Unpolarised light through two polaroids at 60°: I get I0/4 but the key says I0/8. What am I missing?

7 answers & replies

• [+][–]Tanvi Deshmukh12thSample· 16d ago

  You missed the first polaroid! Unpolarised light has all directions equally, so a polaroid transmits exactly half of $I_0$, no matter the orientation. Polarised light of intensity $I_0/2$ then falls on the second one, and only then does Malus' law apply:

  $I = \frac{I_0}{2}\cos^2 60° = \frac{I_0}{2}\times\frac14 = \frac{I_0}{8}$

  Your second answer is right too: $\frac{I_0}{2}\cos^245°\cos^245° = \frac{I_0}{2}\times\frac12\times\frac12 = \frac{I_0}{8}$.

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  • [+][–]Farhan ShaikhDropperSample· 16d agoAuthor

    the "half for unpolarised, always" was the piece missing. also the cos vs cos² cleared up, thanks everyone

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• [+][–]Pooja Iyer12thSample· 16d ago

  Calculator-free checks for Malus' law (these multiply the intensity of the polarised light falling on the analyser): $\theta = 0°$ gives full, $90°$ gives zero, $45°$ gives half, $30°$ gives $\cos^230° = 3/4$, and $60°$ gives $\cos^260° = 1/4$. Mnemonic I use (dramatically): "half the unpolarised, then cos squared". So for your 60° question: $\frac12\times\frac14 = \frac18$.

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• [+][–]Siddharth RaoDropperSample· 16d ago

  On the cos vs cos² question: it's $\cos^2$ for the intensity. The electric field amplitude after the analyser is $E_0\cos\theta$ (only the component along the pass axis gets through) and intensity is proportional to $E^2$, so $I = I_0\cos^2\theta$. I made exactly this error in a test once: used cos instead of cos².

  On the three polaroids: the first polariser makes the light polarised along its axis. The last one, being crossed, would block it all. The middle one rotates the polarisation direction by 45°, so the light now has a component along the last axis. So polaroids don't just "block", they project.

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  • [+][–]Pooja Iyer12thSample· 16d ago

    oof, noted: amplitude has cos, intensity has cos². writing it on my hand 🙈

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• [+][–]Rahul DubeyDropperSample· 16d ago

  Brewster. At the Brewster angle the reflected light is completely plane polarised (electric vector perpendicular to the plane of incidence) and the reflected and refracted rays are at 90° to each other, so $i_B + r = 90°$. The refracted light is partially polarised, not fully, because only a part of the perpendicular component is reflected.

  Numerical: glass of $n = 1.5$:

  $i_B = \tan^{-1}(1.5) \approx 56.3°$

  You can derive $\tan i_B = n$ from Snell's law with $r = 90° - i_B$: $n = \sin i_B/\sin r = \sin i_B/\cos i_B$.

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• [+][–]Meera Patel11thSample· 16d ago

  A useful remark for the "why do we study this" part: polarisation proves that light is a transverse wave. Sound waves in air are longitudinal and can't be polarised. Polaroid sunglasses cut glare because light reflected from a road or water is partially polarised, and the lenses block that polarisation direction.

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