{"id":3066,"date":"2018-06-08T02:23:51","date_gmt":"2018-06-08T09:23:51","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=3066"},"modified":"2023-08-06T14:12:54","modified_gmt":"2023-08-06T14:12:54","slug":"diffraction-by-a-single-slit-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/diffraction-by-a-single-slit-problems-and-solutions.htm","title":{"rendered":"Diffraction by a single slit \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Diffraction by a single slit \u2013 problems and solutions<\/p>\n

1. Light with wavelength<\/a> of 500 nm passes through a slit 0.2 mm wide. The diffraction<\/a> pattern on a screen 60 cm away. Determine the distance<\/a> between the central maximum and the second minimum.<\/span><\/span><\/p>\n

\"Diffraction<\/p>\n

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

\u03bb = 500 nm = 500 x 10-9 <\/sup>m = 5 x 10-7<\/sup> m<\/span><\/span><\/p>\n

d = 0.2 mm = 0.2 x 10-3<\/sup> m = 2 x 10-4<\/sup> m<\/span><\/span><\/p>\n

l = 60 cm = 0.6 m<\/span><\/span><\/p>\n

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

Wanted <\/u>:<\/u> y ?<\/span><\/span><\/p>\n

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

The width of the slit is minimal compared to the distance between the slit and the screen so that the angle is minimal (the width of the slit in the figure above is enlarged). The angle is so small that the sin \u03b8 \u2248 tan \u03b8.<\/span><\/span><\/span><\/p>\n

sin \u03b8 \u2248 tan \u03b8 = y \/ l = y \/ 0.6<\/span><\/span><\/span><\/p>\n

Equation of d<\/span>iffraction by a single slit <\/span>(min<\/span>ima<\/span>) :<\/span><\/span><\/span><\/p>\n

d sin \u03b8 = n \u03bb<\/span><\/span><\/span><\/p>\n

(2 x 10-4<\/sup>)(y\/0,6) = (2)(5 x 10-7<\/sup>)<\/span><\/span><\/p>\n

(2 x 10-4<\/sup>) y = (0.6)(10 x 10-7<\/sup>)<\/span><\/span><\/p>\n

(2 x 10-4<\/sup>) y = 6 x 10-7<\/sup><\/span><\/span><\/p>\n

y = (6 x 10-7<\/sup>) \/ (2 x 10-4<\/sup>)<\/span><\/span><\/p>\n

y = 3 x 10-3<\/sup><\/span><\/span><\/p>\n

y = 0.003 m<\/span><\/span><\/span><\/p>\n

y = 3 m<\/span>m<\/span><\/span><\/span><\/p>\n

2. <\/span>Monochromatic light with wavelength of 5000 <\/span>\u00c5 (1 \u00c5 = 10<\/span>\u221210<\/sup><\/span> m) passes through the single slit, produces diffraction pattern the first maximum as shown in figure. Determine the wide of slit.<\/span><\/span><\/span><\/p>\n

\"Diffraction<\/p>\n

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

\u03bb = <\/span>5000 \u00c5 = 5000 x 10-10<\/sup> m = 5 x 10-7<\/sup> m <\/span><\/span><\/p>\n

sin 30o <\/sup>= 0,5<\/span><\/span><\/p>\n

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

Wanted :<\/u> wide of slit (d) ?<\/span><\/span><\/p>\n

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

d sin \u03b8 = n \u03bb<\/span> <\/span><\/span><\/p>\n

d (0.5) = (1)(5 x 10-7<\/sup>)<\/span><\/span><\/p>\n

d = (5 x 10-7<\/sup>) \/ (0.5) <\/span><\/span><\/p>\n

d = 10 x 10<\/span>-7 <\/sup><\/span>m<\/span><\/span><\/span><\/p>\n

d = 1 x 10<\/span>-6<\/sup><\/span> m<\/span><\/span><\/span><\/p>\n

d = 1 x 10-3<\/sup> mm<\/span><\/span><\/p>\n

d = 0.001 mm<\/span><\/span><\/p>\n

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Diffraction refers to the phenomenon in which waves spread out when they encounter an obstacle or pass through an aperture. When monochromatic light (light of a single wavelength) passes through a single slit, it doesn’t just travel in a straight line; instead, it spreads out and creates a diffraction pattern on a screen placed behind the slit.<\/p>\n

For a single slit, the primary feature of the diffraction pattern is a central bright maximum, flanked on both sides by a series of alternating dark and bright fringes (minima and maxima). Here’s how to understand and describe the diffraction pattern from a single slit:<\/p>\n

    \n
  1. Central Maximum<\/strong>: The central bright fringe is the most intense and broadest. The intensity decreases as one moves away from the central maximum.<\/li>\n
  2. Minima<\/strong>: The dark fringes or minima occur at angles \ufffd<\/span>\u03b8<\/span><\/span><\/span><\/span><\/span> such that: \ufffdsin\u2061(\ufffd)=\ufffd\ufffd<\/span>a<\/span>sin<\/span>(<\/span>\u03b8<\/span>)<\/span>=<\/span><\/span>m\u03bb<\/span><\/span><\/span><\/span><\/span> where:<\/li>\n<\/ol>\n
      \n
    • \ufffd<\/span>a<\/span><\/span><\/span><\/span><\/span> is the width of the slit.<\/li>\n
    • \ufffd<\/span>\u03bb<\/span><\/span><\/span><\/span><\/span> is the wavelength of the light.<\/li>\n
    • \ufffd<\/span>m<\/span><\/span><\/span><\/span><\/span> is an integer, excluding zero (i.e., \u00b11, \u00b12, \u00b13, …).<\/li>\n<\/ul>\n
        \n
      1. Maxima<\/strong>: Between these minima, there are secondary maxima, but they are less bright than the central maximum and decrease in intensity further away from the center.<\/li>\n
      2. Wide Slit vs. Narrow Slit<\/strong>: The width of the central maximum is inversely proportional to the slit width. That is, a narrower slit will produce a wider central maximum and vice versa.<\/li>\n
      3. Longer Wavelength vs. Shorter Wavelength<\/strong>: The angular positions of the minima and maxima depend on the wavelength. Longer wavelengths will produce more spread-out patterns compared to shorter wavelengths.<\/li>\n
      4. Comparison with Double Slit<\/strong>: A single-slit diffraction pattern is distinct from a double-slit interference pattern, though they are related phenomena. If you had a double slit, you’d see an interference pattern of multiple bright and dark fringes. However, if the slits are wide enough, each slit would also produce its diffraction pattern, leading to an “envelope” effect where the intensity of the interference fringes changes due to the single-slit diffraction.<\/li>\n<\/ol>\n

        The mathematical understanding of single-slit diffraction uses the Huygens principle, which states that every point on a wavefront can be thought of as a source of secondary spherical wavelets that spread out in the forward direction. By integrating the effect of all these wavelets, one can derive the diffraction pattern.<\/p>\n

        In practical applications and labs, observing single-slit diffraction patterns can be used to determine the wavelength of light or the size of the slit, given the other parameter.<\/p>\n

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

        Diffraction by a single slit \u2013 problems and solutions 1. Light with wavelength of 500 nm passes through a slit 0.2 mm wide. The diffraction pattern on a screen 60 cm away. Determine the distance between the central maximum and the second minimum. 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":"Diffraction by a single slit \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-3066","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\/3066","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=3066"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3066\/revisions"}],"predecessor-version":[{"id":8526,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3066\/revisions\/8526"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=3066"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=3066"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=3066"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}