Using the expression 2 d   sin   θ = λ , one calculates the values of d by measuring the…

Using the expression 2d sin θ=λ, one calculates the values of d by measuring the corresponding angle θ in the range 0 to 90°. The wavelength λ is exactly known and the error in θ is constant for all the values of θ. As θ increases from 0°,
  1. the absolute error in d remains constant.
  2. the absolute error in dincreases.
  3. the fractional error in d remains constant.
  4. the fractional error in d decreases.

Solution

2d sin θ=λ
d=λ2 sin θ
Differentiate on both the sides,
∂d=λ2∂ cosec θ
∂d=λ2-cosec θ cot θ∂θ
∂d=-λ cos  θ2 sin2 θ∂θ
 θ=increases and λ cos θ2 sin2 θ decreases.
Alternate solution
d=λ2 sin θ
ℓn d=ℓn λ-ℓn 2-ℓn sin θ
Δdd=0-0-1sin θ×cos θΔθ
Fractional error, +d=cot θΔθ
Absolute error, Δd=d cot θΔθ
d2sin θ×cos θsin θ
Δd=cos θsin2 θ

Asked in: JEE Advanced 2013 (Paper 2)

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