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iragen [17]
3 years ago
14

Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter

n is observed on a screen located at a distance of 3.0 m from the slits. Where will the second order dark fringe above the center of central bright fringe will form?
Physics
1 answer:
goblinko [34]3 years ago
7 0

Answer:

134.77 mm

Explanation:

Wave length of light λ = 599 x 10⁻⁹ m

Slit separation d = 20 x 10⁻⁶ m

Screen distance D = 3 m

Distance of second dark fringe from centre

= 1.5 x λ D / d  

Putting the  values given above

distance = \frac{1.5\times599\times10^{-9}\times 3}{20\times10^{-6}}

= 134.77 x 10⁻³ m

= 134.77 mm.

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Hi!


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1. <u>Chunking </u>involves organizing and breaking down information into easier groups to expand capacity.

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2. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of <u>short-term</u> memory.

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