Question

In a $YDSE$, a thin film $(\mu =1.6)$ of thickness $0.01\text{mm}$ is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the ${10}^{th}$ bright fringe earlier. The wave length of the wave is

• A. $600\dot{\text{A}}$
• B. $6000\dot{\text{A}}$
• C. $60\dot{\text{A}}$
• D. $660\dot{\text{A}}$

Answer 1

The correct option is B $6000\dot{\text{A}}$


Shift of fringes due to introduction of glass slab is,
$\Delta y=\frac{(\mu -1)tD}{d}$

Initially the central maxima was at $y=0$ and due to shift its new position will be,

$y=\frac{(\mu -1)tD}{d}$

According to question,

$\Delta y=y_{10}$

$\Rightarrow \frac{(\mu -1)tD}{d}=\frac{10\lambda D}{d}$

Here, $y_{10}$ is the position of ${10}^{th}$ bright fringe when slab was not present.

$\Rightarrow (\mu -1)t=10\lambda$

$\Rightarrow \lambda =\frac{(1.6-1)(0.01\times {10}^{-3})}{10}=0.6\times {10}^{-6}$

$\Rightarrow \lambda =6000\times {10}^{-10}\text{m}=6000\dot{A}$

Hence, option $\left(B\right)$ is correct.