1
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Question

# Light of wavelength 5000 ˚A is incident normally on a slit of width 2.5×10−4 cm. The angular position of second minimum from the central maximum is

A
sin1(15)
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B
sin1(25)
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C
(π3)
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D
(π6)
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E
(π4)
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Solution

## The correct option is B sin−1(25)We know that, Angular width of central maximum =2λawhere, λ =wavelength of light, a =width of single slit, So, sinθ=2λawhere λ=5000˚A=5000×10−10ma=2.5×10−6m⇒sinθ=2×5000×10−102.5×10−6=sinθ=10000×10−102.5×10−6=sinθ=1025=sinθ=25θ=sin−1(25)

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