Question

In a double slit experiment light has a frequency of 6 $\times 10^{14} s ^{-1} .$ The distance between the centres of adjacent bright fringes is $0.75 mm$. What is the distance between the slits if the screen is $1.5 m$ away?

• A. $10^{-1} \,m$
• B. $10^{-3} \,m$
• C. $10^{-5}\, m$
• D. $1 \,m$

Answer 1

Answer: (B)

$v=6 \times 10^{14} \,s ^{-1}$
$\lambda=c / v=\frac{3 \times 10^{8}}{6 \times 10^{14}}=5 \times 10^{-7} m$
$\beta=0.75 mm =0.75 \times 10^{-3} m$
$D =1.5 m$
$\beta=\frac{\lambda D }{ d }=\frac{5 \times 10^{-7} \times 1.5}{0.75 \times 10^{-3}}=10^{-3} m$
$\beta=10^{-3} \,m$