Optics 6 Question 58
60. In Young’s double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the 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 monochromatic light used in the experiment.
(1983, 6M)
Show Answer
Solution:
- Shifting of fringes due to introduction of slab in the path of one of the slits, comes out to be,
$$ \Delta y=\frac{(\mu-1) t D}{d} $$
Now, the distance between the screen and slits is doubled.
Hence, the new fringe width will become.
$$ \begin{aligned} & \omega^{\prime}=\frac{\lambda(2 D)}{d} \\ & \text { Given, } \Delta y=\omega^{\prime} \text { or } \frac{(\mu-1) t D}{d}=\frac{2 \lambda D}{d} \\ & \therefore \quad \lambda=\frac{(\mu-1) t}{2}=\frac{(1.6-1)\left(1.964 \times 10^{-6}\right)}{2} \\ & =0.5892 \times 10^{-6} m=5892 \AA \end{aligned} $$
