In Youngs double slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is

Question

In Youngs double slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is (A)  sin -1(0.01)  (B)  sin -1(0.0001)  (C)  sin -1(0.001)  (D)  sin -1(0.1)

Tardigrade Admin · Accepted Answer

In Youngs double slit experiment half angular width is given by sin θ =(λ /d) sin θ =(589× 10-9/0.589× 10-3)=10-3 θ = sin -1(0.001)

Q. In Youngs double slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is

$sin^{- 1} (0.01)$

$sin^{- 1} (0.0001)$

$sin^{- 1} (0.001)$

$sin^{- 1} (0.1)$

In Youngs double slit experiment half angular width is given by
$sin θ = \frac{λ}{d}$
$sin θ = \frac{589 \times 10 ^{- 9}}{0.589 \times 10 ^{- 3}} = 10^{- 3}$
$θ = sin^{- 1} (0.001)$

Q. In Youngs double slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is

sin−1(0.01)

sin−1(0.0001)

sin−1(0.001)

sin−1(0.1)

In Youngs double slit experiment half angular width is given by sinθ=dλ​ sinθ=0.589×10−3589×10−9​=10−3 θ=sin−1(0.001)

$sin^{- 1} (0.01)$

$sin^{- 1} (0.0001)$

$sin^{- 1} (0.001)$

$sin^{- 1} (0.1)$

In Youngs double slit experiment half angular width is given by
$sin θ = \frac{λ}{d}$
$sin θ = \frac{589 \times 10 ^{- 9}}{0.589 \times 10 ^{- 3}} = 10^{- 3}$
$θ = sin^{- 1} (0.001)$