Hershel experiment regarding IR radiation

[mildpiranha]

I have been struggling with the following; (apologies if this is a simple question)

After reading the Hershel experiment where with a prism light was used to split white light into a spectrum of colours and the red light was warmer than the violet light, then he placed a detector past the red and discovered infrared where his detector registered a higher temperature.

My question is why does the red light have a higher temperature than the violet, when the violet light has a higher energy?

Also if he placed a detector past the violet section in the ultraviolet region would he have detected a colder temperature?

Thank you for your help

 

[mfb]

Light does not have a temperature.

The temperature of the thermometer depends on the light intensity (and the absorption coefficient of the surface). Sunlight has a strong infrared component, the intensity in the ultraviolet range is low (and below a certain wavelength, the prism will just absorb the light).

The energy per photon is larger for ultraviolet light.

 

[mildpiranha]

Thank you for your response.

To clarify does that mean that in the case of this experiment that the red light increased the temperature of the thermometer more than the violet light did as its intensity was greater?

So, if I had equal intensities of red light on one thermometer and violet light on another thermometer would the thermometer which had the violet light on have a greater intensity due to the greater energy of violet light?

 

[mfb]

To clarify does that mean that in the case of this experiment that the red light increased the temperature of the thermometer more than the violet light did as its intensity was greater?

Right.

So, if I had equal intensities of red light on one thermometer and violet light on another thermometer would the thermometer which had the violet light on have a greater intensity due to the greater energy of violet light?

Not in general. Intensity already accounts for this difference, its unit is W/m^2 (power per area). The area of the thermometer stays the same, and the absorbed power determines the temperature reading.

A quantative analysis has to be more careful: The thermometer will always receive light in some wavelength range, where the size of this range depends on the wavelength-dependency of the refractive index of the prism, and the received power depends on this size. Also, not the total intensity hitting the thermometer will get absorbed, and this fraction can depend on the wavelength. But those are technical details - the interesting result of the experiment was the notable heating effect where no light was visible.