Given ellipse

$\frac{x^2}{4}+\frac{y^2}{2}=1$

Compare with the standard form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

so,

$a^2=4, \qquad b^2=2.$

Use the midpoint form of a chord of a conic

For the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$

the chord whose midpoint is $(x_1,y_1)$ is given by

$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}.$

Here midpoint is

$(x_1,y_1)=\left(1,\frac12\right).$

Therefore the chord is

$\frac{x\cdot 1}{4}+\frac{y\cdot (1/2)}{2}=\frac{1^2}{4}+\frac{(1/2)^2}{2}.$

Simplify:

$\frac{x}{4}+\frac{y}{4}=\frac14+\frac18=\frac38.$

Hence,

$x+y=\frac32.$

Find the endpoints of the chord

Solve the system:

$x+y=\frac32 \implies y=\frac32-x.$

Substitute into the ellipse:

$\frac{x^2}{4}+\frac{(\frac32-x)^2}{2}=1.$

Multiply by 4:

$x^2+2\left(\frac32-x\right)^2=4.$

Expand:

$x^2+2\left(x^2-3x+\frac94\right)=4$ $x^2+2x^2-6x+\frac92=4$ $3x^2-6x+\frac12=0.$

Multiply by 2:

$6x^2-12x+1=0.$

So the two roots are the $x$-coordinates of the endpoints. Their difference is

$|x_1-x_2|=\frac{\sqrt{(-12)^2-4\cdot 6\cdot 1}}{6}=\frac{\sqrt{144-24}}{6}=\frac{\sqrt{120}}{6}=\frac{2\sqrt{30}}{6}=\frac{\sqrt{30}}{3}.$

Since the chord lies on the line $x+y=\frac32$, we have slope $-1$, so for endpoints,

$\Delta y=-\Delta x.$

Therefore chord length is

Final answer

$\boxed{\frac{2}{3}\sqrt{15}}$

This corresponds to Option A.