Young's Double Slit Interference

Question

Monochromatic light of wavelength 600 nm illuminates two narrow slits separated by 0.2 mm, and the interference pattern is observed on a screen 1 m away. Calculate the width of one bright fringe in the pattern.

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In Young's double-slit experiment evenly spaced bright and dark fringes arise from the superposition of two coherent waves, and the fringe width is $\beta=\frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ the slit-to-screen distance and $d$ the slit separation. Converting to SI units, $\lambda=600	imes10^{-9}$ m, $D=1$ m and $d=0.2	imes10^{-3}$ m. Then $\beta=\frac{600	imes10^{-9}	imes1}{0.2	imes10^{-3}}=\frac{6	imes10^{-7}}{2	imes10^{-4}}=3	imes10^{-3}$ m $=3$ mm. The fringe width represents the constant spacing between successive maxima, arising because each bright fringe corresponds to a path difference that is a whole-number multiple of the wavelength. Since the path difference grows uniformly across the screen, the bright and dark bands stay equally spaced, the hallmark of two-source interference that distinguishes it from the unequal spacing seen in single-slit diffraction. The value 1.5 mm is wrong because it doubles the slit separation by mistake. The value 6 mm is wrong as it halves $d$ instead. The value 0.3 mm is wrong since it misplaces a power of ten in the unit conversion. This NCERT relation shows fringe width grows with wavelength and screen distance but shrinks with slit separation. A check confirms the millimetre-scale spacing is realistic for visible light and a 0.2 mm slit gap.

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