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

Tardigrade Admin · Accepted Answer

Fringe width, β=(d λ/d) d=(D λ/β) =(6000 × 10-10 ×(40 × 10-2)/0.012 × 10-2) =0.2 cm

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

0.24 cm

0.2 cm

2.4 cm

0.024 cm

Fringe width, $β = \frac{d λ}{d}$
$d = \frac{D λ}{β}$
$= \frac{6000 \times 1 0 ^{- 10} \times ( 40 \times 1 0 ^{- 2} )}{0.012 \times 1 0 ^{- 2}}$
$= 0.2 c m$

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