Hydrolysis of sucrose gives:

We begin by recalling the structure of sucrose. Sucrose is a disaccharide containing two monosaccharide units:

$$\text{sucrose} = \alpha\text{-D-Glucopyranosyl-(1}\rightarrow 2\text{)-}\beta\text{-D-Fructofuranoside}$$

This description tells us two important facts. First, the glucose part is in the $$\alpha$$-pyranose form, because the anomeric carbon of glucose has the same configuration as the anomeric carbon of $$\alpha$$-D-glucose. Second, the fructose part is in the $$\beta$$-furanose form, because the anomeric carbon of fructose has the opposite configuration from $$\alpha$$-D-fructose and therefore matches $$\beta$$-D-fructose.

Now we look at what happens in hydrolysis. Hydrolysis means cleavage of the glycosidic bond in the presence of water (and usually an acid or the enzyme invertase):

$$\text{sucrose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+\;\text{or invertase}} \text{glucose} + \text{fructose}$$

Because the bond is broken exactly at the glycosidic junction, no other configuration changes occur at the anomeric carbons during the cleavage itself. Therefore, the two products keep the configurations that existed in the original disaccharide linkage:

$$\alpha\text{-D-Glucose} \;+\; \beta\text{-D-Fructose}$$

Next we attach their optical rotation signs. The D/L notation fixes only the configuration around the penultimate carbon, not the sign of optical rotation. For naturally occurring sugars we know:

$$\alpha\text{-D-Glucose} \;:\; [\alpha]_{D}^{20} = +52.5^\circ \text{ (positive, written as } (+) \text{)}$$

$$\beta\text{-D-Fructose} \;:\; [\alpha]_{D}^{20} = -92.4^\circ \text{ (negative, written as } (-) \text{)}$$

So, attaching the optical signs, we obtain:

$$\alpha\text{-D-(+)-Glucose} \;+\; \beta\text{-D-(-)-Fructose}$$

This pair exactly matches what is stated in Option B.

Hence, the correct answer is Option B.