In Youngs double slit experiment, the slits are 0.5 mm apart and interference pattern is observed on a screen placed at a distance of 1 m from the place containing the slits. If wavelength of the incident light is 6000 mathringA then the separation between the third bright fringe and central maxima is :

Question

In Youngs double slit experiment, the slits are 0.5 mm apart and interference pattern is observed on a screen placed at a distance of 1 m from the place containing the slits. If wavelength of the incident light is 6000 mathringA then the separation between the third bright fringe and central maxima is : (A) 4 mm (B) 3.6 mm (C) 3 mm (D) 2.5 mm

Tardigrade Admin · Accepted Answer

Separation between, third-bright fringe and central maxima =(3 D λ/d)=(3 × 1 × 6 × 10-7/0.5 × 10-3) =3.6 × 10-3 m =3.6 mm

Q. In Youngs double slit experiment, the slits are 0.5mm apart and interference pattern is observed on a screen placed at a distance of 1m from the place containing the slits. If wavelength of the incident light is 6000A˚ then the separation between the third bright fringe and central maxima is :

4 mm

3.6 mm

3 mm

2.5 mm

Separation between, third-bright fringe and central maxima =d3Dλ​=0.5×10−33×1×6×10−7​ =3.6×10−3m=3.6mm

Separation between, third-bright fringe and central maxima
$= \frac{3 D λ}{d} = \frac{3 \times 1 \times 6 \times 1 0 ^{- 7}}{0.5 \times 1 0 ^{- 3}}$
$= 3.6 \times 1 0^{- 3} m = 3.6 mm$