Question:
In a Young's double slit experiment, the separation between the slits is $0.15 \mathrm{~mm}$. In the experiment, a source of light of wavelength $589 \mathrm{~nm}$ is used and the interference pattern is observed on a screen kept $1.5 \mathrm{~m}$ away. The separation between the successive bright fringes on the screen is:
Correct Option: , 3
Solution:
(3) Given, distance between screen and slits, $D=1.5 \mathrm{~m}$
Seperation between slits, $\mathrm{d}=0.15 \mathrm{~mm}$
Wavelength of source of light, $\lambda=589 \mathrm{~nm}$
Fringe-width, $w=\frac{D}{d} \lambda=\frac{1.5}{0.15 \times 10^{-3}} \times 589 \times 10^{-9} \mathrm{~m}$
$=589 \times 10^{-2} \mathrm{~mm}=5.89 \mathrm{~mm} \approx 5.9 \mathrm{~mm}$