Question

When a mica sheet of thickness 7 microns and $\mu =1.6$ is placed in the path of one of interfering beams in the $YDSE$, then the central fringe gets at the position of the seventh bright fringe. What is the wavelength of light used ?

• A. $6000\stackrel{\circ }{A}$
• B. $3000\stackrel{\circ }{A}$
• C. $4000\stackrel{\circ }{A}$
• D. $2000\stackrel{\circ }{A}$

Answer 1

The correct option is A $6000\stackrel{\circ }{A}$


Optical path length of light in the glass slab $=t\mu$
Thus, extra path length introduced in the light coming from $S_1$ $=t\mu -t=t(\mu -1)$



If CBF is formed at P, then extra optical path length for light from $S_1$ due to glass slab should match the excess path length of light from $S_2$.
i.e $t(\mu -1)=n\lambda$
$\Rightarrow \lambda =\frac{t(\mu -1)}{n}$
Since P is the 7th bright fringe location (before), $n=7$
$\therefore \lambda =\frac{7\times {10}^{-6}(1.6-1)}{7}=6\times {10}^{-7}m=6000\stackrel{\circ }{A}$