Question

Young’s double slit experiment is carried with two thin sheets of thickness 10.4 $\mu m$ 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 400 nm to 780 nm is used then which wavelength will form maxima exactly at point O, the centre of the screen?

• A. 416 nm only
• B. 624 nm only
• C. 416 nm and 624 nm only
• D. 408 nm only

Answer 1

The correct option is C 416 nm and 624 nm only

Total shift = $t({\mu }_1-{\mu }_2)\frac{D}{d}=10.4\times {10}^{-6}\left(0.12\right)\frac{D}{d}$
$=1.248\times {10}^{-6}\left(\frac{D}{d}\right)$
Now for maxima to be at centre, this shift has to be $n\beta$
$\Rightarrow \frac{n\lambda D}{d}=1.248\times {10}^{-6}\frac{D}{d}$
$\Rightarrow \lambda =\frac{1.248\times {10}^6}{n}=\frac{1248}{n}nm$
So, $\lambda$ = 624 nm, 416 nm
So, option(C)