Wave Optics

JEE Main level, single correct, medium

89 of 191

Question 89 of 191

In a Young's double slit experiment two slits are separated by 2mm2mm and the screen is placed one meter away. When a light of wavelength 500nm500nm is used, the fringe separation will be :

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Answer: (D) 0.25mm0.25mm

β=λDd=500×10−9×12×10−3\beta =\dfrac{\lambda D}{d}=\dfrac{500\times {10}^{-9}\times 1}{2\times {10}^{-3}}
β=52×10−4m=2.5×10−1mm\beta =\dfrac{5}{2}\times {10}^{-4}m=2.5\times {10}^{-1}mm
b=0.25mmb=0.25mm

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