Question

2. An unpolarized beam of light with intensity 10 = 98.8 lux is incident on three ideal polarizing filters (lux = 2.). The axis of the first filter (filter A) makes an angle of 18° with the vertical and the axis of the second filter (filter B) makes an angle of 37° with the vertical. The third filter (filter C) is oriented so that its axis is horizontal. a. What is the intensity and polarization direction of the light after it passes through the first filter? b. What is the intensity of the light after it passes through the first and second polarizing filters? c. What is the intensity of the light after it passes through all three polarizing filters? Note: Light passes polarizing filters in the following order A, B, and C. d. At what two angles, with respect to the vertical, would you need to orient the axis of filter C so that the intensity of the light after it passes all three filters is 0.0 lux? Note: Light passes polarizing filters in the following order A, B, and C.

Answer 1

• When unpolarized light beam passes through a polarizing filter, Putensity of transmitted light will be a where I. is the ineensity of incident light · Transmitted light will be polarized in the direction of axis of filter. • When polarized light passes through polarizing filter intensity of transmitted light is given by I= Is Costa ( Law of MALUS) Here Io = Intensity of incident light between plane of polarization of incident light and axis of filter. a) The incident light is unpolarized, when it passes through its intensity will be halved. Thus intensity of transmitted light I = To 98.8 lux =149.4 lux Transmitted light will be polarized along the axis of filter. Thus polarization direction at 18° from vertical b). light transmitted through the first filter is polarized. Thus Law of Malus can be applied. Put Io=49.4 lux and o=37-18° = 119° in expression for Malus law I= 49.4 lux Cosigo = (49.4 lux) (0.894) = 144.16 lux) c) Put Io = 44.16 luy and 8 = 90°- 379= 153') in expression for I= 44.16 cog53º = (44.16 lux ) (0.362) = 115.98 lux Malus law

1) Put I=0, Io = 44.16 lux in the expression for malus law 0 = I. Cosa → Cose = 0 => 1 Cose = 0 = [@=+90°, -9.0 / Here 0 = Angle between axis of polarizing filter B and C But second filter makes an angle of 37° with vertical Then angle between vertical and axis offilter c will be : 90+37° - 1137° || and (90°) +37° = 753° )