Question

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

Answer: (C)

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

$\sin \theta =\sqrt{\frac{I}{I_0}}$

Differentiating the above equation,

$\cos \theta d\theta =\frac{1}{2}\frac{1}{I^{3/2}{I}_0^{1/2}}$

Thus $d\theta =\frac{1}{2I}\tan \theta dI$

$⟹\frac{d\theta }{\theta }=\frac{\tan \theta dI}{2\theta I}$

Put $\theta =30\times \frac{\pi }{180}radians$, $dI=0.002,I=5$,

Percentage error in angle $=\frac{d\theta }{\theta }\times 100$%

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