Using the expression $2d \sin \theta = \lambda$, one calculates the values of $d$ by measuring the corresponding angles $\theta$ in the range $0$ to $90^\circ$. The wavelength $\lambda$ is exactly known and the error in $\theta$ is constant for all values of $\theta$. As $\theta$ increases from $0^\circ$

Step 1: Express $d$ in terms of $\theta$.

Rearrange the given Bragg's relation to solve for $d$.

$$2d \sin \theta = \lambda$$ $$d = \frac{\lambda}{2\sin\theta}$$

Step 2: Determine the fractional error in $d$.

Take the natural logarithm of both sides.

$$\ln d = \ln\left(\frac{\lambda}{2}\right) - \ln(\sin\theta)$$

Differentiate both sides to find the relation for the fractional error.

$$\left|\frac{\Delta d}{d}\right| = \cot\theta\,\Delta\theta$$

Step 3: Analyze the behavior of the fractional error.

The error in angle measurement $\Delta\theta$ is constant.

$$\theta \uparrow \implies \cot\theta \downarrow$$

As $\theta$ increases from $0^\circ$ to $90^\circ$, $\cot\theta$ decreases, which means the fractional error decreases.

Step 4: Analyze the behavior of the absolute error.

Express the absolute error $\Delta d$ by multiplying the fractional error by $d$.

$$\Delta d = d\cot\theta\,\Delta\theta$$ $$\Delta d = \frac{\lambda\,\Delta\theta}{2}\cdot\frac{\cos\theta}{\sin^{2}\theta}$$

The derivative of the trigonometric term is negative in the interval $0^\circ < \theta < 90^\circ$.

$$\frac{d}{d\theta}\left(\frac{\cos\theta}{\sin^{2}\theta}\right) < 0$$

Since the derivative is negative, the absolute error also decreases as $\theta$ increases.

Step 5: Evaluate the limits at the boundaries.

Analyze the behavior at the boundaries to confirm the trend.

$$\theta \to 0^\circ \implies \Delta d \to \infty, \quad \frac{\Delta d}{d} \to \infty$$ $$\theta \to 90^\circ \implies \Delta d \to 0, \quad \frac{\Delta d}{d} \to 0$$

Both the absolute error and the fractional error decrease as $\theta$ increases.

Answer: The fractional error in $d$ decreases.