a. A beam of red light (? = 660 nm) b. A sound wave playing an \"A\" (f = 440 Hz

Question

a. A beam of red light (? = 660 nm)

b. A sound wave playing an "A" (f = 440 Hz).

Treat these two waves as plane waves passing through a slit whose width equals the width of the door. Find the angle that gives the position of the first dark diffraction fringe. From that, assuming you are 2 m back from the door, estimate how far outside the door you could be and still detect the wave. (See the picture for a clarification. The distance x is desired.)

Note: The speed of light in air is about 3 x 108 m/s and the speed of sound in air is about 330 m/s.

Please offer a lengthy explanation. Thank You

Explanation / Answer

             When a light and sound waves are passing through a doorway, whose width should be on the order of thier wavelenth.

               for light, wavelength Ll= 660 nm,         velocity Cl = 3 x 108 = 324 m/s

               for sound, frequency f = 440 Hz             velocity Cs = 330 m/s

                               wavelenth Ll = Cs/ f   = 330/ 440    = 3/4    

                                                Ll= 0.75m

             distance between doorway and observer D= 2 m,            width = a

             The doorway is acts as a slit and it is wider. When light and sound waves are passing through the wider doorway, sound wave gets diffracted because its wavelength is larger and nearly equals to the width of doorway. But wavelenth of light is very smaller than the width of the doorway, so light wave can travel without any diffraction. This is the reason for that we can hear sound arround corners but can't see arround corners.

The position of first order fringe for sound

                          y = m Ls D/a

                                = 1 x 0.75 x 2/a

                          y = 1.5/a

       angle makes by first order fringe

                                theeta = y/D

                                           = 1.5 x 2/a

                             theta = 3/a

      i) here the positon of fringe y = 1.5/a

                     therefore   y is indirectly proportional to the width a.

           if width increases, the position of the fringes should be near to its centre.

             width decreases, the position should be far away from centre.

ii)         tan theeta = 3/a , when theeta is 45 degree, value should be maximum 1.

                                      tan 45 = 3/a                   1= 3/a

                                            a = 3 m

      When width of the door is 3m, sound waves could be detected upto 45 degree

                         y= 1.5/a = 1.5/ 3

                          y = 0.5 m

we can detect sound upto 0.5 m

Here only sound wave gets diffracted but light can't be diffracted.

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