In Young's double slit experiment, when wavelength used is 6000 mathringA and the screen is 40 cm from the slits, then the fringes are 0.012 cm apart. What is the distance between the slits?

Question

In Young's double slit experiment, when wavelength used is 6000 mathringA and the screen is 40 cm from the slits, then the fringes are 0.012 cm apart. What is the distance between the slits? (A) 0.024 cm (B) 2.4 cm (C) 0.24 cm (D) 0.2 cm

Tardigrade Admin · Accepted Answer

Given D=40 cm =40 × 10-2 m λ=600 A=6000 × 10-10 m W=0.012 cm =0.012 × 10-2 m ∴ Fringe width, W=( D λ/d) ⇒ d=(D λ/W) where, d = distance between the slits d=(40 × 10-2 × 6000 × 10-10/0.012 × 10-2) =2 × 10-3 m =2 mm =0.2 cm

Q. In Young's double slit experiment, when wavelength used is 6000A˚ and the screen is 40cm from the slits, then the fringes are 0.012cm apart. What is the distance between the slits?

0.024 cm

2.4 cm

0.24 cm

0.2 cm

Given D=40cm=40×10−2m λ=600A=6000×10−10m W=0.012cm=0.012×10−2m ∴ Fringe width, W=dDλ​ ⇒d=WDλ​ where, d= distance between the slits d=0.012×10−240×10−2×6000×10−10​ =2×10−3m=2mm=0.2cm

Q. In Young's double slit experiment, when wavelength used is $6000 \overset{˚}{A}$ and the screen is $40 c m$ from the slits, then the fringes are $0.012 c m$ apart. What is the distance between the slits?