Question

Using the expression $2dsin\theta =\lambda$, one calculates the values of d by measuring the corresponding angles $\theta$ in the range $0^{\circ }to{90}^{\circ }$. The wavelength $\lambda$ 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

The correct option is D The fractional error in d decreases

$d=\frac{\lambda 2sin\theta }{=}\frac{\lambda }{2}cosec\theta \frac{\Delta d}{\Delta \theta }=\frac{\lambda }{2}(-cosec\theta cot\theta )[\because \lambda exactlyknown]$
$\Delta d=-\frac{\lambda }{2}cosec\theta cot\theta \Delta \theta$
$\Rightarrow$ Absolute error in d decreases with the increase in $\theta$ as both cosec $\theta$ and cot $\theta$ decrease with increase in $\theta$ from $0^{\circ }to{90}^{\circ }.$
Clearly, $\frac{\Delta d}{d}=-cot\theta .\Delta \theta$
$\Rightarrow$ Fractional error decreases with the increase in $\theta$ from $0^{\circ }to{90}^{\circ }$.