NCERT Exemplar · Class 12 Physics Ray Optics and Optical Instruments

(a) $\displaystyle 7.5$°. Normal emergence gives \(\displaystyle r_2=0\), so \(\displaystyle r_1=A\). \[r_1+r_2=A \;\Rightarrow\; r_1=5^\circ \] \[\sin i=n\sin r_1=1.5\times\sin 5^\circ=1.5\times 0.0872=0.131 \] \[i=\sin^{-1}(0.131)\approx 7.5^\circ \]

(d) Red. Normal incidence means no bending, so every colour crosses the same thickness \(\displaystyle d\); the colour with the smallest \(\displaystyle n\) is fastest. \[v=\frac{c}{n},\qquad t=\frac{d}{v}=\frac{nd}{c} \] \[n_{\mathrm{red}}<n_{\mathrm{green}}<n_{\mathrm{blue}}<n_{\mathrm{violet}} \;\Rightarrow\; t_{\mathrm{red}}<t_{\mathrm{green}}<t_{\mathrm{blue}}<t_{\mathrm{violet}} \]

(c) The image moves away from the lens with non-uniform acceleration. Let the object distance be \(\displaystyle x=|u|>f\) (\(\displaystyle f\), \(\displaystyle x\) in m), \(\displaystyle u=-x\). \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\;\Rightarrow\; \frac{1}{v}+\frac{1}{x}=\frac{1}{f}\;\Rightarrow\; v=\frac{xf}{x-f} \] \[v_I=\frac{dv}{dt}=-\frac{f^{2}}{(x-f)^{2}}\frac{dx}{dt},\qquad \frac{dx}{dt}=-5\ \mathrm{m\,s^{-1}} \] \[v_I=\frac{5f^{2}}{(x-f)^{2}}\ \mathrm{m\,s^{-1}}\ (>0),\qquad x\downarrow\Rightarrow v_I\uparrow \] \[a_I=\frac{dv_I}{dt}=\frac{50f^{2}}{(x-f)^{3}}\ \mathrm{m\,s^{-2}}>0,\ \text{not constant} \]

(b) Both bows are cones about the antisolar axis, at about \(\displaystyle 42^\circ\) (primary) and \(\displaystyle 51^\circ\) (secondary). \[\theta_{\mathrm{primary}}\approx 42^\circ,\qquad \theta_{\mathrm{secondary}}\approx 51^\circ \] Ground observer: horizon cuts the cone \(\displaystyle \Rightarrow\) arcs. Aeroplane: nothing cuts it \(\displaystyle \Rightarrow\) full concentric circles.

(c) Yellow is at the critical angle, so a colour of larger \(\displaystyle n\) has a smaller critical angle and is totally reflected. Red refracts away from the normal; green also reflects totally. \[\sin i_c=\frac{1}{n},\quad n=\frac{n_{\mathrm{denser}}}{n_{\mathrm{rarer}}},\quad \sin i=\sin i_c(\text{yellow})=\frac{1}{n_{\mathrm{yellow}}} \] \[n_{\mathrm{red}}<n_{\mathrm{yellow}}<n_{\mathrm{green}}<n_{\mathrm{blue}} \] \[\text{red: } i_c\uparrow\Rightarrow i<i_c\Rightarrow \text{refracts},\ r>i \] \[\text{green, blue: } i_c\downarrow\Rightarrow i>i_c\Rightarrow \text{total internal reflection} \]

(c) Convex either way: \(\displaystyle f\) is positive and the same for both orientations. \[\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] Curved side first, \(\displaystyle R_1=+20\ \mathrm{cm}\), \(\displaystyle R_2=\infty\): \[\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-0\right)=\frac{1}{40}\ \mathrm{cm^{-1}}\;\Rightarrow\; f=+40\ \mathrm{cm} \] Plane side first, \(\displaystyle R_1=\infty\), \(\displaystyle R_2=-20\ \mathrm{cm}\): \[\frac{1}{f}=0.5\left(0-\frac{1}{-20}\right)=\frac{1}{40}\ \mathrm{cm^{-1}}\;\Rightarrow\; f=+40\ \mathrm{cm} \]

(b) In both, the refractive index falls along the path, so the ray bends away from the normal until it is totally reflected.\[n\sin\theta=\text{constant along the ray} \] \[n\downarrow\ \Rightarrow\ \sin\theta\uparrow\ \Rightarrow\ \theta\to90^\circ\ \Rightarrow\ \text{ray turns back} \]Ionosphere: \(\displaystyle n\) falls with height. Mirage: \(\displaystyle n\) falls toward the hot ground.

(b) Ray 2. The incident ray passes through the focus, so it reflects parallel to the principal axis, away from the mirror (ray $\displaystyle 4$ runs into it).Source at \(\displaystyle F\): \(\displaystyle u=f\) \[\frac1v=\frac1f-\frac1u=\frac1f-\frac1f=0\ \Rightarrow\ v=\infty \]

(b) Ray $\displaystyle 2$ bends toward the normal entering turpentine, then away from it into water. Ray $\displaystyle 1$ crosses the normal, ray $\displaystyle 3$ is undeviated, ray $\displaystyle 4$ bends away on entering turpentine.\[\sin i=n_t\sin r_t=n_w\sin r_w,\qquad n_t>n_w>1 \] \[\Rightarrow\ r_t<r_w<i \]

(a) The refracted ray lies on the same side of the normal as the incident ray.\[n_1\sin i=n_2\sin r,\qquad n_1=1,\ n_2<0 \] \[\sin r=\frac{\sin i}{n_2}<0\ \Rightarrow\ r<0 \]