Exactly. I ran some back-of-the-envelope calculations on this 3.5 years ago in another Slashdot thread. https://slashdot.org/comments....
And because we're presumably too lazy to click that link, I'll paste it below for your reading pleasure...
This is why: http://what-if.xkcd.com/17/ [xkcd.com]
There simply isn't enough solar power delivered to the surface of the aircraft, even at 100% conversion efficiency, to move people and luggage using only available sunlight.
Google tells me direct illumination to a surface perpendicular to incoming full intensity sunlight is about 1.4 kW per square meter. Google also tells me that the wing surface area of a 747 is around 5500 square feet. Only half of the 747 wing is directly illuminated by sunlight at any given moment, but the surface of the fuselage could be covered with photocells as well, so 5500 square feet overall is probably a decent estimate for the directly illuminated surface area of the aircraft as a whole. And for hand-wavy purposes lets assume that the entire surface of the 747 is perpendicular to the incoming sunlight (i.e. a planar plane... pun totally intended). And that we have perfectly efficient photocells giving us 100% conversion efficiency. Running the math, this gives us around 715kW under bright direct sunlight, or about 959 horsepower -- the equivalent of 1.5 2012 Ford Shelby GT500's.
Each engine of a 747 generates around 15,000 horsepower at cruise, and around 30,000 at takeoff, and a 747 has four engines. So you need around 125 times the power generated by a perfectly efficient perfectly illuminated solar-powered 747 to get said plane off the ground, and around 65 times the power for cruising. And then you could only fly it in the middle of the day near the equator.