Unpolarized light of intensity $I_0$is incident on a stack of polarizing filters, each with its axis rotated $15˚cw$with respect to the previous filter. What light intensity emerges from the last filter?

Electromagnetic waves, or EM waves, are waves that are formed when an electric field and a magnetic field vibrate together. EM waves, in other words, are made up of oscillating magnetic and electric fields.

The strength or amount of light produced by a certain lighting source is referred to as light intensity. It is a measurement of a light source's wavelength-weighted power.

$$\begin{gathered} Givenvalues: \\ \theta =15˚ \\ Weknowthat: \\ {I}_1=\left(1/2\right)I_0 \\ Weuseformula: \\ {I}_2=I_1{\cos }^2\theta \\ Frompreviousformula,wehavetofindvaluefor{I}_7: \\ First,wefindvaluefor{I}_2: \\ {I}_2=I_1{\cos }^2\theta \\ {I}_2=\raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\cos }^2\left(15˚\right) \\ {I}_2=5\times 0.933 \\ {I}_2=4.66(Substitutevaluesinequation.) \end{gathered}$$

$$\begin{gathered} I_3=I_2{\cos }^2\theta \\ {I}_3=4.66{\cos }^2(15˚) \\ {I}_3=4.66\times 0.933 \\ {I}_3=4.34 \end{gathered}$$

$$\begin{gathered} I_4=I_3{\cos }^2\theta \\ {I}_4=4.34{\cos }^2(15˚) \\ {I}_4=4.34\times 0.933 \\ {I}_4=4.05 \end{gathered}$$

$$\begin{gathered} I_5=I_4{\cos }^2\theta \\ {I}_5=4.05{\cos }^2(15˚) \\ {I}_5=4.05\times 0.933 \\ {I}_5=3.78 \end{gathered}$$

$$\begin{gathered} I_6=I_5{\cos }^2\theta \\ {I}_6=3.78{\cos }^2(15˚) \\ {I}_6=3.78\times 0.933 \\ {I}_6=3.53 \end{gathered}$$

$$\begin{gathered} I_7=I_6{\cos }^2\theta \\ {I}_7=3.53{\cos }^2(15˚) \\ {I}_7=3.53\times 0.933 \\ {I}_7=3.29 \end{gathered}$$

The result for the light of intensity $I_0$is incident on a stack of $7$polarizing filters and each axis rotated $15˚cw$from previous filter to last filter is .