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A double slit arrangement produces interference fringes for sodium light $(λ = 589 nm)$ that have an angular separation of $3.50 \times 10^{- 3} radian.$ For what wavelength would the angular separation be 10% greater?

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527 nm

648 nm

722 nm

449 nm

answer is B.

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Detailed Solution

Angular separation $= \frac{λ}{d}$

For angular separation to be 10% greater $λ$ should be 10% greater.

New wavelength is $(589 + \frac{589}{10}) nm$

or (589 + 58.9) nm, i.e., 647.9 nm, i.e., 648 nm.

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