Question

Using the expression $2d\sin \theta =\lambda$, one calculates the values of $d$ by measuring the corresponding angles $\theta$ in the range $\theta$ to ${90}^o$. The wavelength $\lambda$ is exactly known and the error in $\theta$ is constant for all values of $\theta$. As $\theta$ increases from $0^o$,

• 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

We have $2dsin\theta =\lambda$
Let $2d=y$
Thus we get, $ysin\theta =\lambda$
$\Rightarrow lny+lnsin\theta =ln\lambda$

Differentiating both the sides we get
$\Rightarrow \frac{dy}{y}=-cot\theta$
$\Rightarrow \left|\frac{dy}{y}\right|=cot\theta$
As $y=2d$, we see that the fractional error in d decreases as $\theta$ increases.