Question

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

• A. $600 \,\mathring{A}$
• B. $6000\,\mathring{A}$
• C. $60 \,\mathring{A}$
• D. $660\,\mathring{A}$

Answer 1

Answer: (B)

Shift of fringes is given by $\frac{(\mu-1) t D}{d}$
This is equal to position of $10^{\text {th }}$ bright fringe $\frac{10 \lambda D}{d}$
$ \frac{0.6 \times 1 \times 10^{-5} D}{d}=\frac{10 \lambda D}{d} $
$ \lambda=0.6 \times 10^{-6} $
$\therefore \lambda=6 \times 10^{-7} 4$
or $ \lambda=6000 \,\mathring{A}$