Question

ʜᴇʏᴀ ᴍᴀᴛᴇ ᴘʀᴏᴠᴇ ʟᴇɴs ᴍᴀᴋɪɴɢ ғᴏʀᴍᴜʟᴀ.​

Answer 1

We know that -

$\implies{\sf \dfrac{\mu}{v}-\dfrac{1}{u}=\dfrac{\mu-1}{R}\: \: \:→(1)}$

D' is virtual object of ${\sf R_2}$ Surface

$\implies{\sf \dfrac{\dfrac{1}{\mu}}{v}-\dfrac{1}{v'}=\dfrac{\dfrac{1}{\mu}-1}{R_2} }$

$\implies{\sf \dfrac{1}{v}-\dfrac{\mu'}{v'}=\dfrac{1-\mu}{R_2}\:\:\:\:→(2)}$

$\underline{\rm Adding \: equation \: 1\: and \:2}$

$\implies{\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{\mu-1}{R_1}+\dfrac{1-\mu}{R_2} }$

$\implies{\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{\mu-1}{R_2} }$

$\implies{\sf \dfrac{1}{v}-\dfrac{1}{u}=(\mu -1)\left[\dfrac{1}{R_1}-\dfrac{1}{R_2}\right] }$

When image kept at $\infty$ then image formed at focus

u = $\infty$ , v = f

$\implies{\sf \dfrac{1}{f}-\dfrac{1}{\infty}=(\mu-1)\left[\dfrac{1}{R_1}-\dfrac{1}{R_2}\right]}$

$\color{red}\bullet\underline{\boxed{\sf \dfrac{1}{f}=(\mu-1)\left[\dfrac{1}{R_1}-\dfrac{1}{R_2}\right]}}$

Answer 2

Answer:

$adii \: ole \: ole \: kaiche \: ho \:$