Question

Monochromatic light of wavelength 600 nm is used in a Young's double slit experiment. One of the slits is covered by a transparent sheet of thickness 1.8 x ${10}^{-5}$m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet then?

Answer 1

Consider two points $P_1$ and $P_2$ on the screen at $y_1=0.2mm$ and $y_2=1.6mm$ respectively.

Te optical path difference at $P_1$ is

$P_1=\frac{d{y}_1}{d}=\frac{\left(10/4000\right)}{0.2}=5\times {10}^{-4}mm=5000\mathring{A}$

In the visible range $4000-7000\mathring{A}$

$n_1=\frac{5000}{4000}=1.25$ and $n_2=\frac{5000}{7000}=0.714$

$\therefore Thewavelengthwhichformsmaximaat$P$is$I=5000$\mathring{A}$

For the point $P_2$

$P_2=\frac{d{y}_2}{d}=\frac{\left(10/4000\right)}{1.6}=4\times {10}^{-4}mm=4000\mathring{A}$

$n_1=\frac{40000}{4000}=10$ and $n_2=\frac{40000}{7000}=5.71$

The integers between 5.71 and 10 are 6,7,8,9 and 10

$\therefore$ The wavelength which form maxima at $P_2$ are

${\lambda }_1=4000\mathring{A}$ $Forn=10$

${\lambda }_2=4444\mathring{A}$ $Forn=9$

${\lambda }_3=5000\mathring{A}$ $Forn=8$

${\lambda }_4=5714\mathring{A}$ $Forn=7$

${\lambda }_5=6666\mathring{A}$ $Forn=6$