f 720-nm and 620-nm light passes through two slits 0.68 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.0 m away

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Answer:

0.0003 m = 0.3 mm

Explanation:

For constructive interference in the Young's experiment.

The position of the mth fringe from the central fringe is given by

y = L(mλ/d)

λ = wavelength = 720 nm = 720 × 10⁻⁹ m

L = distance between slits and screen respectively = 1.0 m

d = separation of slits = 0.68 mm = 0.68 × 10⁻³ m

m = 2

y = 1(2 × 720 × 10⁻⁹/(0.68 × 10⁻³) = 0.00212 m = 2.12 mm

For the 620 nm light,

y = 1(2 × 620 × 10⁻⁹/(0.68 × 10⁻³) = 0.00182 m = 1.82 mm

Distance apart = 2.12 - 1.82 = 0.3 mm = 0.0003 m

The distance apart should be 0.0003 m = 0.3 mm

Calculation of the distance apart:

Since

The position of the mth fringe from the central fringe is provided by

y = L(mλ/d)

Here,

λ = wavelength = 720 nm = 720 × 10⁻⁹ m

L = distance between slits and screen respectively = 1.0 m

d = separation of slits = 0.68 mm = 0.68 × 10⁻³ m

m = 2

So,

y = 1(2 × 720 × 10⁻⁹/(0.68 × 10⁻³) = 0.00212 m = 2.12 mm

Now For the 620 nm light,

y = 1(2 × 620 × 10⁻⁹/(0.68 × 10⁻³) = 0.00182 m = 1.82 mm

Now finally

Distance apart should be

= 2.12 - 1.82

= 0.3 mm

= 0.0003 m

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