Exercises — Thin lenses — lens equation, lens maker's equation

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\text{(single surface)}$ $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\text{(lens maker)}$ $\frac{1}{v}-\frac{1}{u}=\frac{1}{f},m=\frac{v}{u}\text{(lens equation \& magnification)}$ Sign rules (Cartesian, light travels →): real object $u<0$; real image $v>0$; converging $f>0$; a surface has $R>0$ when its centre of curvature is on the outgoing side.

What we are asked: the sign of $f$ from shape alone. First surface: centre on outgoing side $\Rightarrow R_1>0$. Second surface: centre on incoming side $\Rightarrow R_2<0$. Then $\frac{1}{R_1}-\frac{1}{R_2}=(\text{positive})-(\text{negative})=\text{positive}$, and $(n-1)>0$. So $\frac{1}{f}>0\Rightarrow f>0\Rightarrow$ converging. This is a biconvex lens — the classic magnifying-glass shape.

Object shape/glass is irrelevant here — we already have $f$. Use the lens equation $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. "40 cm in front" = real object on the incoming side $\Rightarrow u=-40$ cm. (We are not asked to solve yet — just to recognise the tool and the sign.)

$\frac{1}{R_1}=0$ and $\frac{1}{R_2}=0$, so $\frac{1}{f}=(n-1)(0-0)=0$. Zero power, $f=\infty$: it shifts light sideways but does not focus. This is the sanity check that flat glass "does nothing."

$\frac{1}{f}=(1.5-1)\left(\frac{1}{15}-\frac{1}{-25}\right)=0.5\left(\frac{1}{15}+\frac{1}{25}\right).$ $\frac{1}{15}+\frac{1}{25}=\frac{5+3}{75}=\frac{8}{75}$, so $\frac{1}{f}=0.5\cdot \frac{8}{75}=\frac{4}{75}{\text{cm}}^{-1}$. $f=\frac{75}{4}=+18.75$ cm. Positive $\Rightarrow$ converging. ✔

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{4}{75}+\frac{1}{-30}.$ Common denominator $150$: $\frac{4}{75}=\frac{8}{150}$, $\frac{1}{30}=\frac{5}{150}$. $\frac{1}{v}=\frac{8-5}{150}=\frac{3}{150}=\frac{1}{50}\Rightarrow v=+50$ cm (real, outgoing side). $m=\frac{v}{u}=\frac{50}{-30}=-\frac{5}{3}\approx -1.67$: inverted, magnified $1.67\times$. ✔ Look at the ray diagram below — the object is beyond $f$, so we get a real, inverted, enlarged image.

$\frac{1}{v}=\frac{1}{-20}+\frac{1}{-12}=\frac{-3-5}{60}=-\frac{8}{60}=-\frac{2}{15}\Rightarrow v=-7.5\text{cm}.$ Negative $v\Rightarrow$ virtual image on the incoming side. $m=\frac{v}{u}=\frac{-7.5}{-12}=+0.625$: erect and diminished — exactly what a diverging lens always does for a real object. ✔

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}+\frac{1}{-20}=0.$ $\frac{1}{v}=0\Rightarrow v=\infty$: the rays leave parallel and never meet. No image forms at any finite distance. This is the reversal of "parallel in → focus": here we put the source at the focus, so it comes out collimated.

$\frac{1}{v}=\frac{1}{20}+\frac{1}{-10}=\frac{1-2}{20}=-\frac{1}{20}\Rightarrow v=-20\text{cm}.$ Negative $\Rightarrow$ virtual, on the incoming side. $m=\frac{v}{u}=\frac{-20}{-10}=+2$: erect, magnified $2\times$. This is the magnifying-glass mode of a converging lens. ✔

Symmetric biconvex: $\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{R}-\frac{1}{-R}=\frac{2}{R}$. $\frac{1}{f}=(n-1)\frac{2}{R}\Rightarrow R=2(n-1)f=2(0.6)(40)=48\text{cm}.$ Each face has radius $48$ cm. ✔ Notice the two faces add through the minus sign — the recurring trap of the maker's equation.

Surface 1 (air→glass, $n_1=1,n_2=1.5$), parallel input $u\to -\infty$ so $\frac{1}{u}\to 0$: $\frac{1.5}{v_1}-0=\frac{1.5-1}{20}=\frac{0.5}{20}=\frac{1}{40}\Rightarrow \frac{1.5}{v_1}=\frac{1}{40},v_1=60\text{cm}.$ Surface 2 (glass→air, $n_1=1.5,n_2=1$), object is the surface-1 image at $u_2=v_1=60$ cm (thin lens, no gap): $\frac{1}{v}-\frac{1.5}{60}=\frac{1-1.5}{-20}=\frac{-0.5}{-20}=\frac{1}{40}.$ $\frac{1.5}{60}=\frac{1}{40}$, so $\frac{1}{v}=\frac{1}{40}+\frac{1}{40}=\frac{1}{20}\Rightarrow v=20$ cm. Parallel in → focus at $20$ cm, so $f=+20$ cm — matching the maker's equation directly. ✔ The intermediate image $v_1$ did its job and cancelled out.

Real, same-size image $\Rightarrow m=-1$ (real images through a single lens are inverted). $m=\frac{v}{u}=-1\Rightarrow v=-u$. Plug into the lens equation: $\frac{1}{-u}-\frac{1}{u}=\frac{1}{f}\Rightarrow -\frac{2}{u}=\frac{1}{12}\Rightarrow u=-24$ cm. Then $v=-u=+24$ cm. Both distances equal $2f$: the "$2f$–$2f$" symmetric imaging condition. ✔

(a) $f$: $\frac{1}{f}=(0.5)\left(\frac{1}{10}-\frac{1}{-10}\right)=0.5\cdot \frac{2}{10}=0.1\Rightarrow f=+10$ cm. (b) $v$: $\frac{1}{v}=\frac{1}{10}+\frac{1}{-15}=\frac{3-2}{30}=\frac{1}{30}\Rightarrow v=+30$ cm (real, outgoing). (c) $m$: $m=\frac{v}{u}=\frac{30}{-15}=-2$. (d) height: $h^{\prime }=mh=(-2)(2)=-4$ cm, i.e. $4$ cm tall, inverted. (e) $v>0\Rightarrow$ real; $m<0\Rightarrow$ inverted, magnified $2\times$. ✔

Algebra. Real object: $u=-|u|$. From $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{f}-\frac{1}{|u|}=\frac{|u|-f}{f|u|}$. The denominator $f|u|>0$, so the sign of $\frac{1}{v}$ (hence of $v$) equals the sign of $|u|-f$.

• If $|u|>f$: numerator $>0\Rightarrow v>0\Rightarrow$ real.
• If $|u|<f$: numerator $<0\Rightarrow v<0\Rightarrow$ virtual.
• If $|u|=f$: numerator $=0\Rightarrow v=\infty$ (the L3.1 boundary). Numeric check, $f=20$: $u=-25$ ($|u|=25>20$): $\frac{1}{v}=\frac{25-20}{20\cdot 25}=\frac{5}{500}=\frac{1}{100}\Rightarrow v=+100$ cm, real. ✔ $u=-15$ ($|u|=15<20$): $\frac{1}{v}=\frac{15-20}{20\cdot 15}=\frac{-5}{300}=-\frac{1}{60}\Rightarrow v=-60$ cm, virtual. ✔

Radius. Flat face: $\frac{1}{R_2}=0$. So $\frac{1}{f}=(n-1)\frac{1}{R_1}\Rightarrow R_1=(n-1)f=0.5\cdot 25=12.5$ cm ($>0$, centre on outgoing side, as required). Image at $u=-50$: $\frac{1}{v}=\frac{1}{25}+\frac{1}{-50}=\frac{2-1}{50}=\frac{1}{50}\Rightarrow v=+50$ cm. Magnification: $m=\frac{50}{-50}=-1$: real, inverted, same size ($2f$–$2f$ again). ✔