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…

Question

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°,

MARKS Content Team · Accepted Answer

Correct answer: (D) the fractional error in d decreases.. 2d sin θ=λ d=(λ)/(2 sin θ) Differentiate on both the sides, ∂d=(λ)/2∂ cosec θ ∂d=(λ)/2-cosec θ cot θ∂θ ∂d=(-λ cos θ)/(2 sin^2 θ)∂θ θ=increases and (λ cos θ)/(2 sin^2 θ) decreases. Alternate solution d=(λ)/(2 sin θ) ℓn d=ℓn λ-ℓn 2-ℓn sin θ (Δd)/d=0-0-1/(sin θ)×cos θΔθ Fractional error, +d=cot θΔθ Absolute error, Δd=d cot θΔθ d/(2sin θ)×(cos θ)/(sin θ) Δd=(cos θ)/(sin^2 θ)

Using the expression $2 d \sin θ = λ$ , 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 $0 °$ ,

Solution

Relevant Questions