Question

In an interference experiment, $3^{rd}$ bright fringe is obtained on the screen with a light of $700\text{nm}$. What should be the wavelength of light source in order to obtain $5^{th}$ bright fringe at the same point -

• A. $420\text{nm}$
• B. $500\text{nm}$
• C. $750\text{nm}$
• D. $630\text{nm}$

Answer 1

The correct option is A $420\text{nm}$

Given, ${\lambda }_1=700\text{nm}$
As we known that $y=\frac{\Delta xD}{d}$
For the bright fringe, $\Delta x=n\lambda$
Here, $\Delta x_1=\Delta x_2$
$n_1{\lambda }_1=n_2{\lambda }_2$
$\frac{3D{\lambda }_1}{d}=\frac{5D{\lambda }_2}{d}$
$3{\lambda }_1=5{\lambda }_2$
$\therefore {\lambda }_2=\frac{3{\lambda }_1}{5}=\frac{3\times 700}{5}=420\text{nm}$

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