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In a Young's double slit experiment, the slits are separated by $d = 0.5\,\text{mm}$ and the screen is $D = 1.0\,\text{m}$ away. With light of wavelength $600\,\text{nm}$, the fringe width on the screen is:
A$0.3\,\text{mm}$
B$0.6\,\text{mm}$
C$1.2\,\text{mm}$
D$3.0\,\text{mm}$
Answer & Solution
Correct answer: C. $1.2\,\text{mm}$
Fringe width $\beta = \dfrac{\lambda D}{d} = \dfrac{600 \times 10^{-9} \times 1.0}{0.5 \times 10^{-3}} = 1.2 \times 10^{-3}\,\text{m} = 1.2\,\text{mm}$.
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