Using the expression

PHYSICS · Single correct · CBSE · Physics · Units and Measurements · Units and Measurements

Problem

Using the expression 2dsin⁡θ=λ2d \sin \theta = \lambda , one calculates the values of dd by measuring the corresponding angles θ\theta in the range 0 to 90°. The wavelength λ\lambda is exactly known and the error in θ\theta is constant for all values of θ\theta . As θ\theta increases from 0° (1) The absolute error in dd remains constant (2) The absolute error in dd increases (3) The fractional error in dd remains constant (4) The fractional error in dd decreases

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