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← JEE Advanced 2013 · Paper 2
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JEE Advanced 2013 · Paper 2, Physics question 1

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°,

Choose one or more options
Answer and worked solution

Official answer: D

Error propagation. d=λ2sin⁡θd=\dfrac\lambda{2\sin\theta}, so
Δd=λcos⁡θ2sin⁡2θΔθ,Δdd=cot⁡θ Δθ.\Delta d=\frac{\lambda\cos\theta}{2\sin^2\theta}\Delta\theta,\qquad\frac{\Delta d}d=\cot\theta\,\Delta\theta.

As θ increases towards 90°, both cos⁡θsin⁡2θ\frac{\cos\theta}{\sin^2\theta} and cot⁡θ\cot\theta decrease: the absolute and the fractional errors both fall. Only (D) is offered among the true statements.

(D)

Topics:Errors in measurement
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