Question

In Young's double-slit experiment with slit separation $0.1\text{m}$, one observes a bright fringe at angle $1/40\text{rad}$ by using the light of wavelength ${\lambda }_1$. When the light of wavelength ${\lambda }_2$ is used a bright fringe is seen at the same angle in the same setup. Given that ${\lambda }_1$ and ${\lambda }_2$ are in the visible range $(380\text{nm}$ to $740\text{nm})$, their values are

• A. $400\text{nm},500\text{nm}$
• B. $625\text{nm},500\text{nm}$
• C. $380\text{nm},525\text{nm}$
• D. $380\text{nm},500\text{nm}$

Answer 1

The correct option is B $625\text{nm},500\text{nm}$

Given:-
$d=0.1\text{mm};\theta =1/40\text{rad}$

Path difference between the waves at $\theta =1/40\text{rad}$ is

$\Delta x=d\sin \theta$

$\Delta x=d\theta$

Here $\theta$ is small, so $\sin \theta \approx \theta$

$\Delta x=0.1\times {10}^{-3}\times \frac{1}{40}=2.5\times {10}^{-6}\text{m}$

For bright fringe, $\Delta x=n\lambda$

$\Rightarrow \lambda =\frac{\Delta x}{n}$

$\lambda =\frac{2.5\times {10}^{-6}}{n}\text{meter}=\left(\frac{2500}{n}\right)\text{nm}$

For $\lambda$ to lie in the visible range, $\left(380\text{nm}\text{to}740\text{nm}\right)$ the value of $n$ can be $4\text{and}5$.

$\therefore {\lambda }_1=\frac{2500}{4}=625\text{nm}$

and ${\lambda }_2=\frac{2500}{5}=500\text{nm}$

Hence, option $\left(\text{B}\right)$ is the correct answer.

Note :- The wavelengths of light $625\text{nm}$ and $500\text{nm}$, will form a bright fringe which will overlap with each other. But the difference between them is that for the wavelength of light i.e. $625\text{nm}$ it will be its $4^{th}$ bright fringe. And for the wavelength of light i.e. $500\text{nm}$, it will be it's $5^{th}$ bright fringe.