In the Young's double slit experiment with sodium light, the slits are 0.589 m apart. The angular separation of the third maximum from the central maximum will be (given λ=589 mm )

Question

In the Young's double slit experiment with sodium light, the slits are 0.589 m apart. The angular separation of the third maximum from the central maximum will be (given λ=589 mm ) (A)  sin -1(0.33 × 108) (B)  sin -1(0.33 × 10-6) (C)  sin -1(3 × 10-8) (D)  sin -1(3 × 10-6)

Tardigrade Admin · Accepted Answer

Using relation, d sin θ=n λ ⇒ sin θ=(n λ/d) For n=3, sin θ =(3 λ/d)=(3 × 589 × 10-9/0.589) =3 × 10-6 text or θ= sin -1(3 × 10-6)

Q. In the Young's double slit experiment with sodium light, the slits are $0.589 m$ apart. The angular separation of the third maximum from the central maximum will be (given $λ = 589 mm)$

$sin^{- 1} (0.33 \times 1 0^{8})$

$sin^{- 1} (0.33 \times 1 0^{- 6})$

$sin^{- 1} (3 \times 1 0^{- 8})$

$sin^{- 1} (3 \times 1 0^{- 6})$

Using relation, $d sin θ = nλ \Rightarrow sin θ = \frac{nλ}{d}$
For $n = 3, sin θ = \frac{3 λ}{d} = \frac{3 \times 589 \times 1 0 ^{- 9}}{0.589}$
$= 3 \times 1 0^{- 6} or θ = sin^{- 1} (3 \times 1 0^{- 6})$

Q. In the Young's double slit experiment with sodium light, the slits are 0.589m apart. The angular separation of the third maximum from the central maximum will be (given λ=589mm)

sin−1(0.33×108)

sin−1(0.33×10−6)

sin−1(3×10−8)

sin−1(3×10−6)