{"id":2214,"date":"2018-04-28T10:22:57","date_gmt":"2018-04-28T02:22:57","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2214"},"modified":"2023-08-08T14:19:00","modified_gmt":"2023-08-08T14:19:00","slug":"compound-microscope-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/compound-microscope-problems-and-solutions.htm","title":{"rendered":"Compound microscope \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Compound microscope \u2013 problems and solutions<\/span><\/p>\n

1.<\/span><\/p>\n

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

Based on the figure above, what is the overall magnification? Assume a normal eye so the near point (N) = 25 cm.<\/span><\/p>\n

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

The focal length of the objective lens (fo<\/sub>) = 2 cm<\/span><\/p>\n

The distance between the object and the objective lens (do<\/sub>) = 2.2 cm<\/span><\/p>\n

The near point of the normal eye (N) = 25 cm<\/span><\/p>\n

The focal length of the eyepiece lens (fe<\/sub>) = 25 cm <\/span><\/p>\n

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

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

If the final image formed by the eyepiece lens at infinity then the eye is relaxed. If the final image not at infinity then the eye is not relaxed.<\/span><\/p>\n

Based on the figure above, the distance of the virtual image formed by the eyepiece lens at infinity. The final image at infinity so that the eye is relaxed. <\/span><\/p>\n

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

First, calculate the image distance from the objective lens (do<\/sub>‘). The objective lens is the converging lens so that the image distance calculated using the equation of the converging lens<\/a>. <\/span><\/p>\n

The focal length of the converging lens is positive.<\/span><\/p>\n

1\/do<\/sub>‘ = 1\/fo<\/sub> \u2013 1\/do <\/sub>= 1\/2 \u2013 1 \/ 2.2 = 11 \/ 22 \u2013 10\/22 = 1\/22<\/span><\/p>\n

do<\/sub>‘ = 22\/1 = 22 cm<\/span><\/p>\n

The overall magnification of the microscope<\/a> :<\/span><\/p>\n

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

2.<\/span><\/p>\n

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

Based on the figure above, what is the distance between the objective lens and the eyepiece lens of the microscope? <\/span><\/p>\n

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

The microscope has both objective and eyepiece (ocular) lenses. A microscope is used to view objects that are very close. The image formed by the objective lens is the real image. The image is magnified by the eyepiece lens into a very large virtual image. <\/span><\/p>\n

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

The focal length of the objective lens (fo<\/sub>) = 1.8 cm (The focal length is positive because the objective lens is the converging lens)<\/span><\/p>\n

The focal length of the eyepiece lens (fe<\/sub>) = 6 cm (The focal length is positive because the ocular lens is the converging lens)<\/span><\/p>\n

The distance between the object and the objective lens (do<\/sub>) = 2 cm (plus sign indicates that image is real)<\/span><\/p>\n

Wanted: The distance between the objective lens and the eyepiece lens (length of microscope = d)<\/span><\/p>\n

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

The distance between the image and the objective lens <\/u>(<\/u>d<\/u>o<\/u><\/sub>\u2019)<\/u> :<\/span><\/p>\n

1\/do <\/sub>+ 1\/do<\/sub>\u2019 = 1\/fo<\/sub><\/span><\/p>\n

1\/do<\/sub>\u2019 = 1\/fo<\/sub> – 1\/do <\/sub><\/span><\/p>\n

1\/do<\/sub>\u2019 = 1 \/ 1.8 – 1\/2 = 10\/18 – 9\/18 = 1\/18<\/span><\/p>\n

do<\/sub>\u2019 = 18 cm <\/span><\/p>\n

The real image formed by the objective lens located at the focal point of the eyepiece lens, as shown on the figure above.<\/span><\/p>\n

The distance between the objective lens and the ocular lens :<\/u><\/span><\/p>\n

l = do<\/sub>\u2019 + fe <\/sub>= 18 cm + 6 cm = 24 cm <\/span><\/p>\n

3.<\/span><\/p>\n

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

The distance between the object and the objective lens is 1.1 cm, the focal length of the objective lens is 1 cm and the focal length of the ocular lens is 5 cm. What is the overall magnification of the microscope?<\/span><\/p>\n

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

The distance between the object and the objective lens (do<\/sub>) = 1.1 cm <\/span><\/p>\n

The focal length of the objective lens (fo<\/sub>) = 1 cm (The focal length is positive because the objective lens is the converging lens)<\/span><\/p>\n

The focal length of the ocular lens (fe<\/sub>) = 5 cm (The focal length is positive because the ocular lens is the converging lens)<\/span><\/p>\n

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

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

The overall magnification of a microscope is the product of the magnification of the objective lens (mo<\/sub>) times the angular magnification (Me<\/sub>) of the eyepiece lens.<\/span><\/p>\n

The magnification of the objective lens :<\/u><\/span><\/p>\n

The equation of the linear magnification of the objective lens <\/u>(m<\/u>ob<\/u><\/sub>)<\/u> :<\/span><\/p>\n

mo<\/sub> = ho<\/sub>\u2019\/ho <\/sub>= do<\/sub>\u2019\/do <\/sub>= (l \u2013 fe<\/sub>)\/do<\/sub><\/span><\/p>\n

where<\/span><\/p>\n

ho<\/sub>\u2019 = the height of image formed by the objective lens<\/span><\/p>\n

ho<\/sub> = the height of object<\/span><\/p>\n

do<\/sub>\u2019 = the distance between the real image formed by the objective lens and the objective lens<\/span><\/p>\n

do <\/sub>= distance between object and the objective lens<\/span><\/p>\n

fe <\/sub>= the focal length of the ocular lens<\/span><\/p>\n

l = the length of microscope = distance between the two lenses = the distance between the real image formed by the objective lens (do<\/sub>\u2019) + the focal length of the ocular lens fe<\/sub>) <\/span><\/p>\n

The distance between the real image formed by the objective lens and the objective lens <\/u>(<\/u>d<\/u>o<\/u><\/sub>\u2019)<\/u> :<\/span><\/p>\n

First, determine the distance between the real image and the objective lens :<\/span><\/p>\n

1\/do<\/sub> + 1\/do<\/sub>\u2019 = 1\/fo<\/sub><\/span><\/p>\n

1\/do<\/sub>\u2019 = 1\/fo<\/sub> – 1\/do <\/sub><\/span><\/p>\n

1\/do<\/sub>\u2019 = 1\/1 \u2013 1 \/ 1.1 = 11\/11 – 10\/11 = 1\/11<\/span><\/p>\n

do<\/sub>\u2019 = 11 cm <\/span><\/p>\n

The real image formed by the objective lens is at the ocular focal point, as shown in the figure above.<\/span><\/p>\n

The magnification of the objective lens :<\/u><\/span><\/p>\n

mo <\/sub>= do<\/sub>\u2019 \/ do<\/sub> = 11 cm \/ 1.1 cm = 10<\/span><\/p>\n

The magnification of the ocular lens :<\/u><\/span><\/p>\n

If the eye is relaxed, then the magnification of the ocular lens (Me<\/sub>) :<\/span><\/p>\n

Me<\/sub> = N \/ fe<\/sub><\/span><\/p>\n

Where<\/span><\/p>\n

N = the near point of eye (25 cm) <\/span><\/p>\n

fe <\/sub>= the focal length of the ocular lens = 5 cm<\/span><\/p>\n

The angular magnification of the ocular lens <\/u>(M<\/u>e<\/u><\/sub>)<\/u> :<\/span><\/p>\n

Me <\/sub>= N \/ fe <\/sub>= 25 cm \/ 5 cm = 5<\/span><\/p>\n

The overall magnification of the microscope :<\/u><\/span><\/p>\n

M = mo <\/sub>x Me<\/sub> = 10 x 5 = 50<\/span><\/p>\n

4.<\/span><\/p>\n

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

The distance between the objective lens and the ocular lens of the microscope.<\/span><\/p>\n

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

The distance between object and the objective lens (do<\/sub>) = 2 cm<\/span><\/p>\n

The focal length of the objective lens (fo<\/sub>) = 1.8 cm<\/span><\/p>\n

The distance between the real image and the ocular lens (de<\/sub>) = 6 cm<\/span><\/p>\n

The focal length of the ocular lens (fe<\/sub>) = 6 cm <\/span><\/p>\n

Wanted:<\/u> The distance between the objective lens and the ocular lens (the length of the microscope)<\/span><\/p>\n

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

If the eye is relaxed, the final image formed by the ocular lens is at infinity. The final image is at infinity if the real image formed by the objective lens is at the focal point of the ocular lens.<\/span><\/p>\n

The distance between the objective lens and the ocular lens (l) = the distance between the real image formed by the objective lens (do<\/sub>\u2019) + the focal length of the ocular lens (fe<\/sub>).<\/span><\/p>\n

The distance between the real image and the objective lens (do<\/sub>\u2019) :<\/span><\/p>\n

1\/do<\/sub> + 1\/do<\/sub>\u2019 = 1\/fo<\/sub><\/span><\/p>\n

1\/do<\/sub>\u2019 = 1\/fo<\/sub> – 1\/do <\/sub><\/span><\/p>\n

1\/do<\/sub>\u2019 = 1 \/ 1.8 – 1\/2 <\/span><\/p>\n

1\/do<\/sub>\u2019 = 10\/18 – 9\/18 = 1\/18<\/span><\/p>\n

do<\/sub>\u2019 = 18 cm <\/span><\/p>\n

The distance between the objective lens and the ocular lens (the length of the microscope) :<\/span><\/p>\n

l = do<\/sub>\u2019 + fe <\/sub><\/span><\/p>\n

l = 18 cm + 6 cm <\/span><\/p>\n

l = 24 cm<\/span><\/p>\n

    \n
  1. What is a compound microscope?<\/strong><\/span>\n
      \n
    • Answer:<\/strong> A compound microscope is an optical instrument that uses multiple lenses to magnify small objects, enabling the viewer to observe details that are too small to be seen with the naked eye.<\/span><\/li>\n<\/ul>\n<\/li>\n
    • How does a compound microscope differ from a simple microscope?<\/strong><\/span>\n
        \n
      • Answer:<\/strong> A simple microscope uses just one lens for magnification (like a magnifying glass), whereas a compound microscope uses two sets of lenses \u2013 the objective lens near the specimen and the eyepiece lens through which the viewer looks.<\/span><\/li>\n<\/ul>\n<\/li>\n
      • What are the primary parts of a compound microscope, and what are their functions?<\/strong><\/span>\n
          \n
        • Answer:<\/strong> The primary parts include:<\/span>\n
            \n
          • Objective Lenses:<\/strong> These are a series of lenses with varying magnifications placed closest to the specimen.<\/span><\/li>\n
          • Eyepiece (Ocular) Lens:<\/strong> This is the lens through which the viewer looks.<\/span><\/li>\n
          • Stage:<\/strong> The flat platform where the specimen is placed.<\/span><\/li>\n
          • Condenser:<\/strong> Focuses light on the specimen.<\/span><\/li>\n
          • Illuminator\/Light Source:<\/strong> Provides light to view the specimen.<\/span><\/li>\n
          • Coarse and Fine Focus Knobs:<\/strong> Adjust the focus.<\/span><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n
          • How is the total magnification of a compound microscope determined?<\/strong><\/span>\n
              \n
            • Answer:<\/strong> The total magnification is determined by multiplying the magnification of the objective lens by the magnification of the eyepiece lens.<\/span><\/li>\n<\/ul>\n<\/li>\n
            • Why is immersion oil sometimes used with high-power objective lenses?<\/strong><\/span>\n
                \n
              • Answer:<\/strong> Immersion oil is used to increase the resolution of the microscope. It has the same refractive index as glass, which reduces the scattering of light and allows more light to enter the objective lens.<\/span><\/li>\n<\/ul>\n<\/li>\n
              • What does the term “resolution” mean in microscopy?<\/strong><\/span>\n
                  \n
                • Answer:<\/strong> Resolution refers to the ability of the microscope to clearly distinguish two points that are very close together. It defines the clarity and level of detail that can be seen.<\/span><\/li>\n<\/ul>\n<\/li>\n
                • Why is staining often used when viewing specimens under a compound microscope?<\/strong><\/span>\n
                    \n
                  • Answer:<\/strong> Staining enhances the contrast between different parts of a specimen, making certain structures or components more visible under the microscope.<\/span><\/li>\n<\/ul>\n<\/li>\n
                  • What is a “parfocal” lens system in a compound microscope?<\/strong><\/span>\n
                      \n
                    • Answer:<\/strong> A parfocal lens system means that when one objective lens is in focus, the other objective lenses will also be approximately in focus. It allows for easier switching between objective lenses without significant refocusing.<\/span><\/li>\n<\/ul>\n<\/li>\n
                    • What is the difference between a monocular and binocular compound microscope?<\/strong><\/span>\n
                        \n
                      • Answer:<\/strong> A monocular compound microscope has one eyepiece for viewing, while a binocular compound microscope has two eyepieces, providing more comfortable viewing and depth perception.<\/span><\/li>\n<\/ul>\n<\/li>\n
                      • Why might a specimen need to be very thin when viewed under a compound microscope?<\/strong><\/span><\/li>\n<\/ol>\n
                          \n
                        • Answer:<\/strong> A thin specimen allows more light to pass through, resulting in a clearer image. Additionally, thick specimens might not be in focus throughout their depth due to the limited depth of field of the microscope.<\/span><\/li>\n<\/ul>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

                          Compound microscope \u2013 problems and solutions 1. Based on the figure above, what is the overall magnification? Assume a normal eye so the near point (N) = 25 cm. Known :<\/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":"Compound microscope \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-2214","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\/2214","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=2214"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2214\/revisions"}],"predecessor-version":[{"id":8624,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2214\/revisions\/8624"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2214"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2214"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2214"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}