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@`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `0@`

Options

A. (a)the absolute error in `d` remains constant.
B. (b)the absolute error in `d` increases.
C. (c)the fractional error in `d` remain constant.
D. (d)the fractional error in `d` decreases.

Explanation

To solve the problem using the expression \(2d \sin \theta = \lambda\), we will derive the relationship between the error in \(d\) and the angle \(\theta\). Here’s a step-by-step solution:

### Step 1: Rearranging the Expression
We start with the equation:
\[ 2d \sin \theta = \lambda \]
From this, we can express \(d\) in terms of \(\theta\):
\[ d = \frac{\lambda}{2 \sin \theta} \]

### Step 2: Differentiating the Expression
Next, we differentiate both sides with respect to \(\theta\) to find the relationship between the errors:
\[ \frac{dd}{d\theta} = \frac{d}{d\theta} \left(\frac{\lambda}{2 \sin \theta}\right) \]
Using the quotient rule, we get:
\[ \frac{dd}{d\theta} = -\frac{\lambda \cos \theta}{2 \sin^2 \theta} \]

### Step 3: Introducing Errors
Now, we introduce the errors. Let \(\delta D\) be the error in \(d\) and \(\delta \theta\) be the error in \(\theta\):
\[ \frac{\delta D}{D} = \frac{\delta \lambda}{\lambda} + \frac{\cos \theta}{\sin \theta} \delta \theta \]

### Step 4: Considering Known Wavelength
Since the wavelength \(\lambda\) is exactly known, we have:
\[ \delta \lambda = 0 \]
Thus, the equation simplifies to:
\[ \frac{\delta D}{D} = \frac{\cos \theta}{\sin \theta} \delta \theta \]

### Step 5: Analyzing the Behavior of Errors
As \(\theta\) increases from \(0^\circ\) to \(90^\circ\):
- \(\cos \theta\) decreases from \(1\) to \(0\).
- \(\sin \theta\) increases from \(0\) to \(1\).

This means that the term \(\frac{\cos \theta}{\sin \theta}\) will also decrease as \(\theta\) increases.

### Step 6: Conclusion
Since \(\frac{\delta D}{D}\) is proportional to \(\frac{\cos \theta}{\sin \theta}\), and this term decreases as \(\theta\) increases, we conclude that:
- The fractional error in \(d\) decreases as \(\theta\) increases.

Thus, the correct option is:
**Option D: The fractional error in \(D\) decreases.**