Question

Unpolarized light has intensity of 100 W/m2, this light goes through three filters. First filter- angle of 15 degrees clockwise to vertical, second filter- angle of 15 degrees counterclockwise to vertical, third filter- angle of 5 degrees counterclockwise to vertical. 1) What is final intensity and also the new polarized light has what overall orientation?

2) Once the light has gone through the filter, what is the electric and magnetic field rms values of the light

3) If wavelength equals 100 nm, if the pupil has a diameter of 5mm how many photons would hit a human eye in 2 secs.

Answer 1

In first part we used malus law. The we used intensity basic defination to find number of photons.

Answer: Formulae we will use : et eis when unpolarised Light passes through filter , it gets polarized and intensity becomes half of initial intensity. (i) when polarised light falls on filter the intensity of outcoming eight is given. by malus eight formula - [I = Io cos?p] angle b/w Plane of polarization of Lase light and principal axis of filter. ; Livintensity of light coming from filter 1 Z; = Io = 50 W/m2 (ii) Intensity from filter 2: Ig = I, cos? = 50 cos? 300 o filter I is 150 clockwise and filter 2 is 15° anticlockwise. So net angle b/w them is 30°, s on Iz = 50% (13,4 = 50x 3 = 37:5W/m2 ose 4 - cliis intensity after Filter 3 : I3 = Iz cos? 100 filter 2 = 15° a clock sl filter 3 = 5° a. clock not = 36.37 W/m2 net angle =100 Final I = 36.37 w/ m2 = M

We know, I = k Eo? (Eo= max. E'F) 5. I = Eo Eo = 136.37 = 6:03 Now, Erms = Eo. = 4.26 Erms = 4. 26 N/C = tinoto Brms - Eyms /c (C = 3x108)

© intensity = Energy per unit area persee. energy of I Photon = hc = 6.626x3x103478 210 -7 (x = loonm = 107m) = 19.878 x 10-19 J e Intensity = (No. 06 photons) Cenergy of Photon) COST Area x Time... No. of photon - I XAXT E = 36.37x78.5x10x2. x 19.88 X10-19 No. of photons & 1013 Here A = 4 T7R? - 4x 3.14 x(2.57x10-6 TERSEY = 78.5 x 10-6 m2 T = a sec. -