Q. We get $60$ fringes in the field of view of monochromatic light of wavelength $4000\overset{^\circ }{A}$ in Young's double-slit experiment. What will be the number of fringes obtained in the same field of view, if we use monochromatic light of wavelength $6000\overset{^\circ }{A}$ .
NTA AbhyasNTA Abhyas 2020
Solution:
$\beta =\frac{\lambda D}{d}$
When monochromatic light of wavelength $4000A$ is used:
$60th$ fringe is obtained at $=60\beta $
$=60\frac{\lambda _{1} D}{d}$ ......eq 1
When, monochromatic light of wavelength $6000A$ is used,
$\beta _{2}=\frac{\lambda _{2} D}{d }$
$nth$ fringe is at distance = $n\left(\right.\frac{\left(\lambda \right)_{2} D}{d}\left.\right)$ ....eq2
Both monochromatic lights have same fringe width area,
$60\lambda _{1}\frac{D}{d}=n\lambda _{2}\frac{D}{d}60\times 4000=n\times 6000n=40$
When monochromatic light of wavelength $4000A$ is used:
$60th$ fringe is obtained at $=60\beta $
$=60\frac{\lambda _{1} D}{d}$ ......eq 1
When, monochromatic light of wavelength $6000A$ is used,
$\beta _{2}=\frac{\lambda _{2} D}{d }$
$nth$ fringe is at distance = $n\left(\right.\frac{\left(\lambda \right)_{2} D}{d}\left.\right)$ ....eq2
Both monochromatic lights have same fringe width area,
$60\lambda _{1}\frac{D}{d}=n\lambda _{2}\frac{D}{d}60\times 4000=n\times 6000n=40$