JEE Challenger
More from Units and Measurements
PhyMedAdv 2013 • Paper 2

Variation of Error in Lattice Spacing with Angle of Incidence

Using the expression 2dsin⁡θ=λ2d \sin \theta = \lambda, the value of dd is determined by measuring the corresponding angle θ\theta over the interval 00 to 90∘90^{\circ}. If the wavelength λ\lambda is precisely known and the measurement uncertainty in θ\theta is constant for all values of θ\theta, then as θ\theta increases from 0∘0^{\circ}:

Options

A

the absolute error in dd remains constant.

B

the absolute error in dd increases.

C

the fractional error in dd remains constant.

D

the fractional error in dd decreases.

Correct

Topics & Concepts

Units and MeasurementsMeasuring InstrumentsDiffraction

Step-by-Step Solution

To determine how the absolute and fractional errors in the lattice spacing dd vary with the angle θ\theta, we start with the Bragg diffraction relation: 2dsin⁡θ=λ2d \sin\theta = \lambda

Expressing the lattice spacing dd as a function of θ\theta: d=λ2sin⁡θd = \frac{\lambda}{2\sin\theta}

Taking the natural logarithm on both sides: ln⁡d=ln⁡(λ2)−ln⁡(sin⁡θ)\ln d = \ln\left(\frac{\lambda}{2}\right) - \ln(\sin\theta)

Differentiating both sides to find the fractional error: Δdd=∣−cot⁡θ Δθ∣=cot⁡θ Δθ\frac{\Delta d}{d} = |-\cot\theta \, \Delta\theta| = \cot\theta \, \Delta\theta

We are given that λ\lambda is precisely known (i.e., Δλ=0\Delta\lambda = 0) and the error in θ\theta, Δθ\Delta\theta, is constant for all values of θ\theta.

  1. Fractional Error in dd: Fractional Error=Δdd=cot⁡θ⋅Δθ\text{Fractional Error} = \frac{\Delta d}{d} = \cot\theta \cdot \Delta\theta As θ\theta increases from 0∘0^\circ to 90∘90^\circ, cot⁡θ\cot\theta decreases monotonically from ∞\infty to 00. Since Δθ\Delta\theta is constant, the fractional error Δdd\frac{\Delta d}{d} decreases.

  2. Absolute Error in dd: Δd=dcot⁡θ⋅Δθ=(λ2sin⁡θ)cot⁡θ⋅Δθ=λcos⁡θ2sin⁡2θ⋅Δθ\Delta d = d \cot\theta \cdot \Delta\theta = \left(\frac{\lambda}{2\sin\theta}\right) \cot\theta \cdot \Delta\theta = \frac{\lambda \cos\theta}{2\sin^2\theta} \cdot \Delta\theta As θ\theta increases from 0∘0^\circ to 90∘90^\circ, cos⁡θ\cos\theta decreases while sin⁡2θ\sin^2\theta increases. Consequently, cos⁡θsin⁡2θ\frac{\cos\theta}{\sin^2\theta} decreases, meaning the absolute error Δd\Delta d also decreases (and does not increase or remain constant).

Therefore, as θ\theta increases from 0∘0^\circ to 90∘90^\circ, the fractional error in dd decreases.

Correct Option: D

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