If the maximum distance of normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1,b<2,$ from the origin is 1, then the eccentricity of the ellipse is:

Equation of the normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ is $\frac{2x}{\cos \theta }-\frac{by}{\sin \theta }=4-b^2 \perp r$ distance from (0,0) is 

$=\left|\frac{4-b^2}{\sqrt{4{\sec }^2\theta +b^2\cos e{c}^2\theta }}\right|=\left|\frac{4-b^2}{\sqrt{(4+b^2)+4{\tan }^2\theta +b^2{\cot }^2\theta }}\right|\le \frac{4-b^2}{\sqrt{4+b^2+4b}}(\text{using\hspace{0.17em}\hspace{0.17em}}AM\ge GM)$  

$\Rightarrow b=1 \therefore e=\sqrt{\frac{4-1}{4}}=\frac{\sqrt{3}}{2}$