Question

In a young's double slit experiment, the slits are separated by $0.3\text{mm}$ and the screen is $1.5\text{m}$ away from the plane of slits. Distance between fourth bright fringe on both sides of central bright fringe is $2.4\text{cm}$. The frequency of light used is
$\times {10}^{14}\text{Hz}$

• A. 5.0
• B. 5
• C. 5.00

Answer 1

Answer: (A), (B), (C)

The position of fourth bright fringe is given as $x_4=\frac{4\lambda D}{d}$
Distance between fourth bright fringe on other both sides of central bright fringe is $x=\frac{8\lambda D}{d}$
Given that,
$x=2.4\text{cm}$
$D=1.5\text{m}$
$d=0.3\text{mm}$
On substitution, we get

$2.4\times {10}^{-2}=\frac{8\lambda \left(1.5\right)}{0.3\times {10}^{-3}}$
$\lambda =6\times {10}^{-7}\text{m}$

Frequency of light is given by $\nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{6\times {10}^{-7}}=5\times {10}^{14}\text{Hz}$

Ans: $5,5.0,5.00$