Question
Intensity observed in an interference pattern is $$I = {I_0}{\sin ^2}\theta .$$ At $$\theta = {30^ \circ }$$ intensity $$I = 5 \pm 0.0020\,W/{m^2}.$$ Find percentage error in angle if $${I_0} = 20\,W/{m^2}.$$
A.
$$\frac{4}{\pi }\sqrt 3 \times {10^{ - 2}}\% $$
B.
$$\frac{2}{\pi }\sqrt 3 \times {10^{ - 2}}\% $$
C.
$$\frac{1}{\pi }\sqrt 3 \times {10^{ - 2}}\% $$
D.
$$\frac{3}{\pi }\sqrt 3 \times {10^{ - 2}}\% $$
Answer :
$$\frac{4}{\pi }\sqrt 3 \times {10^{ - 2}}\% $$
Solution :
$$\eqalign{
& I = {I_0}{\sin ^2}\theta \Rightarrow \theta = {\sin ^{ - 1}}\sqrt {\frac{I}{{{I_0}}}} \cr
& d\theta = \frac{1}{2}\frac{{\sqrt {{I_0}} }}{{\sqrt {{I_0} - I} }}\frac{{dI}}{{\sqrt {{I_0}} \sqrt I }} \cr
& \frac{{d\theta }}{\theta } = \frac{1}{2}\frac{{dI}}{{\sqrt {I\left( {{I_0} - I} \right)} {{\sin }^{ - 1}}\sqrt {\frac{I}{{{I_0}}}} }} \cr
& \therefore \% \,{\text{error}} = \frac{4}{\pi }\sqrt 3 \times {10^{ - 2}}\% \cr} $$