Question

if 1.0 mm diameter argon ion laser beam 514.5nm propagation through the air at reflective index n =1.000 passes through a series of three 10.0mm thick pieces of SI-UV suprasil fused silica reflective index (n_{= 1.46156) at normal incidence , what is the intensity of the beam that exits the attenuator? suppose that the intensity of light passing through this attenuator is used a measurment of} p₀ for absorption experment . if two of three quartiz flats are mistakenly replaced with flats made of BK-7 glass reflective index ( n= 1.52049) and calculate the relative error introduced into transmission measurment assuming that P₀ remain constant. in this question the attenuator is modified, where the intensity of light is attenuated by passage through only two pieces of suprasil SI-UV suprasil fused silica however the angle of incidence in the modified attenuator of the beam is now variable, calculate the intensity of light varies with angle of incidence of the light beam and draw the diagram of the modified attenuator and show how the argon ion laser beam is affected by its passage through the device?

Answer 1

According to Malus law, the intensity (I) of light transmitted by the analyzer is given by,

I = I₀cos²θ

where I₀ is the intensity of the light incident on the analyzer.

When θ = 90°, I = I₀ cos²90° = 0

So, the intensity of light transmitted by the analyzer is minimum when the transmission axes of the analyzer and polarizer are perpendicular to each other. Same is this case, the argon ion laser beam propagate through the air at reflective index n =1.000 and passes through a series of three 10.0mm thick pieces of SI-UV suprasil fused silica reflective index (n= 1.46156) at normal incidence. Thus the intensity of exit beam will be near to zero.