Correct option: A.Infrared photon has more energy than the photon of visible light
a
Photographic plates uses photoelectric effect to produce the spectrum. So, if a electromagnetic ray is projected on it, it will be sensitive towards it.
Hence, its sensitive to both ultraviolet and infrared rays.
The infrared rays do cast shadows like visible rays as they are not able to pass through most opaque objects just like visible rays.
Now, we know that the energy of a electromagnetic wave is given by, $E =\frac{ hc }{\lambda}$
or, we can say that $E \propto \frac{1}{\lambda}$
Hence, if the wavelength decreases the energy will increase. Or, the lesser the wavelength the higher the energy. Infrared photons will have higher wavelength than that of visible light. Hence, its energy should be lesser than than visible light.