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Physics Question on Youngs double slit experiment

In Young's double slit experiment, monochromatic light of wavelength 5000 , \text{Å} is used. The slits are 1.0 mm apart and the screen is placed 1.0 m away from the slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ×10−6 m.\\_ \times 10^{-6} \, \text{m}.

Answer

Given: - Wavelength of light: λ=5000 A˚=5000×10−10 m\lambda = 5000 \, \text{\AA} = 5000 \times 10^{-10} \, \text{m} - Distance between slits: d=1.0 mm=1.0×10−3 md = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} - Distance between the slits and the screen: D=1.0 mD = 1.0 \, \text{m}

Step 1: Calculating the Fringe Width

The fringe width β\beta in Young’s double-slit experiment is given by:

β=λDd\beta = \frac{\lambda D}{d}

Substituting the given values:

β=5000×10−10×1.01.0×10−3 m\beta = \frac{5000 \times 10^{-10} \times 1.0}{1.0 \times 10^{-3}} \, \text{m} β=5×10−3 m=5 mm\beta = 5 \times 10^{-3} \, \text{m} = 5 \, \text{mm}

Step 2: Finding the Distance Where Intensity is Half of Maximum

The intensity becomes half of the maximum intensity at the position of the first secondary maximum. The position yy where this occurs is given by:

y=β4y = \frac{\beta}{4}

Substituting the value of β\beta:

y=5×10−34 my = \frac{5 \times 10^{-3}}{4} \, \text{m} y=1.25×10−3 m=125×10−6 my = 1.25 \times 10^{-3} \, \text{m} = 125 \times 10^{-6} \, \text{m}

Conclusion:

The distance from the centre of the screen where the intensity becomes half of the maximum intensity for the first time is 125×10−6 m125 \times 10^{-6} \, \text{m}.