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puteri [66]
3 years ago
8

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

e difference (in degrees) between the waves from the two slits arriving at a point 5 mm from the central maximum when the wavelength is 400 nm? (Convert your result so the angle is between 0 and 360°.)
Physics
1 answer:
Anni [7]3 years ago
4 0

Answer:

The phase difference is  \Delta \phi  = 180^o

Explanation:

From the question we are told that

     The distance between the slits is d = 0.2 \ mm = \frac{0.2}{1000}  = 0.2 *10^{-3} \ m

     The distance to the screen is D = 100 cm = \frac{100}{100} = 1 \ m

      The wavelength is  \lambda = 400nm

 The distance of the wave from the  central maximum is L =  5mm = 5*10^{-3} m

   

Generally the path difference of this  waves is mathematically represented as

              y = d sin \theta

Here \theta is the angle between the the line connecting the mid-point of the slits with  the screen and the line  connecting the mid-point of the slits to the central maximum

  This implies that

              tan \theta  = \frac{L}{D}

     =>     \theta = tan ^{-1} \frac{L}{D}

             \theta = tan ^{-1} [\frac{5*10^{-3}}{1}]

           \theta =0.2865

Substituting values into the formula for path difference

       y = 0.2 *10^{-3} sin(0.2864)  

       y = 9.997*10^{-7} \ m  

The phase difference is mathematically represented as

          \Delta \phi = \frac{2 \pi }{\lambda }  * y

Substituting values        

         \Delta \phi = \frac{2 \pi }{400 *10^{-9} }  \ * 9.997*10^{-7}

         \Delta \phi =5 \pi

Converting to degree

         \Delta \phi =5 \pi radians = 5 (180^o)  =  180^o  

the solution is subtracted by 360° in order to get the actual angle

 

             

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