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 → direction mein travel karta hai): real object $u<0$; real image $v>0$; converging $f>0$; ek surface ka $R>0$ hota hai jab uska centre of curvature outgoing side par ho.

Hum kya pooch rahe hain: sirf shape se $f$ ka sign. Pehli surface: centre outgoing side par $\Rightarrow R_1>0$. Doosri surface: centre incoming side par $\Rightarrow R_2<0$. Tab $\frac{1}{R_1}-\frac{1}{R_2}=(\text{positive})-(\text{negative})=\text{positive}$, aur $(n-1)>0$. Isliye $\frac{1}{f}>0\Rightarrow f>0\Rightarrow$ converging. Yeh ek biconvex lens hai — classic magnifying-glass shape.

Object ki shape/glass yahan irrelevant hai — hamare paas pehle se $f$ hai. Lens equation use karo: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. "40 cm saamne" = real object incoming side par $\Rightarrow u=-40$ cm. (Hum abhi solve nahi kar rahe — sirf tool aur sign pehchanna hai.)

$\frac{1}{R_1}=0$ aur $\frac{1}{R_2}=0$, isliye $\frac{1}{f}=(n-1)(0-0)=0$. Zero power, $f=\infty$: yeh light ko sideways shift karta hai lekin focus nahi karta. Yeh sanity check hai ki flat glass "kuch nahi karta."

$\frac{1}{f}=(1.5-1)\left(\frac{1}{15}-\frac{1}{-25}\right)=0.5\left(\frac{1}{15}+\frac{1}{25}\right).$ $\frac{1}{15}+\frac{1}{25}=\frac{5+3}{75}=\frac{8}{75}$, isliye $\frac{1}{f}=0.5\cdot \frac{8}{75}=\frac{4}{75}{\text{cm}}^{-1}$. $f=\frac{75}{4}=+18.75$ 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 $1.67\times$. ✔ Neeche wala ray diagram dekho — object $f$ se aage hai, isliye hame real, inverted, enlarged image milti hai.

$\frac{1}{v}=\frac{1}{-20}+\frac{1}{-12}=\frac{-3-5}{60}=-\frac{8}{60}=-\frac{2}{15}\Rightarrow v=-7.5\text{cm}.$ Negative $v\Rightarrow$ virtual image incoming side par. $m=\frac{v}{u}=\frac{-7.5}{-12}=+0.625$: erect aur diminished — yahi ek diverging lens real object ke liye hamesha karta hai. ✔

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}+\frac{1}{-20}=0.$ $\frac{1}{v}=0\Rightarrow v=\infty$: rays parallel hokar nikalti hain aur kabhi nahi milti. Kisi finite distance par koi image nahi banti. Yeh "parallel in → focus" ka reversal hai: yahan hum source ko focus par rakhte hain, toh woh collimated hokar nikalti hai.

$\frac{1}{v}=\frac{1}{20}+\frac{1}{-10}=\frac{1-2}{20}=-\frac{1}{20}\Rightarrow v=-20\text{cm}.$ Negative $\Rightarrow$ virtual, incoming side par. $m=\frac{v}{u}=\frac{-20}{-10}=+2$: erect, magnified $2\times$. Yeh ek converging lens ka magnifying-glass mode hai. ✔

Symmetric biconvex: $\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{R}-\frac{1}{-R}=\frac{2}{R}$. $\frac{1}{f}=(n-1)\frac{2}{R}\Rightarrow R=2(n-1)f=2(0.6)(40)=48\text{cm}.$ Har face ka radius $48$ cm hai. ✔ Dhyan do ki dono faces minus sign ke through add hote hain — maker's equation ka baar baar aane wala trap.

Surface 1 (air→glass, $n_1=1,n_2=1.5$), parallel input $u\to -\infty$ isliye $\frac{1}{u}\to 0$: $\frac{1.5}{v_1}-0=\frac{1.5-1}{20}=\frac{0.5}{20}=\frac{1}{40}\Rightarrow \frac{1.5}{v_1}=\frac{1}{40},v_1=60\text{cm}.$ Surface 2 (glass→air, $n_1=1.5,n_2=1$), object surface-1 image hai $u_2=v_1=60$ cm par (thin lens, koi gap nahi): $\frac{1}{v}-\frac{1.5}{60}=\frac{1-1.5}{-20}=\frac{-0.5}{-20}=\frac{1}{40}.$ $\frac{1.5}{60}=\frac{1}{40}$, isliye $\frac{1}{v}=\frac{1}{40}+\frac{1}{40}=\frac{1}{20}\Rightarrow v=20$ cm. Parallel in → focus $20$ cm par, isliye $f=+20$ cm — maker's equation se directly match karta hai. ✔ Intermediate image $v_1$ ne apna kaam kiya aur cancel ho gaya.

Real, same-size image $\Rightarrow m=-1$ (ek single lens se real images inverted hoti hain). $m=\frac{v}{u}=-1\Rightarrow v=-u$. Lens equation mein daalo: $\frac{1}{-u}-\frac{1}{u}=\frac{1}{f}\Rightarrow -\frac{2}{u}=\frac{1}{12}\Rightarrow u=-24$ cm. Tab $v=-u=+24$ cm. Dono distances $2f$ ke barabar hain: "$2f$–$2f$" 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, matlab $4$ cm lamba, inverted. (e) $v>0\Rightarrow$ real; $m<0\Rightarrow$ inverted, magnified $2\times$. ✔

Algebra. Real object: $u=-|u|$. $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{f}-\frac{1}{|u|}=\frac{|u|-f}{f|u|}$ se. Denominator $f|u|>0$ hai, isliye $\frac{1}{v}$ (hence $v$) ka sign $|u|-f$ ke sign ke barabar hai.

• Agar $|u|>f$: numerator $>0\Rightarrow v>0\Rightarrow$ real.
• Agar $|u|<f$: numerator $<0\Rightarrow v<0\Rightarrow$ virtual.
• Agar $|u|=f$: numerator $=0\Rightarrow v=\infty$ (L3.1 wali 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: $\frac{1}{R_2}=0$. Isliye $\frac{1}{f}=(n-1)\frac{1}{R_1}\Rightarrow R_1=(n-1)f=0.5\cdot 25=12.5$ cm ($>0$, centre outgoing side par, jaisa required hai). $u=-50$ par image: $\frac{1}{v}=\frac{1}{25}+\frac{1}{-50}=\frac{2-1}{50}=\frac{1}{50}\Rightarrow v=+50$ cm. Magnification: $m=\frac{50}{-50}=-1$: real, inverted, same size (phir se $2f$–$2f$). ✔