Youngs double slit experiment is performed with light of wavelength 550 nm. The separation between the slits is 1-10 mm and the screen is placed at a distance of 1 m. The distance between the consecutive bright and dark fringes is

Question

Youngs double slit experiment is performed with light of wavelength 550 nm. The separation between the slits is 1-10 mm and the screen is placed at a distance of 1 m. The distance between the consecutive bright and dark fringes is (A) 0.5 mm (B) 1.0 mm (C) 1.5 mm (D) None of the above.

Tardigrade Admin · Accepted Answer

β = (λ D/d) = (550 × 10-9 × 1/1.1 × 10-3) = 500 × 10-6 m ≅ 0.5 mm This is the separation between the consecutive bright fringes. The separation between the consecutive bright and dark fringes will be β/2

Q. Youngs double slit experiment is performed with light of wavelength 550 nm. The separation between the slits is 1-10 mm and the screen is placed at a distance of 1 m. The distance between the consecutive bright and dark fringes is

0.5 mm

1.0 mm

1.5 mm

None of the above.

$β = \frac{λ D}{d} = \frac{550 \times 1 0 ^{- 9} \times 1}{1.1 \times 1 0 ^{- 3}} = 500 \times 1 0^{- 6} m ≅ 0.5 mm$ This is the separation between the consecutive bright fringes. The separation between the consecutive bright and dark fringes will be $β /2$

Q. Youngs double slit experiment is performed with light of wavelength 550 nm. The separation between the slits is 1-10 mm and the screen is placed at a distance of 1 m. The distance between the consecutive bright and dark fringes is

0.5 mm

1.0 mm

1.5 mm

None of the above.

β=dλD​=1.1×10−3550×10−9×1​=500×10−6m≅0.5mm This is the separation between the consecutive bright fringes. The separation between the consecutive bright and dark fringes will be β/2

$β = \frac{λ D}{d} = \frac{550 \times 1 0 ^{- 9} \times 1}{1.1 \times 1 0 ^{- 3}} = 500 \times 1 0^{- 6} m ≅ 0.5 mm$ This is the separation between the consecutive bright fringes. The separation between the consecutive bright and dark fringes will be $β /2$