In a Youngs double slit experiment, the slits are separated by 0.5 mm and the screen is placed 1.0 m away. If the light source emits two wavelengths 600 nm and 450 nm, find the minimum distance from the central maximum where the bright fringes of both wavelengths coincide.

B. 3.0 mm — The condition for coincidence is n_1 _1 = n_2 _2. Solving for the smallest integers n_1 and n_2 gives n_1 = 3 and n_2 = 4, leading to y = n_1 _1 D / d = 3.6 10^-3 m.

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$In a Youngs double slit experiment, the slits are separated by d = 0.5 mm and the screen is placed D = 1.0 m away. If the light source emits two wavelengths λ_{1} = 600 nm and λ_{2} = 450 nm, find the minimum distance y from the central maximum where the bright fringes of both wavelengths coincide.$

A. $1.5 mm$
B. $3.0 mm$
C. $4.5 mm$
D. $6.0 mm$

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Correct answer: B — $3.0 mm$

Solution

The condition for coincidence is n_{1} λ_{1} = n_{2} λ_{2} . Solving for the smallest integers n_{1} and n_{2} gives n_{1} = 3 and n_{2} = 4, leading to y = \frac{n _{1} λ _{1} D}{d} = 3.6 \times 1 0^{- 3} m .

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In a Youngs double slit experiment, the slits are separated by d=0.5 mm and the screen is placed D=1.0 m away. If the light source emits two wavelengths λ1​=600 nm and λ2​=450 nm, find the minimum distance y from the central maximum where the bright fringes of both wavelengths coincide.

Solution

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