$V$-$I$ characteristic of a solar cell please explain the VI characteristics of a solar cell. The characteristics is given in my book without any explanation. How can the Voltage decrease on increasing current shouldn't it be opposite.
Solar Cell I-V characteristics

(Image  from Electrical 4 U - Characteristics of a Solar Cell and Parameters of a Solar Cell)
 A: If I invert your graph and change the direction of the current axis I get the graph in the middle.  

That is an $IV$ characteristic which looks very like that of a forward biassed diode and in fact the solar cell is a diode which is designed to act as a current source when illuminated with light, the current delivered by the current source being proportional to the solar radiance (intensity) falling on it.
If I flip the centre graph back to the orientation of your graph you will see that the only difference is the direction of positive current and an offset of $I_{\rm sc}$.  
The simple equivalent circuit os shown on the left with a current source $i_{\rm ph}$ and a forward biassed diode with a current $i_{\rm D}$ passing though it with $i=i_{\rm ph} -i_{\rm D}$ and the current $i_{\rm D}$ is very small.  
This model can be improved to better represent a real solar cell by the introduction of two resistor $R_{\rm p} \approx 1000 \,\Omega$ and $R_{\rm s} \approx 0.5\,\Omega$ as shown in the right hand diagram.

A: VI characteristic graph of diode with and without illumination (breakdown not shown):

When light falls on the depletion region of the diode, electron-hole pairs are generated that try to reach their parent nuclei. The electrons move towards the n-side and the holes to the p-side. This leads to a photovoltage that tries to make current flow in reverse direction.
Let $V_p$ be the photovoltage and $V$ be the external applied voltage. Then under illuminated conditions:
• If $V=0$, there's a negative current (i.e., a reverse current) in the circuit due to the photovoltage.
• In forward-biased, if $V_p>V$ then the current is negative and if $V>V_p$ then the current is positive.
• In reverse-bias ($V<0$), the photovoltage is largely responsible for the negative current; the current does not increase much on increasing the applied voltage (until reaching breakdown).
OP's graph just seems concerned with the magnitude of current in case of solar cell (quadrant 4 in my diagram).
A: A simple (but sufficient) model is to consider the sunlight as a stream of photons, each with their own energy depending on the colors in the sunlight. Each photon can give its energy to a single electron.
If you try to get out a small current from the solar cell, there are more than enough photons, but you can't change the energy of the photons. That limits the voltage.
If you try to get a high current out, you need a high voltage, but that gives a problem. Red photons only carry a limited amount of energy. Purple photons carry only a little bit more. So as you try to achieve a higher voltage, you quickly lose usable colors of photons. That's the sharp drop in current.
A: The characteristic is measured under illumination, when the solar cell is a source of electrical power, connected to a load. When the resistance of the load varies, one can get different points on the curve.
Compare with a battery: voltage decreases when a load is connected.
