Question

In Young's double-slit experiment, using monochromatic light of wavelength $\lambda$, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index $1.6$ and thickness $2.0\mu \text{m}$ is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the plane of slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the light.

• A. $4500\mathring{\text{A}}$
• B. $5700\mathring{\text{A}}$
• C. $6000\mathring{\text{A}}$
• D. $4000\mathring{\text{A}}$

Answer 1

The correct option is C $6000\mathring{\text{A}}$

When mica sheet of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the interfering beams, optical path increases by $(\mu -1)t$. Hence, the shift on the screen,
$y_0=\frac{D}{d}(\mu -1)t$
When the distance between the plane of slits and screen is changed from $D$ to $2D$, then the distance between successive maxima is,
$\beta =\frac{2\lambda D}{d}$

According to the given information,

$\frac{D}{d}(\mu -1)t=\frac{2\lambda D}{d}$

$\Rightarrow \lambda =\frac{(\mu -1)t}{2}$

$=\frac{(1.6-1)}{2}\times 2.0\times {10}^{-6}=6000\mathring{\text{A}}$

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