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 fring

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 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 :

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Answer:

A

Solution:

$Δ P = d \sin θ$

$= d θ$

$= \frac{d y}{D} = \frac{10^{- 3} \times 1.270 mm}{1 m} = 1.27 μ m$

$1.27 μ m$

TestHub · Accepted Answer

Δ P = d sin θ = d θ = d y D = 10 - 3 × 1 . 270 mm 1 m = 1 . 27 μ m 1 . 27 μ m

Physics - Wave Optics Question with Solution | TestHub

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