In a Young's double slit exp. light of 500nm ia used to produce an interference pattern. When the distance b/w the slits is 0.05mm, the angular width(in degree) of the fringes formed on the distance screen close to :   Option: 1 0.17 Option: 2 0.57 Option: 3 1.7   Option: 4 0.07

$\\ given\ \lambda=500nm\\ d=0.05 mm\\ \theta(in \ radian)=\frac{\lambda}{d}=\frac{500 \times 10^{-9}}{0.05 \times 10^-3}=0.01\\ Now, \ \theta(in degree)=0.01 \times \frac{180}{\pi}=0.57^o$

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