Question

Intensity observed in an interference pattern is given by $I=I_0{\sin }^2\theta$. At $\theta ={30}^{\circ }$, intensity $I=5\pm 0.002$. The percentage error in $\theta$ if $I_0=20\text{W}/{\text{m}}^2$ 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

The correct option is C $\frac{4\sqrt{3}}{\pi }\times {10}^{-2}$

Given,
Intensity, $I=I_0{\sin }^2\theta$

$I=\frac{I_0}{2}(1-\cos 2\theta )$

Differentiating $I$ with respect to $\theta$, we get,

$\frac{dI}{d\theta }=I_0\sin 2\theta$

$\Rightarrow dI=I_0\sin 2\theta d\theta$

Further, percentage error in angle,

$\frac{d\theta }{\theta }\times 100=\left(\frac{dI}{I_0\sin 2\theta }\right)\frac{1}{\theta }\times 100$

$=\frac{0.002}{20\times \sin {60}^{\circ }}\times \frac{1}{{30}^{\circ }\times \frac{\pi }{{180}^{\circ }}}\times 100$

$=\frac{4\sqrt{3}}{\pi }\times {10}^{-2}$

Hence, option $\left(\text{C}\right)$ is correct.