Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ=632.8 nm ) . The distance between the screen and the slits is 100 cm . If a bright fringe is observed on a screen at a distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :

Question

Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ=632.8 nm ) . The distance between the screen and the slits is 100 cm . If a bright fringe is observed on a screen at a distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to : (A) 1.27 μ m (B) 2 nm (C) 2.87 nm (D) 2.05 μ m

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8b7c0e8f6cc18225ce0a23631c8ce9e4-.png" /> y=(n D λ/d) n=(y d/D λ)=(1.27 × 10-3 × 10-3/1 × 632.8 × 10-9)=2 Path difference Δ x=n λ =2 × 632.8 nm =1265.6 nm =1.27 μ m

$1.27 μ m$

$2 nm$

$2.87 nm$

$2.05 μ m$

$y = \frac{n D λ}{d}$
$n = \frac{y d}{D λ} = \frac{1.27 \times 1 0 ^{- 3} \times 1 0 ^{- 3}}{1 \times 632.8 \times 1 0 ^{- 9}} = 2$
Path difference $Δ x = nλ$
$= 2 \times 632.8 nm$
$= 1265.6 nm$
$= 1.27 μ m$

1.27μm

2nm

2.87nm

2.05μm

y=dnDλ​ n=Dλyd​=1×632.8×10−91.27×10−3×10−3​=2 Path difference Δx=nλ =2×632.8nm =1265.6nm =1.27μm

$1.27 μ m$

$2 nm$

$2.87 nm$

$2.05 μ m$

$y = \frac{n D λ}{d}$
$n = \frac{y d}{D λ} = \frac{1.27 \times 1 0 ^{- 3} \times 1 0 ^{- 3}}{1 \times 632.8 \times 1 0 ^{- 9}} = 2$
Path difference $Δ x = nλ$
$= 2 \times 632.8 nm$
$= 1265.6 nm$
$= 1.27 μ m$