Question

Using the expression $2dsin\theta =\lambda$, one calculates the values of d by measuring the corresponding angles $\theta$ in the range $0^0$ to ${90}^0$. The wavelength $\lambda$ is exactly known and the error in $\theta$ is constant for all values of $\theta$. As $\theta$ increases from ${10}^0$.

• 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.

$2d\sin \theta =\lambda$

$\Rightarrow d=\frac{d}{2\sin \theta }$

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.