Exercises — Optical instruments — human eye, simple microscope, compound microscope, telescope

KYA chahiye: woh number jo batata hai "angle kitni baar bada dikhta hai." YEH formula KYU: relaxed eye ka matlab hai image infinity par baithti hai, jo tab hota hai jab object exactly focus par ho. Upar derive kiya gaya tha ki angular power tab yeh hoti hai: $M_{\infty }=\frac{D}{f}.$ Plug in: $M=\frac{25}{5}=\boxed{5}$. Object naked-eye near point se angle mein $5\times$ bada dikhta hai.

YEH formulas KYU: ek telescope infinity par objects dekhta hai, isliye reference angle woh real angle hai jo object subtend karta hai — actual angle $\alpha$ jo uske top aur bottom se aane wali rays ke beech hai, roughly $h/f_o$ focus hone par — $h/D$ nahi (kisi star ko near point par laane ka koi sawaal nahi). Yahan $h$ intermediate image ki height hai aur $\alpha ,\beta$ aankh par incoming aur outgoing angles hain. Normal adjustment ⇒ dono foci coincide karte hain. $M=\frac{f_o}{f_e}=\frac{120}{3}=\boxed{40}.$ Tube length $f_o+f_e$ KYU hoti hai (black box nahi): parallel starlight objective ke paas pahunchti hai; objective use objective ke focal plane mein focus karta hai, objective ke peeche $f_o$ doori par. Relaxed aankh ke liye eyepiece ko rays parallel bahar bhejna chahiye, jo tab hota hai jab uska focal point us image par ho — yaani eyepiece tube mein aur $f_e$ neeche ho. Toh lens-to-lens gap hai $L=f_o+f_e=120+3=\boxed{123\text{ cm}}.$

Magnification — YEH formula KYU: virtual image ko exactly $D$ par rakhne se object woh sabse bada angle subtend karta hai jis par aankh abhi bhi focus kar sakti hai, isliye yeh magnifier ki maximum setting hai. Derived result use karke, $M_D=1+\frac{D}{f}=1+\frac{25}{5}=\boxed{6}.$ Relaxed value $5$ par extra "$+1$" aankh ko strain karne ka reward hai.

Object distance — lens equation KYU: hum jaante hain image kahan hai ($v=-25$, virtual ⇒ hamare convention ke hisaab se negative) aur object kahan rakhein ($u$) yeh jaanna chahte hain. Lens equation exactly un dono ko $f$ ke saath link karta hai. $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ use karo $v=-25\text{ cm}$, $f=+5\text{ cm}$ ke saath: $\frac{1}{-25}-\frac{1}{u}=\frac{1}{5}\Rightarrow \frac{1}{u}=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}.$ Toh $u=-\frac{25}{6}\approx \boxed{-4.17\text{ cm}}$. Minus real object confirm karta hai; $|u|=4.17<f=5$, isliye object focal length ke andar baithta hai, exactly jaisa ek magnifier require karta hai.

Objective — $m_o\approx L/f_o$ KYU (approximation derive karna): object just outside $f_o$ ke bahar baithta hai, isliye $|u_o|\approx f_o$. Real image tube ke far focus ke paas banti hai, $v_o\approx L$ par. Phir linear magnification magnitude ($m_o=v_o/u_o$, height ratio) hai $|m_o|=\frac{|v_o|}{|u_o|}\approx \frac{L}{f_o}=\frac{15}{1.5}=10.$ Image inverted hai, isliye sign ke saath $m_o\approx -10$. Eyepiece — $D/f_e$ KYU: eyepiece ek simple magnifier ki tarah us intermediate image par act karta hai, relaxed eye ⇒ image at infinity. Iska factor magnifier ka angular power $M_{\infty }=D/f_e$ hai; hum isse $m_e$ likhte hain lekin yaad rakho yeh capital-$M$ (angular) quantity hai: $m_e=\frac{D}{f_e}=\frac{25}{5}=5.$ Yeh multiply karte hain (eyepiece already-enlarged picture ko aur badata hai), aur signs bhi multiply karte hain: $M=m_o{m}_e\approx (-10)(5)=-50\Rightarrow |M|=\boxed{50}.$ Minus batata hai ki final image inverted hai — jo cells dekhne ke liye theek hai.

KYU: $M=f_o/f_e$ tab badhta hai jab objective lamba ho aur eyepiece chhota ho. Correct choice: objective $f_o=80\text{ cm}$, eyepiece $f_e=2\text{ cm}$: $M=\frac{80}{2}=\boxed{40}.$ Swap karne par (objective $2$, eyepiece $80$): $M=\frac{2}{80}=0.025(<1),$ jo angle ko chhota karta hai — ek "ulta" telescope, exactly wahi jo tum binoculars mein ulti taraf se dekhne par dekhte ho.

$M_{\infty }=D/f$ se kyun shuru karein: yeh wahi ek relation hai jo given magnification ko unknown focal length se link karta hai. Isko invert karo: $f=\frac{D}{M_{\infty }}=\frac{25}{4}=6.25\text{ cm}=0.0625\text{ m}.$ Power $P=1/f$ hai jahan $f$ metres mein ho (dioptre ki definition): $P=\frac{1}{0.0625}=\boxed{16\text{D}}.$

YEH formula KYU: relaxed-eye compound power $M=\frac{L}{f_o}\cdot \frac{D}{f_e}$ hai (objective size factor × eyepiece angular factor). Hum $f_e$ ko chhodkar sab fix karte hain aur uske liye solve karte hain — aise hi tum target ke liye design karte ho. $300=\frac{20}{1}\cdot \frac{25}{f_e}=\frac{500}{f_e}.$ $f_e$ ke liye invert karo: $f_e=\frac{500}{300}=\boxed{1.67\text{ cm}}(=5/3\text{ cm}).$ Sanity check: $m_o=20/1=20$, $m_e=25/(5/3)=15$, aur $20\times 15=300$. ✓

Pehle ray picture par symbols fix karo. Maano $h$ objective ke focal plane mein bane intermediate image ki height hai. Maano $\alpha$ woh angle hai jo door object aankh par subtend karta hai ("without instrument" angle), aur $\beta$ woh angle hai jo final image subtend karta hai ("with instrument" angle). Dono chhote angles hain jo axis par measure hote hain.

(a) Normal adjustment — $f_o/f_e$ KYU: door object se parallel rays angle $\alpha$ par aati hain; objective unhe apne focal plane mein height $h$ ki image par focus karta hai, toh us right triangle ki geometry deti hai $\alpha \approx \frac{h}{f_o}$ (opposite $h$ over adjacent $f_o$). Eyepiece ki uski focus par yahi image hai, toh woh rays parallel angle $\beta \approx \frac{h}{f_e}$ par bahar bheji jaati hain. Isliye $M=\frac{\beta }{\alpha }=\frac{h/f_e}{h/f_o}=\frac{f_o}{f_e}=\frac{100}{5}=\boxed{20}.$ $h$ cancel ho jaata hai — exactly isliye object ki real height answer mein kabhi nahi aati.

(b) Near point — extra factor $\left(1+\frac{f_e}{D}\right)$ KYU, derived: ab eyepiece ko final image $D$ par daalni hai ($\infty$ par nahi). Eyepiece par apply magnifier near-point rule se, uska angular factor $\frac{1}{f_e}$-type se badhke "$+$" term include karta hai: exit angle ${\beta }^{\prime }=\frac{h}{f_e}\left(1+\frac{f_e}{D}\right)$ ho jaata hai. Same $\alpha =h/f_o$ se divide karke, $M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=20\left(1+\frac{5}{25}\right)=20\times 1.2=\boxed{24}.$ Fractional gain: $(24-20)/20=0.2=20\%$. Note karo yeh microscope ke "$+1$" se chhota relative boost hai, kyunki telescope ke liye extra factor $f_e/D$ hai, aur $f_e\ll D$.

(a) Objective image (Figure 1 mein yellow-ray crossing). Real object ⇒ $u_o=-2.5\text{ cm}$, $f_o=+2\text{ cm}$. $v_o$ ke liye lens equation solve karo rearranged form $\frac{1}{v_o}=\frac{1}{f_o}+\frac{1}{u_o}$ use karke (same equation, dono sides mein $\frac{1}{u_o}$ add karo): $\frac{1}{v_o}=\frac{1}{2}+\frac{1}{-2.5}=0.5-0.4=0.1,$ toh $v_o=+10\text{ cm}$. Positive sign matlab hai yeh wahan hai jahan Figure 1 mein red arrow baithta hai: objective ke $10\text{ cm}$ peeche real image. ✓

(b) Objective magnification: $m_o=\frac{v_o}{u_o}=\frac{10}{-2.5}=-4$. Minus woh inversion hai jo tum Figure 1 mein dekhte ho (red arrow neeche point karta hai). Size factor $|m_o|=4$.

(c) Eyepiece (Figure 1 mein green rays). Red image eyepiece ke liye object hai, jo use $D$ par throw karta hai. Magnifier near-point rule use karke, $m_e=1+\frac{D}{f_e}=1+\frac{25}{6.25}=1+4=5.$

(d) Total — signs multiply karte hain: $M=m_o{m}_e=(-4)(5)=-20\Rightarrow |M|=\boxed{20}.$ Minus batata hai ki final image inverted hai, jo green stage ko feed karne wale neeche jaate red arrow se match karta hai.

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ se shuru karo, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$ mein rearrange karo, $u<0$, $f>0$ ke saath. Figure 2 se match karne ke liye $f=5\text{ cm}$ lo.

$M=D/|u|$ KAHAN se aata hai aur iska domain KYU hai. Woh angle jo object lens ke through subtend karta hai $\theta \approx h/|u|$ hai (object height $h$ uski distance $|u|$ par); naked-eye reference ${\theta }_0=h/D$ hai. Toh angular power hai $M=\frac{\theta }{{\theta }_0}=\frac{h/|u|}{h/D}=\frac{D}{|u|}.$ Yeh tabhi meaningful hai jab virtual image wahan land kare jahan aankh use focus kar sake, yaani $|v|$ $D$ aur $\infty$ ke beech ho. Us window ke bahar image ya toh un-focusable hai (closer than $D$) ya real hai (object outside $f$), aur formula ab ek usable magnifier describe nahi karta. Woh window exactly Figure 2 ki blue curve hai green dot aur red line ke beech.

(i) Jab $u\to -f^{-}$ (object andar se focus ke paas aata hai, $|u|<f$): $\frac{1}{u}\to -\frac{1}{f}$, toh $\frac{1}{v}\to 0$ aur $|v|\to \infty$ — yeh Figure 2 ki blue curve hai jo red dashed line $|u|=5$ ke paas seedhi neeche jaati hai. Virtual image infinity ki taraf bhaagti hai: relaxed-eye limit. Concretely, $|u|=4.9$ par formula deta hai $v=1/(0.2-1/4.9)\approx -245\text{ cm}$; $|u|=4.99$ par, $v\approx -2490\text{ cm}$ — $-\infty$ ki taraf bhaag raha hai.

(ii) Magnification range (Figure 2 mein do marked lines padhna). $|u|=f$ par (image at $\infty$, red line) hum sabse chhoti usable value paate hain $M_{\infty }=D/f=25/5=5$. Object ko thoda inside $f$ kheenchne se image $\infty$ se $D$ ki taraf aati hai; $v=-D$ par (green dot yellow $v=-25$ line par baithta hai) hum sabse badi usable value paate hain $M_D=1+D/f=6$, jo $|u|=25/6\approx 4.17\text{ cm}$ par milti hai. Toh usable magnification closed range $[D/f,1+D/f]=[5,6]$ mein span karta hai — ek hayraan kar dene wali narrow band, jo green dot ke paas flat blue segment se visual hoti hai.

(iii) Exactly $u=-f$ par: image at infinity, eye relaxed, $M=D/f=5$. Just beyond, $|u|>f$ (object outside focus): ab $\frac{1}{f}+\frac{1}{u}$ positive ho jaata hai, toh $v>0$ — image far side par jump kar jaati hai aur real aur inverted ho jaati hai (isliye Figure 2 ki blue curve sirf $|u|<f$ ke liye draw ki gayi hai; red line ke daayein taraf device ab magnifier nahi hai). $|u|=0$ par doosri extreme mein, $M=D/|u|\to \infty$ formally, lekin image tab $D$ se closer baithti hai aur focus nahi ho sakti — L5 trap dekho. Saare cases cover: inside focus → usable virtual magnifier ($M\in [5,6]$ near-point end ke paas, larger lekin un-focusable jab $|u|\to 0$); at focus → image at $\infty$, $M=5$; outside focus → real inverted image, magnifier nahi.

Numeric check $f=5$ ke liye: $u=-25/6\approx -4.17\text{ cm}$ par, $v=-25$ (image at $D$), $M=6$; $u=-5$ par, $v\to \infty$, $M=5$. ✓

Simple microscope, relaxed $M$ for $f=5$ ::: $D/f=5$ Simple microscope, near-point $M$ for $f=5$ ::: $1+D/f=6$ Telescope $M$ for $f_o=120,f_e=3$ ::: $40$ Compound scope $|M|$ for $L=15,f_o=1.5,f_e=5$ ::: $50$ (inverted) Dioptres of a $6.25\text{ cm}$ lens ::: $16\text{D}$