Question

Sodium light has two wavelengths 589nm and 589.6nm as the path difference increases when the visibility of fringes minimum

Answer 1

Answer:

0.289 mm ≈ 0.29mm OR(to be exact) 289395.333* 10^(-6)m

Explanation:

First let derive this formula

P/λ1 - P/λ2 = 1/2

Phase difference = xd/D     ,where d=distance between slits & D= distance between slit and screen & x= distance of maximum/minimum from the centre of the screen.

• Pb=Phase difference of nth bright fringe=nλ
• Pd=Phase difference of nth dark fringe=(2n-1)λ/2
• so to obtain a fringe of minimum visibility, if the light of both wavelength form maxima's and minima's at half the distance from each other then the area will have minimum visibility.
• Understand it like this, first imagine the double slit experiment with  minima's and maxima's right next to each other & now imagine first a bright fringe of fringe width='x' and now the first dark fringe starts from 'x/2' of first bright fringe and now the the second bright fringe starts from'x/2' of first dark fringe. This area will have minimum visibility thus.
• n=Pb/λ1       &        (2n-1)/2 = (n-1/2) = Pd/λ2
• subtract both the equation to get a phase difference upon wavelength difference equal to half the distance from both fringes i.e.
• (n)-(n-1/2) = Pb/λ1 - Pd/λ2
• as we consider Pb≈Pd≈very small difference between maxima's and minima's so
• 1/2 = p(1/λ1 - 1/λ2)  Put values λ1=589nm and λ2 = 589.6nm
• Phase difference = 0.289mm ≈ 0.29mm

Hope it helps (plz mark as brainliest if it helps)