Question

$\text{YDSE}$ is carried with two thin sheets of thickness $10.4\mu \text{m}$ each and refractive index ${\mu }_1=1.52$ and $\mu =1.40$ covering the slits $S_1$ and $S_2$ respectively. If white light of range $400\text{nm}$ to $780\text{nm}$ is used, then which wavelength will form maxima exactly at point $O$, the centre of the screen

• A. $416\text{nm}$ only
• B. $624\text{nm}$ only
• C. $416\text{nm}$ and $624\text{nm}$ only
• D. $400\text{nm}$ and $316\text{nm}$ only

Answer 1

The correct option is C $416\text{nm}$ and $624\text{nm}$ only

The path differences between interfering waves at $O$ will be,

$\Delta x=[S_{1}O+({\mu }_1-1\left)t\right]-[S_{2}O+({\mu }_2-1\left)t\right]$

Here, ${\mu }_1>{\mu }_2$, therefore the optical path of wave from slit $S_2$ will be higher compared to that from slit $S_2$.

From geometry $S_{1}O=S_{2}O$

$\Rightarrow \Delta x={\mu }_{1}t-{\mu }_{2}t=({\mu }_1-{\mu }_2)t$

$=(1.52-1.40)\times 10.4\mu \text{m}$

$=0.12\times 10400\text{nm}=1248\text{nm}$

In order to obtain maxima at centre $O$ of screen, the path difference,
$\Delta x=1248=n\lambda [n=-1,2,3,...]$

For $n=2$ we get $\lambda =624\text{nm}$

For $n=3$ we get $\lambda =416\text{nm}$

Hence, $\left(C\right)$ is the correct answer.