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Light of wavelength 6000 A° is incident normally on a slit of width 24×10^-5 cm. The angular position of second mininum from central maximum will be ----

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30°

60°

15°

0°

answer is C.

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Detailed Solution

$\begin{array}{l} asin θ = nλ; \sin θ = \frac{2 \times 6000 \times 10^{- 10}}{24 \times 10^{- 5} \times 10^{- 2}} = \frac{1}{2} \\ \Rightarrow θ = 30^{0} \end{array}$

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