Question

The intensity observed in an interference pattern is, $I=I_{0}sin^{2}\theta $ . At $\theta =30^\circ $ , intensity is, $I=\left(5 \pm 0 . 002\right)Wm^{- 2}$ . The percentage error in angle if $I_{0}=20Wm^{- 2}$ is,

• A. $4\sqrt{3}\times 10^{- 2}\%$
• B. $\frac{4}{\pi }\times 10^{- 2}\%$
• C. $\frac{4 \sqrt{3}}{\pi }\times 10^{- 2}\%$
• D. $\sqrt{3}\times 10^{- 2}\%$

Answer 1

Answer: (C)

$I=I_{0}sin^{2}\theta $
Differentiating both sides we get,
$dI=I_{0}\left(2 sin \theta cos \theta d \theta \right)$
or $0.002=20\times 2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}d\theta $
$\Rightarrow d\theta =\frac{2}{\sqrt{3}}\times 10^{- 4}$
Therefore,
$\Rightarrow \frac{d \theta }{\theta }=\frac{\left(2/\left(\sqrt{3}\right) \times \left(10\right)^{- 4}}{\left(\pi \right)/6}=\frac{4 \sqrt{3}}{\pi }\times \left(10\right)^{- 4}$ and $\frac{d \theta }{\theta }\times 100=\frac{4 \sqrt{3}}{\pi }\times 10^{- 2}\%$ .