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)

the fractional error in d decreases

Answer 2

$d=\frac{\lambda}{2 sin \, \theta}$
\hspace10mm ln \, d=ln \, \lambda-ln \, 2-ln \, sin \, \theta
\hspace10mm \frac{\Delta (d)}{d}=0-0-\frac{1}{sin \, \theta} \times cos \theta (\Delta \theta)
Fractional error=$|\frac{\Delta d}{d}|=(cot \, \theta) \Delta \theta$
Absolute error $\Delta d=(d \, cot \theta) \Delta \theta=\big(\frac{\lambda}{2 \, sin \, \theta}\big) \big(\frac{cos \theta}{sin \theta}\big)\Delta \theta$
Now, given that $\Delta \theta$= constant
As $\theta$ increases, sin$\theta$ increases, cos$\theta$and cot$\theta$ decrease.
$\therefore$ Both fractional and absolute errors decrease.