Question

A) You are given a dye with an excitation maximum of 358 nm and an emission maximum of 461 nm, and your fluorescence microscope has fluorescence cubes with the filters below. Do you think you will you be able to see any fluorescence from this dye using any of these fluorescence cubes? Explain your answer. B) Does it make a difference if the barrier filter is a band pass versus a long pass? Why?

Filter cube Excitation (exciter) Emission (barrier) Dichromatic Mirror

WU -> MWU2 330-385 420 400

WB -> MWB2 460-490 520 500

WG -> MWG2 510-550 590 700

Answer 1

A) I think we shall be able to see the fluroscence from the given dye whose excitation is taking place at 358nm and emission is 461nm by using the first dichromatic mirror filter WU-> MWU2 330-385 420 400. The fluroscence filters are strategically oriented within a specialized cube or block that enables the illumination to enter from one side and pass to and from the specimen in defined directions along the microscope optical axis. Now we are showing the transmission spectra as follows in figure 1 and in figure 2

On looking at these figures we can clearly see that the transmission spectra corresponds to the given excitation and emission wavelengths. Fig 1 shows that the filter is Dichoric and 400LP so it meets the specifications for the first option.

B) Yes it will make a difference as the long pass filter allows the transmission for longer wavelengths while bandpass allows the transmission for a certain band of wavelengths. The given wavelengths correspond to blue colour regime. If we want to eliminate the noise that is not spectrally overlapping with the signal then the bandpass filter is a good choice. On the other hand, when the noise spectra is overlapping with the signal than long pass filter is a good option to go with.