A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the center of the poi

Question

A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the center of the point source the z axis. When a sheet of cardboard in which a square hole 0.020 m on a side has been cut is placed between the point source and the screen, 0.50 m from the point source with the hole centered on the z axis, a bright square shows up on the screen. If, instead, a second sheet of cardboard with a similar square hole is placed between the point source and screen, 0.25 m from the point source with the hole centered on the z axis, the bright square it casts on the screen is identical to the bright square from the first sheet. What is the length of the side of the hole in this sheet?

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Maris 3 days 2021-07-20T04:14:56+00:00 1 Answers 2 views 0

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    2021-07-20T04:16:10+00:00

    Answer:

    The length of the side of the hole in the second cardboard sheet is L_2 = 0.01m

    Explanation:

    From the question we are told that

         The distance of the point source  from the screen is  d = 1.0 m

          The length of a side of the  first square hole is  L_1 = 0.020 \ m

          The distance of the cardboard from the point source is D_1 = 0.50\  m

       The distance of the second cardboard from the point source is D_2 = 0.25 \ m

       

    Let take the  \alpha_{max } as  the angle at which the light is passing through the edges of the cardboards square hole

         Since the bright square casted on the screen by both  square holes on the   individual cardboards are then it means that

                  \alpha_{max} __{1}} = \alpha_{max} __{2}}

    This implies that

                 tan (\alpha_{max} __{1}}) = tan (\alpha_{max} __{2}})      

    Looking at this from the SOHCAHTOA concept

                   tan (\alpha_{max} __{1}}) =  \frac{opposite}{Adjacent}

         Here opposite is  the length of the side of the  first cardboard square hole

         and    

          Adjacent is  the  distance of the from the  first cardboard square hole to the point source

    And for  

                tan (\alpha_{max} __{2}}) =  \frac{opposite}{Adjacent}

        Here opposite is  the length of the side of the  second  cardboards square hole (let denote it with L_2)

    and

    Adjacent is the distance of the from the  second  cardboards square hole to the point source

             So

                     tan (\alpha_{max} __{1}}) =  \frac{0.020}{0.50}

             And  

                    tan (\alpha_{max} __{2}}) =  \frac{L_2}{0.25}

    Substituting this into the above equation

                     \frac{0.020}{0.50}  =   \frac{L_2}{0.25}

    Making L_2 the subject

                       L_2 = \frac{0.25 *0.020}{0.50}

                     L_2 = 0.01m

    Since it is a square hole the sides are the same hence

    The length of the side of the hole in the second cardboard sheet is L_2 = 0.01m

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