Engineering
Physics
Physical Quantities
Question

Using the expression 2d sin θ=λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90º. The wavelength λ is exactly known and the error in θ is constant for all values of θ, As θ increases from 0º,

the fractional error in d decreases.

the absolute error in d remains constant

the absolute error in d increases

the fractional error in d remains constant

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Solution
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 d=λ2sinθ

 Δd=λ2sinθcosθdθ

 Δdd=λ2cosθsin2θλ2sinθ

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