Question

In a $YDSE$ experiment using monochromatic light, a mica sheet of refractive index $1.6$ and thickness $1.964\text{micron}$ is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between slits and screen is doubled. It is found that the distance between successive maxima or minima now is the same as the observed fringe shift on introduction of mica sheet. The wavelength of light used is

• A. $5892\dot{\text{A}}$
• B. $6000\dot{\text{A}}$
• C. $6009\dot{\text{A}}$
• D. $4200\dot{\text{A}}$

Answer 1

The correct option is A $5892\dot{\text{A}}$

Shift in the fringe system on introduction of mica sheet,

$\Delta y=\frac{(\mu -1)tD}{d}=\frac{(\mu -1)t\beta }{\lambda }[\because \beta =\frac{\lambda D}{d}]$

When the distance between the slit and screen is doubled, the new fringe width is,

${\beta }^{{}^{\prime }}=\frac{\lambda \left(2D\right)}{d}=2\beta$

According to question:
$\Delta y={\beta }^{{}^{\prime }}$

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

$\Rightarrow \lambda =\frac{(\mu -1)t}{2}=\frac{(1.6-1)\times (1.964\times {10}^{-6})}{2}$

$=0.5892\times {10}^{-6}\text{m}$

$=5892\times {10}^{-10}\text{m}=5892\dot{\text{A}}$

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