Question

Young's double-slit experiment is carried out with two thin sheets of thickness $10.4\mu \text{m}$ each and refractive index ${\mu }_1=1.52$ and ${\mu }_2=1.40$, covering the slits $S_1$ and $S_2$ respectively. If white light of range of wavelength $400\text{nm}$ to $780\text{nm}$ is used, which wavelength will form maxima exactly at point $O$, the center of the screen?

• A. $12480\mathring{\text{A}}$ and $3120\mathring{\text{A}}$
• B. $4160\mathring{\text{A}}$ and $3120\mathring{\text{A}}$
• C. $6240\mathring{\text{A}}$ and $4160\mathring{\text{A}}$
• D. $12480\mathring{\text{A}}$ and $4160\mathring{\text{A}}$

Answer 1

The correct option is C $6240\mathring{\text{A}}$ and $4160\mathring{\text{A}}$

Path difference $=({\mu }_2-{\mu }_1)t=(1.52-1.40)\times 10.4\times {10}^{-6}$
$=1.248\times {10}^{-6}\text{m}=12480\mathring{\text{A}}$

Now, for maxima, the path difference should be integer multiple of $\lambda$

i.e. $n\lambda =12480\mathring{\text{A}}$

${\lambda }_1=12480\mathring{\text{A}}\to n=1$
${\lambda }_2=6240\mathring{\text{A}}\to n=2$
${\lambda }_3=4160\mathring{\text{A}}\to n=3$
${\lambda }_4=3120\mathring{\text{A}}\to n=4$

Therefore, only $6240\mathring{\text{A}}$ and $4160\mathring{\text{A}}$ will form a maxima at point $O$.

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