NCERT Solutions · Class 12 Physics Ray Optics and Optical Instruments

Mirror formula: \(\displaystyle \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\), with \(\displaystyle f=\dfrac{R}{2}\).Concave mirror: \(\displaystyle R=36\ \mathrm{cm}\Rightarrow f=-18\ \mathrm{cm}\) (Cartesian sign convention, pole as origin, light travelling in \(\displaystyle -u\) direction taken negative for real objects). Object distance \(\displaystyle u=-27\ \mathrm{cm}\).\[\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-18}-\frac{1}{-27}=-\frac{1}{18}+\frac{1}{27}=-\frac{3}{54}+\frac{2}{54}=-\frac{1}{54}\]\(\displaystyle v=-54\ \mathrm{cm}\).The screen must be placed $\displaystyle 54$ cm in front of the mirror.Magnification: \(\displaystyle m=-\dfrac{v}{u}=-\dfrac{-54}{-27}=-2\).Image height \(\displaystyle =m\times h=-2\times2.5\ \mathrm{cm}=-5\ \mathrm{cm}\).The image is real, inverted, magnified $\displaystyle 2$×, of size $\displaystyle 5$ cm, formed $\displaystyle 54$ cm in front of the mirror (beyond the centre of curvature, since the object lies between F and C).As the candle is moved closer to the mirror (toward F at $\displaystyle 18$ cm), \(\displaystyle |v|\) increases and the image recedes further from the mirror, so the screen must be moved farther away from the mirror; when the candle reaches F the image forms at infinity, and if it is brought still closer (inside F) the image becomes virtual and no longer falls on a screen.

Convex mirror: \(\displaystyle f=+15\ \mathrm{cm}\) (behind the mirror, positive by convention), object distance \(\displaystyle u=-12\ \mathrm{cm}\).\[\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{-12}=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}=\frac{3}{20}\]\(\displaystyle v=\dfrac{20}{3}=6.67\ \mathrm{cm}\) (positive → behind the mirror, virtual).Magnification: \(\displaystyle m=-\dfrac{v}{u}=-\dfrac{6.67}{-12}=0.556\).Image height \(\displaystyle =0.556\times4.5\ \mathrm{cm}=2.5\ \mathrm{cm}\).The image is virtual, erect, and diminished, located $\displaystyle 6.67$ cm behind the mirror, of height $\displaystyle 2.5$ cm.As the needle is moved farther away, \(\displaystyle |v|\) approaches \(\displaystyle f=15\ \mathrm{cm}\) and \(\displaystyle m\) keeps decreasing — the image stays virtual and erect, moves progressively closer to the focal point behind the mirror, and shrinks further in size.

Apparent-depth relation for near-normal viewing: \(\displaystyle n=\dfrac{\text{real depth}}{\text{apparent depth}}\).\[n_{\text{water}}=\frac{12.5\ \mathrm{cm}}{9.4\ \mathrm{cm}}=1.330\]Refractive index of water \(\displaystyle \approx 1.33\).With the new liquid, \(\displaystyle n=1.63\), same real depth \(\displaystyle =12.5\ \mathrm{cm}\):\[\text{apparent depth}'=\frac{12.5}{1.63}\ \mathrm{cm}=7.67\ \mathrm{cm}\]Shift needed \(\displaystyle =9.4-7.67=1.73\ \mathrm{cm}\).The microscope must be moved up (toward the surface) by about $\displaystyle 1.73$ cm to refocus on the needle, since the image now appears nearer the surface in the denser liquid.

Light escapes the surface only within the cone of the critical angle \(\displaystyle \theta_c\), given by Snell's law at the water-air interface:\[\sin\theta_c=\frac{1}{n}=\frac{1}{1.33}=0.7519\ \Rightarrow\ \theta_c\approx48.75^\circ\]The illuminated circle on the surface has radius \(\displaystyle r=h\tan\theta_c\), where \(\displaystyle h=80\ \mathrm{cm}\) is the depth of the bulb.\[\cos\theta_c=\sqrt{1-\sin^2\theta_c}=\sqrt{1-0.5654}=0.6593,\qquad \tan\theta_c=\frac{0.7519}{0.6593}=1.140\]\[r=80\ \mathrm{cm}\times1.140=91.2\ \mathrm{cm}=0.912\ \mathrm{m}\]\[\text{Area}=\pi r^2=\pi\times(0.912\ \mathrm{m})^2\approx2.61\ \mathrm{m}^2\]The light emerges through a circle of area \(\displaystyle \approx 2.61\ \mathrm{m}^2\) on the water surface.

Prism formula at minimum deviation: \(\displaystyle n=\dfrac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\), with \(\displaystyle A=60^\circ\), \(\displaystyle D_m=40^\circ\).\[n=\frac{\sin 50^\circ}{\sin 30^\circ}=\frac{0.7660}{0.5}=1.532\]Refractive index of the prism glass \(\displaystyle \approx 1.53\).In water (\(\displaystyle n_w=1.33\)), the relevant refractive index is the glass-relative-to-water ratio:\[n_{rel}=\frac{n_{glass}}{n_{water}}=\frac{1.532}{1.33}=1.152\]\[\sin\left(\frac{A+D_m'}{2}\right)=n_{rel}\sin\left(\frac{A}{2}\right)=1.152\times0.5=0.576\]\[\frac{A+D_m'}{2}=\sin^{-1}(0.576)\approx35.2^\circ \ \Rightarrow\ A+D_m'\approx70.4^\circ\]\[D_m'\approx70.4^\circ-60^\circ\approx10.4^\circ\]The new angle of minimum deviation in water is about \(\displaystyle 10.4^\circ\) — much smaller than in air, since the effective refractive index of the glass relative to water is close to 1.

Lensmaker's equation for a double-convex lens with equal radii, \(\displaystyle R_1=+R\), \(\displaystyle R_2=-R\):\[\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(n-1)\left(\frac{1}{R}+\frac{1}{R}\right)=\frac{2(n-1)}{R}\]Given \(\displaystyle n=1.55\), \(\displaystyle f=20\ \mathrm{cm}\):\[\frac{1}{20}=\frac{2(1.55-1)}{R}=\frac{1.1}{R}\]\[R=1.1\times20\ \mathrm{cm}=22\ \mathrm{cm}\]Each face must have a radius of curvature of $\displaystyle 22$ cm.

The beam is convergent toward P, so P acts as a virtual object for the lens: with light travelling toward P, \(\displaystyle u=+12\ \mathrm{cm}\) (virtual object, on the outgoing side of the lens).Thin lens formula: \(\displaystyle \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\).(a) Convex lens, \(\displaystyle f=+20\ \mathrm{cm}\):\[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}=\frac{2}{15}\]\(\displaystyle v=\dfrac{15}{2}=7.5\ \mathrm{cm}\).The beam now converges $\displaystyle 7.5$ cm from the lens — i.e., $\displaystyle 4.5$ cm closer to the lens than P, since the convex lens adds convergence.(b) Concave lens, \(\displaystyle f=-16\ \mathrm{cm}\):\[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48}\]\(\displaystyle v=48\ \mathrm{cm}\).The beam converges $\displaystyle 48$ cm from the lens — farther than P, since the concave lens introduces divergence that delays the convergence point.

Concave lens: \(\displaystyle f=-21\ \mathrm{cm}\); object at \(\displaystyle u=-14\ \mathrm{cm}\), height \(\displaystyle h=3.0\ \mathrm{cm}\).\[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{21}-\frac{1}{14}=-\left(\frac{2}{42}+\frac{3}{42}\right)=-\frac{5}{42}\]\(\displaystyle v=-\dfrac{42}{5}=-8.4\ \mathrm{cm}\).Magnification: \(\displaystyle m=\dfrac{v}{u}=\dfrac{-8.4}{-14}=0.6\).Image height \(\displaystyle =0.6\times3.0\ \mathrm{cm}=1.8\ \mathrm{cm}\).The image is virtual, erect, and diminished, formed $\displaystyle 8.4$ cm from the lens on the same side as the object, of height $\displaystyle 1.8$ cm.As the object is moved farther away (\(\displaystyle |u|\to\infty\)), \(\displaystyle v\to f=-21\ \mathrm{cm}\): the image stays virtual and erect, moves progressively closer to the focus (approaching $\displaystyle 21$ cm from the lens), and keeps shrinking in size, never becoming real.

For thin lenses in contact, powers add: \(\displaystyle \dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}\).Convex lens \(\displaystyle f_1=+30\ \mathrm{cm}\), concave lens \(\displaystyle f_2=-20\ \mathrm{cm}\):\[\frac{1}{f}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\]\(\displaystyle f=-60\ \mathrm{cm}\).The combination has a focal length of \(\displaystyle -60\ \mathrm{cm}\), i.e. it behaves as a diverging lens of focal length $\displaystyle 60$ cm (negative power, since the concave lens's divergence dominates).