In Young's double slit experiment distance between two sources is 0.1 mm. The distance of screen from the source is 20 cm. Wavelength of light used is 5460 Å. Then, angular position of the first dark fringe is

Question

In Young's double slit experiment distance between two sources is 0.1 mm. The distance of screen from the source is 20 cm. Wavelength of light used is 5460 Å. Then, angular position of the first dark fringe is (A) 0.08° (B) 0.16° (C) 0.20° (D) 0.31°

Tardigrade Admin · Accepted Answer

Angular position of first dark fringe θ1 = (2×1-1) (λ/2d) = (λ/2d) = (5460× 10-10 /2 × 0.1 × 10-3) = 2730 × 10-6 rad = 2730 × 10-6 × (180/π) = 0.16°

Q. In Young's double slit experiment distance between two sources is $0.1 mm$ . The distance of screen from the source is $20 c m$ . Wavelength of light used is $5460 \overset{˚}{A}$ . Then, angular position of the first dark fringe is

$0.0 8^{\circ}$

$0.1 6^{\circ}$

$0.2 0^{\circ}$

$0.3 1^{\circ}$

Angular position of first dark fringe
$θ_{1} = (2 \times 1 - 1) \frac{λ}{2 d}$
$= \frac{λ}{2 d} = \frac{5460 \times 1 0 ^{- 10}}{2 \times 0.1 \times 1 0 ^{- 3}}$
$= 2730 \times 1 0^{- 6} r a d$
$= 2730 \times 1 0^{- 6} \times \frac{180}{π}$
$= 0.1 6^{\circ}$

Q. In Young's double slit experiment distance between two sources is 0.1mm. The distance of screen from the source is 20cm. Wavelength of light used is 5460A˚. Then, angular position of the first dark fringe is

0.08∘

0.16∘

0.20∘

0.31∘