Question

Interference fringes are observed on a screen by illuminating two thin slits $1\text{mm}$ apart with a light source $(\lambda =632.8\text{nm})$. The distance between the screen and the slits is $100\text{cm}$. If a bright fringe is observed on a screen at a distance of $1.27\text{mm}$ from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits, is close to :

• A. $1.27\mu \text{m}$
• B. $2.87\mu \text{m}$
• C. $2\mu \text{m}$
• D. $2.05\mu \text{m}$

Answer 1

The correct option is A $1.27\mu \text{m}$


$\text{Path difference,}\Delta x=d\sin \theta =d\theta$

$\text{Also},\theta =\frac{y}{D}$

Given that,
$d=1\text{mm}={10}^{–3}\text{m}$
$D=100\text{cm}=1\text{m}$
$y=1.27\text{mm}=1.27\times {10}^{-3}\text{m}$

$\therefore \Delta x=\frac{dy}{D}$

$\Rightarrow \Delta x=\frac{{10}^{-3}\times 1.27\times {10}^{-3}}{1}$

$\Rightarrow \Delta x=1.27\mu \text{m}$.

Hence, option $\left(A\right)$ is correct.