Question
In Young's double slit experiment, the distance between sources is $1 mm$ and distance between the screen and source is $1 m$. If the fringe width on the screen is $0.06cm$, then $\lambda$ =........$\mathop A\limits^o $
✓
Answer
a
(a)$\beta = \frac{{\lambda \,D}}{d} \Rightarrow (0.06 \times {10^{ - 2}}) = \frac{{\lambda \times 1}}{{1 \times {{10}^{ - 3}}}}$ $ \Rightarrow \lambda = 6000\,Å$
(a)$\beta = \frac{{\lambda \,D}}{d} \Rightarrow (0.06 \times {10^{ - 2}}) = \frac{{\lambda \times 1}}{{1 \times {{10}^{ - 3}}}}$ $ \Rightarrow \lambda = 6000\,Å$
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