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  1. Problems
  2. /A student looks at the reflection of a lamp in a flat sheet of a transparent plastic in air…
MediumPhysicsHigh School (11–12)Optics › Wave OpticsNumericalConceptualNEETBoard ExamAPA-LevelIBAEEE (Amrita)
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A student looks at the reflection of a lamp in a flat sheet of a transparent plastic in air…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A student looks at the reflection of a lamp in a flat sheet of a transparent plastic in air. She finds that the reflected light is completely plane polarised when the reflected ray makes an angle of 36.0∘36.0^\circ36.0∘ with the surface of the plastic. Find the angle of incidence and the refractive index of the plastic. Then find the critical angle of the plastic, and the angle of incidence at which light inside the plastic is completely polarised when it is reflected at the plastic-air surface.

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The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The angle of incidence. The normal is 90∘90^\circ90∘ from the surface. So iB=90∘−36.0∘=54.0∘i_B = 90^\circ - 36.0^\circ = 54.0^\circiB​=90∘−36.0∘=54.0∘.

Step 2. The index. n=tan⁡iB=tan⁡54.0∘=1.38n = \tan i_B = \tan 54.0^\circ = 1.38n=taniB​=tan54.0∘=1.38.

Step 3. The critical angle. sin⁡C=1n=11.38=0.727\sin C = \dfrac{1}{n} = \dfrac{1}{1.38} = 0.727sinC=n1​=1.381​=0.727, so C=46.6∘C = 46.6^\circC=46.6∘.

Step 4. Brewster's angle inside. For light going from the plastic to air, tan⁡iB′=1n=0.727\tan i_B' = \dfrac{1}{n} = 0.727taniB′​=n1​=0.727, so iB′=36.0∘i_B' = 36.0^\circiB′​=36.0∘.

Answer: 54.0∘54.0^\circ54.0∘; 1.381.381.38; 46.6∘46.6^\circ46.6∘; 36.0∘36.0^\circ36.0∘.

Check. The Brewster angle inside, 36.0∘36.0^\circ36.0∘, is smaller than the critical angle, 46.6∘46.6^\circ46.6∘, so it is possible. It also equals the angle of refraction r=90∘−iB=36.0∘r = 90^\circ - i_B = 36.0^\circr=90∘−iB​=36.0∘ of the first part.

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Brewster's law for light that goes from air to a material of index nnn is tan⁡iB=n\tan i_B = ntaniB​=n. At iBi_BiB​ the reflected light is completely plane polarised, and iB+r=90∘i_B + r = 90^\circiB​+r=90∘. Small particles scatter light in proportion to 1/λ41/\lambda^41/λ4. The critical angle has sin⁡C=1/n\sin C = 1/nsinC=1/n.

tan⁡iB=n\displaystyle \tan i_B = ntaniB​=n
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