{"id":15671,"date":"2025-06-10T10:54:47","date_gmt":"2025-06-10T10:54:47","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=15671"},"modified":"2025-06-10T10:54:49","modified_gmt":"2025-06-10T10:54:49","slug":"when-an-electromagnebc-wave-falls-on-a-white-perfectly-reflecbng-surface-it-exerts-a-force-f-on-that-surface","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/when-an-electromagnebc-wave-falls-on-a-white-perfectly-reflecbng-surface-it-exerts-a-force-f-on-that-surface\/","title":{"rendered":"When an electromagneBc wave falls on a white, perfectly reflecBng surface, it exerts a force F on that surface"},"content":{"rendered":"\n

When an electromagneBc wave falls on a white, perfectly reflecBng surface, it exerts a force F on that surface. If the surface is now painted a perfectly absorbing black, what will be the force that the same wave will exert on the surface? A) 4F B) 2F C) F D) F\/2 E) F\/4<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

The correct answer is B) 2F<\/strong>.<\/p>\n\n\n\n

Explanation:<\/strong><\/p>\n\n\n\n

When an electromagnetic wave, such as light, strikes a surface, it exerts a force due to the transfer of momentum. The magnitude of this force depends on how the wave interacts with the surface\u2014whether it is reflected, absorbed, or transmitted.<\/p>\n\n\n\n

    \n
  1. Perfect Reflection (White Surface):<\/strong>
    For a perfectly reflecting surface, the wave bounces back with the same amplitude but opposite direction. The change in momentum per photon (or per wave packet) doubles because the wave’s momentum reverses direction. Since force relates to the rate of change of momentum, the force exerted is twice what it would be if the wave were simply incident without reflection.
    Therefore,<\/strong> for a perfect reflector:
    [ F_{reflection} = 2 \\times \\frac{Power}{c} ]
    where ( c ) is the speed of light, and the power of the wave remains constant.<\/li>\n\n\n\n
  2. Perfect Absorption (Black Surface):<\/strong>
    If the same wave hits a perfectly absorbing surface, it transfers all of its momentum to the surface; the wave’s momentum goes to zero after absorption. The change in momentum per photon is equal to the initial momentum of the photon (since it goes from ( p ) to zero).
    Thus,<\/strong> for perfect absorption:
    [ F_{absorption} = \\frac{Power}{c} ]<\/li>\n\n\n\n
  3. Comparison of Forces:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Reflection (white surface): Force = ( 2 \\times \\frac{Power}{c} ) = ( 2F )<\/li>\n\n\n\n
    • Absorption (black surface): Force = ( \\frac{Power}{c} ) = ( F )<\/li>\n<\/ul>\n\n\n\n

      However, in the problem, the initial force ( F ) was given for the perfectly reflecting surface, which corresponds to ( 2 \\times \\frac{Power}{c} ). The force exerted on the black surface (absorption) will be half<\/strong> of that, because it only involves the transfer of the wave’s initial momentum, not reversed momentum.<\/p>\n\n\n\n

      Correction:<\/strong><\/p>\n\n\n\n

      Actually, the initial force ( F ) on the reflecting surface is ( 2 \\times \\frac{Power}{c} ). The force on the absorbing surface, which only absorbs the momentum, is ( \\frac{Power}{c} ), which is half<\/strong> of the force on the reflecting surface.<\/p>\n\n\n\n

      Final answer:<\/strong><\/p>\n\n\n\n

      Since the initial force on the white surface (reflection) is ( F ), the force on the black surface (absorption) will be half of that<\/strong>, i.e., ( F\/2 )<\/strong>.<\/p>\n\n\n\n

      But wait,<\/strong> the options are:
      A) 4F
      B) 2F
      C) F
      D) F\/2
      E) F\/4<\/p>\n\n\n\n

      Given that the initial force ( F ) was for the perfect reflector, the force for perfect absorption would be ( F\/2 )<\/strong>, which matches option D.<\/p>\n\n\n\n

      Therefore, the correct answer is D) F\/2.<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Summary:<\/strong><\/p>\n\n\n\n

        \n
      • Reflection doubles the momentum transfer, exerting force ( F ).<\/li>\n\n\n\n
      • Absorption transfers half that momentum per photon, exerting force ( F\/2 ).<\/li>\n\n\n\n
      • Hence, when switching from reflection to absorption, the force halves from ( F ) to ( F\/2 ).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

        When an electromagneBc wave falls on a white, perfectly reflecBng surface, it exerts a force F on that surface. If the surface is now painted a perfectly absorbing black, what will be the force that the same wave will exert on the surface? A) 4F B) 2F C) F D) F\/2 E) F\/4 The correct […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-15671","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15671","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=15671"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15671\/revisions"}],"predecessor-version":[{"id":15673,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15671\/revisions\/15673"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=15671"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=15671"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=15671"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}