1. We are given Bragg’s law:

$2d\sin\theta = \lambda$

Since $\lambda$ is exactly known,

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

We need to see how the error in $d$ depends on $\theta$, given that the error in $\theta$ is constant.

2. Use error propagation.

For small errors,

$\Delta d \approx \left|\frac{dd}{d\theta}\right| \Delta\theta$

Now,

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

Differentiate with respect to $\theta$:

$\frac{dd}{d\theta} = \frac{\lambda}{2}(-\csc\theta\cot\theta)$

So,

$\left|\frac{dd}{d\theta}\right| = \frac{\lambda}{2}\csc\theta\cot\theta$

Hence the absolute error is

$\Delta d \approx \frac{\lambda}{2}\csc\theta\cot\theta\,\Delta\theta$

or equivalently,

$\Delta d \propto \csc\theta\cot\theta = \frac{\cos\theta}{\sin^2\theta}$

As $\theta$ increases from $0^\circ$ to $90^\circ$, $\sin\theta$ increases and $\cos\theta$ decreases, so

$\frac{\cos\theta}{\sin^2\theta}$

decreases.

Therefore, the absolute error in $d$ decreases with increasing $\theta$.

So:

• A is false
• B is false

3. Now find the fractional error.

$\frac{\Delta d}{d} \approx \left|\frac{1}{d}\frac{dd}{d\theta}\right|\Delta\theta$

Using $d = \frac{\lambda}{2}\csc\theta$,

$\frac{\Delta d}{d} = \frac{\frac{\lambda}{2}\csc\theta\cot\theta\,\Delta\theta}{\frac{\lambda}{2}\csc\theta}$

This gives

$\frac{\Delta d}{d} = \cot\theta\,\Delta\theta$

Since $\Delta\theta$ is constant,

$\frac{\Delta d}{d} \propto \cot\theta$

As $\theta$ increases from $0^\circ$ to $90^\circ$, $\cot\theta$ decreases.

Therefore, the fractional error in $d$ decreases.

So:

• C is false
• D is true

4. Final option check:

• A: False
• B: False
• C: False
• D: True

Thus the correct answer is:

$\boxed{D}$