To determine how the error in $$d$$ behaves as $$\theta$$ increases, let’s examine the given expression:

$$2d\sin \theta = \lambda$$

We can rearrange this to solve for $$d$$:

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

Given that the wavelength $$\lambda$$ is exactly known, any error in the measurement of $$\theta$$ (denoted as $$\Delta \theta$$) will propagate through to the error in $$d$$. To analyze this, we need to consider how the error propagates in the expression for $$d$$.

Assuming the error propagation formula, the differential of $$d$$ with respect to $$\theta$$ is:

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

Differentiating with respect to $$\theta$$:

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

Using the chain rule, we get the partial derivative:

$$\frac{d(d)}{d(\theta)} = -\frac{\lambda \cos \theta}{2 (\sin \theta)^2}$$

The absolute error in $$d$$, denoted as $$\Delta d$$, is given by:

$$\Delta d = \left| \frac{d(d)}{d(\theta)} \Delta \theta \right| = \left| -\frac{\lambda \cos \theta}{2 (\sin \theta)^2} \right| \Delta \theta = \frac{\lambda \cos \theta}{2 (\sin \theta)^2} \Delta \theta$$

Here, $$\Delta \theta$$ is the constant error in the measurement of $$\theta$$. From the equation above, we observe that $$\Delta d$$ depends on $$\cos \theta / (\sin \theta)^2$$ which is a function of $$\theta$$.

As $$\theta$$ increases from $$0^\circ$$ to $$90^\circ$$, the behavior is as follows:

• When $$\theta$$ is close to $$0^\circ$$, $$\cos \theta \approx 1$$ and $$\sin \theta \approx 0$$, thus $$\left( \frac{\cos \theta}{(\sin \theta)^2} \right)$$ becomes very large, indicating that $$\Delta d$$ is very large.
• As $$\theta$$ approaches $$90^\circ$$, $$\cos \theta \approx 0$$, and $$\sin \theta \approx 1$$, which makes $$\left( \frac{\cos \theta}{(\sin \theta)^2} \right)$$ approach zero, thus making $$\Delta d$$ very small.

Therefore, as $$\theta$$ increases from $$0^\circ$$ to $$90^\circ$$, the absolute error in $$d$$ decreases.

To analyze the fractional error in $$d$$ (denoted as $$\frac{\Delta d}{d}$$), we note:

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

So the fractional error is:

$$\frac{\Delta d}{d} = \frac{\Delta d}{\frac{\lambda}{2 \sin \theta}} = \frac{\frac{\lambda \cos \theta}{2 (\sin \theta)^2} \Delta \theta}{\frac{\lambda}{2 \sin \theta}} = \frac{\cos \theta \Delta \theta}{\sin \theta}$$

As $$\theta$$ increases from $$0^\circ$$ to $$90^\circ$$:

• Initially (at $$\theta \approx 0^\circ$$), $$\cos \theta \approx 1$$ and $$\sin \theta \approx 0$$, making the fractional error large.
• As $$\theta$$ continues to increase, $$\cos \theta$$ decreases and $$\sin \theta$$ increases, thus the fractional error $$\frac{\cos \theta \Delta \theta}{\sin \theta}$$ decreases.

Hence, the fractional error in $$d$$ also decreases as $$\theta$$ increases.

So the correct answer is: