Browse Questions

# Find the width of central maxima given $\lambda = 5 \times 10^{-5}\;cm$ slit separation $d= 0.15 \;cm$ and distance between screen and slit $D= 2.1 \;m$

$(a)\;0.70\;mm \\ (b)\;1.4 \;mm \\ (c)\;14\;mm \\ (d)\;7\;mm$

Width of central maxima is equal to $2 \times$ fringe width.
$\qquad= 2 \times \beta$
$\qquad= 2 \large\frac{\lambda D}{d}$
$\qquad= \large\frac{2 \times 2.1 \times 5 \times 10^{-7}}{0.15 \times 10^{-2}}$
$\qquad= 1.4 \times 10^{-3} \;m$
$\qquad= 1.4 mm$
Hence b is the correct answer.