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A hemispherical light source emits unpolarized light with a sourcepower of 600W.

ID: 1754115 • Letter: A

Question

A hemispherical light source emits unpolarized light with a sourcepower of 600W. The light is incident on a stack of 3 polarizingdiscs (radius 2cm and 0.1mm thickness) located 2m from the source.The polarizing axes of the discs are the y-axis, 45 degrees fromthe origin, and the x-axis, respectively. The discks all touch eachother.
1. Calculate the intensity of the light incident on the firstplate. 2. Derive an expression for the transmitted intensity of thepolarizing stack as a function of time if the middle disc isrotating clockwise with a constant .Assum it startsfrom the y-axis. 3. If each disc has a specific heat similar to Tungsten and adensity similar to carbon, how long will it take to raise thetemperature of the first disc by 50 degree? 4. A container of water is placed justto the right of the third disc. The middle disc rotates with anangular speed of 10s-1. How long will ittake this light source to bring a 0.5L sample of water initially at40 degrees Celcius to a boil? (Consider the average power deliveredover a full period of rotation)
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1. Calculate the intensity of the light incident on the firstplate. 2. Derive an expression for the transmitted intensity of thepolarizing stack as a function of time if the middle disc isrotating clockwise with a constant .Assum it startsfrom the y-axis. 3. If each disc has a specific heat similar to Tungsten and adensity similar to carbon, how long will it take to raise thetemperature of the first disc by 50 degree? 4. A container of water is placed justto the right of the third disc. The middle disc rotates with anangular speed of 10s-1. How long will ittake this light source to bring a 0.5L sample of water initially at40 degrees Celcius to a boil? (Consider the average power deliveredover a full period of rotation)
Please showALL work! I promise to rate you aslifesaver if your answers are correct!

Explanation / Answer

1. To find the intensity some distance from a source of sphericalwaves, we use I=P/(4r2), but this is ahemispherical source so the energy is only being spread over halfas much area: I=P/(2r2) P=600W, r=2m so I=23.87W/m2. Let's call this I0 2. The intensity between the 1st and 2nd polarizers is I1= 0.5I0=11.94W/m^2, and the E field is polarised in the y direction. The intensity after the 2nd polariser isI2=I1cos2. Now the plate is rotating at a rate ofd/dt=, so we can integrate this and write(t)=t+C. The starting angle is 0 rad (it starts at they-axis which is parallel to the polarisation of the incoming light)so C=0 and (t)=t. Hence I2(t)=I1cos2(t), and the angle of the Efield is t. Now there is still a 3rd polarizer! If the absolute angle of theE-field is =t, the angle of the E-field with respect tothe 3rd polariser is =(90-t) SoI3(t)=I2cos2(90-t)=I1cos2(t)cos2(90-t)=11.94cos2(t)cos2(90-t)W/m^2 Now, I'm just going to submit this answer (my connection is alittle dodgy and I don't want to lose all my typing), but don'tworry, there's more to come.

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