 \u2013 1\/40 = -3\/40<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/span><\/sub> = -40\/3 = -13.3<\/span><\/span><\/p>\n

The minus sign indicates that the image is virtual or the image is behind the convex mirror.<\/i><\/span><\/span><\/p>\n

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

m<\/span><\/span> = – <\/span><\/span>d<\/span><\/span>i <\/span><\/sub>\/ <\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = -(-13.3)\/40 = 13.3\/40 = 0.3 <\/span><\/span><\/p>\n

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

The image is 0.3 smaller than the object.<\/i><\/span><\/span><\/p>\n

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

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

h<\/span><\/span>i<\/span><\/sub> = h<\/span><\/span>o<\/span><\/sub> m<\/span><\/span> = (10 cm)(0.3) = 3 cm <\/span><\/span><\/p>\n

d) <\/b><\/u><\/span><\/span>The focal length <\/b><\/u><\/span><\/span>(f) = -20 cm <\/b><\/u><\/span><\/span>and the object distance <\/b><\/u><\/span><\/span>(<\/b><\/u><\/span><\/span>d<\/b><\/u><\/span><\/span>o<\/b><\/u><\/span><\/span><\/sub>) = 50 cm <\/b><\/u><\/span><\/span><\/p>\n

The image distance <\/u><\/span><\/span>(<\/u><\/span><\/span>d<\/u><\/span><\/span>i<\/u><\/span><\/span><\/sub>) :<\/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\/50 = -5\/100 \u2013 2\/100 = -7\/100<\/span><\/span><\/p>\n

d<\/span><\/span>i<\/span><\/sub> = -100\/7 = -14.3<\/span><\/span><\/p>\n

The minus sign indicates that the image is virtual or the image is behind the convex mirror. <\/i><\/span><\/span><\/p>\n

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

m<\/span><\/span> = – <\/span><\/span>d<\/span><\/span>i <\/span><\/sub>\/ <\/span><\/span>d<\/span><\/span>o<\/span><\/sub> = -(-14.3)\/50 = 14.3\/50 = 0.3<\/span><\/span><\/p>\n

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

The image is 0.3 smaller than the object.<\/i><\/span><\/span><\/p>\n

The image height <\/u><\/span><\/span>(h<\/u><\/span><\/span>i<\/u><\/span><\/span><\/sub>)<\/u><\/span><\/span> :<\/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> = h<\/span><\/span>o<\/span><\/sub> m<\/span><\/span> = (10 cm)(0.3) = 3 cm <\/span><\/span><\/p>\n

<\/p>\n

3. <\/span><\/span>On object is 20 cm in front of convex mirror<\/span><\/span>. <\/span><\/span>If the image height is 1\/5 times the object height, determine (<\/span><\/span>a) <\/span><\/span>the image length <\/span><\/span>b) <\/span><\/span>the focal length <\/span><\/span>c)<\/span><\/span> the properties of image<\/span><\/span><\/p>\n

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

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

The image height <\/span><\/span>(h<\/span><\/span>i<\/span><\/sub>) = 1\/5 h = 0.2 h<\/span><\/span><\/p>\n

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

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

a) <\/b><\/span><\/span>the image distance <\/b><\/span><\/span>(<\/b><\/span><\/span>d<\/b><\/span><\/span>i<\/b><\/span><\/sub>)<\/b><\/span><\/span><\/p>\n

Formula of the magnification of image <\/u><\/span><\/span>:<\/u><\/span><\/span><\/p>\n

m<\/span><\/span> = h<\/span><\/span>i <\/span><\/sub>\/ h<\/span><\/span>o<\/span><\/sub> = 0.2h \/ h = 0.2<\/span><\/span><\/p>\n

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

The image is 0.2 smaller than the object.<\/i><\/span><\/span><\/p>\n

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

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

d<\/span><\/span>i<\/span><\/sub> = – <\/span><\/span>m<\/span><\/span> d<\/span><\/span>o<\/span><\/sub> = -(0.2)(20 cm) = -4 cm<\/span><\/span><\/p>\n

The minus sign indicates that the image is virtual or the image is behind the convex mirror.<\/i><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span>The focal length (<\/b><\/span><\/span>f)<\/b><\/span><\/span><\/p>\n

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

1\/f = 1\/<\/span><\/span>d<\/span><\/span>o<\/span> <\/sub>+ 1\/ <\/span><\/span>d<\/span><\/span>i <\/span><\/sub>= 1\/20 \u2013 1\/4 = 1\/20 \u2013 5\/20 = -4\/20<\/span><\/span><\/p>\n

f = -20\/4 = -5 cm <\/span><\/span><\/p>\n

The minus sign indicates that the focal point is virtual.<\/i><\/span><\/span><\/p>\n

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

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

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

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

4. Light strikes a convex mirror parallel to the axis will be reflected ….<\/span><\/span><\/p>\n

A. towards the focal point of the mirror<\/span><\/span><\/p>\n

B. as from the focal point of the mirror<\/span><\/span><\/p>\n

C. through the center of the curvature of the mirror<\/span><\/span><\/p>\n

D. perpendicular to the mirror plane<\/span><\/span><\/p>\n

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

The problem drew in the figure below.<\/span><\/span><\/p>\n

\"Convex<\/p>\n

The correct answer is B.<\/span><\/span><\/p>\n

5. A biker sees the image of a motorcycle behind it 1\/6 times its original size when the distance between the biker and motorcycle is 30 meters. Determine the radius of curvature of the rear view mirror…<\/span><\/span><\/p>\n

A. 7.14 m<\/span><\/span><\/p>\n

B. 8.57 m<\/span><\/span><\/p>\n

C. 12.00 m<\/span><\/span><\/p>\n

D. 24.00 m<\/span><\/span><\/p>\n

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

Magnification of image (M) = 1\/6 times<\/span><\/span><\/p>\n

Object distance (d) = 30 meter<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The radius of curvature of the rear view mirror (R)<\/span><\/span><\/p>\n

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

Calculate the distance of the image (d\u2019)<\/u><\/span><\/span><\/p>\n

Since the magnification of image (M) and the object distance (s) has been known, the image distance can be known using the formula of image magnification :<\/span><\/span><\/p>\n

\"Convex<\/p>\n

A negative sign means the image is virtual. The image is 5 meters behind the convex mirror.<\/span><\/span><\/span><\/p>\n

Calculate the focal length (f)<\/u><\/span><\/span><\/p>\n

Since the object distance (d) and the image distance (d’), then the focal length can be calculated using the formula of the mirror :<\/span><\/span><\/span><\/p>\n

\"Convex<\/p>\n

The radius of curvature (R)<\/u><\/span><\/span><\/p>\n

The radius of curvature of a convex mirror is twice the focal length of a convex mirror.
\nR = 2 f = 2 (6 meters) = 12 meters
\nThe radius of curvature of the convex mirror is 12 meters.
\nThe correct answer is C.<\/span><\/span><\/span><\/p>\n

6. <\/span><\/span>The convex mirror is chosen as the rearview mirror of a motorcycle because the properties of the image produced by the mirror are…<\/span><\/span><\/span><\/p>\n

A. real, upright, minimized<\/span><\/span><\/span><\/p>\n

B. real, upright, enlarged<\/span><\/span><\/span><\/p>\n

C. virtual, upright, minimized<\/span><\/span><\/span><\/p>\n

D. virtual, upright, enlarged<\/span><\/span><\/span><\/p>\n

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

\"Convex<\/span><\/p>\n

Based on the two figure above can concluded that the properties of the image are virtual, upright, <\/span><\/span>minimized<\/span><\/span><\/span>.<\/span><\/span><\/p>\n

The correct answer is C.<\/span><\/span><\/p>\n

7. An <\/span><\/span>object is 12 cm in front of a convex mirror with a radius of 6 cm. The properties of the image are…<\/span><\/span><\/span><\/p>\n

A. real, inverted at a distance of 12 cm<\/span><\/span><\/span><\/p>\n

B. real, upright at a distance of 4 cm<\/span><\/span><\/span><\/p>\n

C. virtual, upright at a distance of 2.4 cm<\/span><\/span><\/span><\/p>\n

D. virtual, inverted at a distance of 6 cm<\/span><\/span><\/span><\/p>\n

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

Object’s distance (d) = 12 cm<\/span><\/span><\/p>\n

Radius of the <\/span><\/span>convex mirror <\/span><\/span><\/span>(r) = 6 cm.<\/span><\/span><\/p>\n

The focal length of the <\/span><\/span>convex mirror<\/span><\/span><\/span> (f) = 6 cm \/ 2 = -3 cm<\/span><\/span><\/p>\n

The focal length of a convex mirror is signed negative because it is virtual. Virtual because it is not passed by light.<\/i><\/span><\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Properties of image<\/span><\/span><\/p>\n

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

Distance of image (d’) :<\/span><\/span><\/p>\n

1\/d’ = 1\/f \u2013 1\/d = -1\/3 \u2013 1\/12 = -4\/12 \u2013 1\/12 = -5\/12<\/span><\/span><\/p>\n

d’ = -12\/5 = -2.4 cm<\/span><\/span><\/p>\n

The image distance signed negative means image is virtual. <\/span><\/span><\/span><\/p>\n

Magnification of image (m):<\/span><\/span><\/span><\/p>\n

m = -d’ \/ d = -(-2.4) \/ 12 = 2.4 \/ 12 = 0.2 times<\/span><\/span><\/p>\n

The magnification of image signed positive means that the image is upright and the magnification of image is 0.2 means the image size is smaller than the object size (reduced).<\/span><\/span><\/span><\/p>\n

The correct answer is C.<\/span><\/span><\/span><\/p>\n

[wpdm_package id=’860′]<\/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. The focal length of a convex mirror is 10 cm and the object distance is 20 cm. Determine (a) the image distance (b) the magnification of image Known : The focal length (f) = -10 cm The minus sign indicates that the focal point of convex mirror is virtual The object distance (do) = … 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":"Convex mirror \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-728","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\/728","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=728"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/728\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=728"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=728"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=728"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}