Answer:
The phase difference is [tex]\Delta \phi = 180^o[/tex]
Explanation:
From the question we are told that
The distance between the slits is [tex]d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m[/tex]
The distance to the screen is [tex]D = 100 cm = \frac{100}{100} = 1 \ m[/tex]
The wavelength is [tex]\lambda = 400nm[/tex]
The distance of the wave from the central maximum is [tex]L = 5mm = 5*10^{-3} m[/tex]
Generally the path difference of this waves is mathematically represented as
[tex]y = d sin \theta[/tex]
Here [tex]\theta[/tex] 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
[tex]tan \theta = \frac{L}{D}[/tex]
=> [tex]\theta = tan ^{-1} \frac{L}{D}[/tex]
[tex]\theta = tan ^{-1} [\frac{5*10^{-3}}{1}][/tex]
[tex]\theta =0.2865[/tex]
Substituting values into the formula for path difference
[tex]y = 0.2 *10^{-3} sin(0.2864)[/tex]
[tex]y = 9.997*10^{-7} \ m[/tex]
The phase difference is mathematically represented as
[tex]\Delta \phi = \frac{2 \pi }{\lambda } * y[/tex]
Substituting values
[tex]\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}[/tex]
[tex]\Delta \phi =5 \pi[/tex]
Converting to degree
[tex]\Delta \phi =5 \pi radians = 5 (180^o) = 180^o[/tex]
the solution is subtracted by 360° in order to get the actual angle
