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)

$<br/>\begin{array}{l}<br/>2d\sin \theta =\lambda \\ <br/>\Rightarrow d=\frac{d}{2\sin \theta }<br/>\end{array}<br/>$
Taking log and differentiating both sides-
$<br/>\frac{\Delta d}{d}=\frac{\Delta \lambda }{\lambda }-\frac{1}{\sin \theta }\cos \theta \Delta \theta <br/>$
For maximum possible error all the errors should be added
$<br/>\Rightarrow \frac{\Delta d}{d}=\frac{\Delta \lambda }{\lambda }+\frac{\cos \theta }{\sin \theta }\Delta \theta <br/>$
As $\lambda$ is exactly known, $\Delta \lambda =0$
$<br/>\Rightarrow \frac{\Delta d}{d}=\cot \theta \Delta \theta <br/>$
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.