Question

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 X is exactly known and the error in $\theta$
is constant for all values of $\theta$. As $\theta$ increases from 0$^{\circ}$

• A. the absolute error in d remains constant
• B. the absolute error in d increases
• C. the fractional error in d remains constant
• D. the fractional error in d decreases

Answer 1

Answer: (D)

$
\begin{array}{l}
2 d \sin \theta=\lambda \\
\Rightarrow d =\frac{ d }{2 \sin \theta}
\end{array}
$
Taking log and differentiating both sides-
$
\frac{\Delta d }{ d }=\frac{\Delta \lambda}{\lambda}-\frac{1}{\sin \theta} \cos \theta \Delta \theta
$
For maximum possible error all the errors should be added
$
\Rightarrow \frac{\Delta d }{ d }=\frac{\Delta \lambda}{\lambda}+\frac{\cos \theta}{\sin \theta} \Delta \theta
$
As $\lambda$ is exactly known, $\Delta \lambda=0$
$
\Rightarrow \frac{\Delta d}{d}=\cot \theta \Delta \theta
$
As $\Delta \theta_{1}$ is constant and on increasing $\theta$, $\cot \theta$ decreasees. So, fractional error in d(i.e $\frac{\Delta d}{d}$ ) will also decrease.