NCERT Exemplar · Class 12 Physics Wave Optics

Polarisation by reflection occurs when the angle of incidence is the Brewster's angle i.e. \(\displaystyle \tan \theta_B=\frac{n_2}{n_1}\) where \(\displaystyle n_2<n_1\). When light travels in such a medium the critical angle is \(\displaystyle \sin \theta_c=\frac{n_2}{n_1}\) where \(\displaystyle n_2<n_1\). As \(\displaystyle \left|\tan \theta_{\mathrm{B}}\right|>\left|\sin \theta_{\mathrm{c}}\right|\) for large angles, \(\displaystyle \theta_{\mathrm{B}}<\theta_{\mathrm{C}}\). Thus, polarisation by reflection shall definitely occur.

\(\displaystyle d_{\text {min }}=\frac{1.22 \lambda}{2 \sin \beta}\) where \(\displaystyle \beta\) is the angle subtended by the objective at the object. For light of \(\displaystyle 5500 \AA\) \[d_{\min }=\frac{1.22 \times 5.5 \times 10^{-7}}{2 \sin \beta} \mathrm{~m} \] For electrons accelerated through 100V the deBroglie wavelength is \[\begin{aligned} & \lambda=\frac{h}{p}=\frac{1.227}{\sqrt{100}}=0.13 \mathrm{~nm}=0.13 \times 10^{-9} \mathrm{~m} \\ & \therefore d_{\min }^{\prime}=\frac{1.22 \times 1.3 \times 10^{-10}}{2 \sin \beta} \end{aligned} \] \[\begin{aligned} & \therefore d_{\min }^{\prime}=\frac{1.22 \times 1.3 \times 10^{-10}}{2 \sin \beta} \\ & \frac{d_{\min }^{\prime}}{d_{\min }}=\frac{1.3 \times 10^{-10}}{5.5 \times 10^{-7}} \sim 0.2 \times 10^{-3} \end{aligned} \]

Without P: \[\begin{aligned} & A=A_{\perp}+A_{11} \\ & A_{\perp}=A_{\perp}^1+A_{\perp}^2=A_{\perp}^0 \sin (k x-\omega t)+A_{\perp}^0 \sin (k x-\omega t+\phi) \\ & A_{11}=A_{11}^{(1)}+A_{11}^{(2)} \\ & A_{11}=A_{11}^0[\sin (k x-w t)+\sin (k x-\omega t+\phi] \end{aligned} \] where \(\displaystyle \mathrm{A}_{\perp}^0, \mathrm{~A}_{11}^0\) are the amplitudes of either of the beam in \(\displaystyle \perp\) and $\displaystyle 11$ polarizations. \[\begin{aligned} & \therefore \text { Intensity }= \\ & =\left\{\left|A_{\perp}^0\right|^2+\left|A_{11}^0\right|^2\right\}\left[\sin ^2(k x-w t)\left(1+\cos ^2 \phi+2 \sin \phi\right)+\sin ^2(k x-\omega t) \sin ^2 \phi\right]_{\text {average }} \\ & =\left\{\left|A_{\perp}^0\right|^2+\left|A_{11}^0\right|^2\right\}\left(\frac{1}{2}\right) . 2(1+\cos \phi) \end{aligned} \] \[=2\left|\mathrm{~A}_{\perp}^0\right|^2 .(1+\cos \phi) \text { since }\left|\mathrm{A}_{\perp}^0\right|_{\text {average }}=\left|\mathrm{A}_{11}^0\right|_{\text {average }} \] With P: Assume \(\displaystyle \mathrm{A}_{\perp}^2\) is blocked: \[\begin{aligned} & \text { Intensity }=\left(A_{11}^1+A_{11}^2\right)^2+\left(A_{\perp}^1\right)^2 \\ & =\left|A_{\perp}^0\right|^2(1+\cos \phi)+\left|A_{\perp}^0\right|^2 . \frac{1}{2} \end{aligned} \] Given: \(\displaystyle I_0=4\left|\mathrm{~A}_{\perp}^0\right|^2=\) Intensity without polariser at principal maxima. Intensity at principal maxima with polariser \[\begin{aligned} & =\left|\mathrm{A}_{\perp}^0\right|^2\left(2+\frac{1}{2}\right) \\ & =\frac{5}{8} \mathrm{I}_0 \end{aligned} \] Intensity at first minima with polariser \[\begin{aligned} & =\left|\mathrm{A}_{\perp}^0\right|^2(1-1)+\frac{\left|\mathrm{A}_{\perp}^0\right|^2}{2} \\ & =\frac{I_0}{8} . \end{aligned} \]