Question

In a YDSE, the screen is rotated by ${60}^{\circ }$ about an axis parallel to the slits. The slit's separation is $3\text{mm}$, slit to screen distance (at central fridge) is $4\text{m}$ and wavelength of light is $450\text{nm}$. The separation between the $3^{rd}$ dark fringe on the both side of the central fringe is

• A. $6\text{mm}$
• B. $8\text{mm}$
• C. $4\sqrt{3}\text{mm}$
• D. $2\sqrt{3}\text{mm}$

Answer 1

The correct option is A $6\text{mm}$

Given, $d=3\text{mm};D=4\text{m}\lambda =450\text{nm};\theta ={60}^{\circ }$


In YDSE, $n^{th}$ position of dark fringe is given by,

$y_n=\frac{(2n-1)}{2}\frac{\lambda D}{d}$

For $3rd$ dark fringe,

$y_3=\frac{5}{2}\frac{\lambda D}{d}=\frac{5}{2}\times \frac{450\times {10}^{-9}\times 4}{3\times {10}^{-3}}=1.5\text{mm}$

From the diagram,

$y^{\prime }=\frac{y}{\cos \theta }=\frac{y}{\cos {60}^{\circ }}=2y$

$\therefore y^{\prime }=2\times 1.5=3\text{mm}$

The total distance between the $3^{3rd}$ dark fridge on both sides is,

$d^{\prime }=2y^{\prime }=6\text{mm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.