In a Young’s double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits
Question:

In a Young’s double slit experiment, the slits are separated by $0.3 \mathrm{~mm}$ and the screen is $1.5 \mathrm{~m}$ away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is $2.4 \mathrm{~cm}$. The frequency of light used

is__________ $\times 10^{14} \mathrm{~Hz}$

Solution:

$8 \beta=2.4 \mathrm{~cm}$

$\frac{8 \lambda \Delta}{\mathrm{d}}=2.4 \mathrm{~cm}$

$\frac{8 \times 1.5 \times c}{0.3 \times 10^{-3} \times f}=2.4 \times 10^{-2}$

$f=5 \times 10^{14} \mathrm{~Hz}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.