Two sources of light 0.6 mm apart and screen is placed at a distance of 1.2 m from them. A light of wavelength 6000 A ˚ used. Then the phase difference between the two light waves interfering on the screen at a point at a distance 3 mm from central bright band is

Fringe width, W = d λ . D = 0.6 × 1 0 − 3 6000 × 1 0 − 10 × 1.2 = 1.2 mm ∴ Number of fringes (n) = 1.2 3 = 2.5 ∴ Phase difference, Δ ϕ = 2 n π = 2 × 2.5 π = 5 π

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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsYDSEMedium2 minPYQ_2023

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Two sources of light $0.6 mm$ apart and screen is placed at a distance of $1.2 m$ from them. A light of wavelength $6000 \overset{˚}{A}$ used. Then the phase difference between the two light waves interfering on the screen at a point at a distance $3 mm$ from central bright band is

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Fringe width, $W = \frac{λ . D}{d} = \frac{6000 \times 1 0 ^{- 10} \times 1.2}{0.6 \times 1 0 ^{- 3}} = 1.2 mm$ $∴$ Number of fringes (n) $= \frac{3}{1.2} = 2.5$ $∴$ Phase difference, $Δ ϕ = 2 n π = 2 \times 2.5 π = 5 π$

Stream:MHT_CETSubject:PhysicsTopic:Wave OpticsSubtopic:YDSE

⏱ 2mℹ️ Source: PYQ_2023

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