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Answer 1

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Answer 2

✒Qᴜᴇsᴛɪᴏɴ :-

Aɴ ᴏʙᴊᴇᴄᴛ 2ᴍ ғʀᴏᴍ ᴛʜᴇ ʟᴇɴs ғᴏʀᴍs ᴀɴ ᴇʀᴇᴄᴛ ɪᴍᴀɢᴇ 1/4ᴛʜ sɪᴢᴇ ᴏғ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ . Dᴇᴛᴇʀᴍɪɴᴇ ᴛʜᴇ ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ʟᴇɴs. Wʜᴀᴛ ᴛʏᴘᴇ ᴏғ ʟᴇɴs ɪᴛ ɪᴛ ?

✒Gɪᴠᴇɴ :-

Aɴ ᴏʙᴊᴇᴄᴛ 2ᴍ ғʀᴏᴍ ᴛʜᴇ ʟᴇɴs ғᴏʀᴍs ᴀɴ ᴇʀᴇᴄᴛ ɪᴍᴀɢᴇ 1/4 ᴛʜ sɪᴢᴇ ᴏғ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ.

✒Tᴏ ғɪɴᴅ :-

Fᴏᴄᴀʟ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ʟᴇɴs

Tʏᴘᴇ ᴏғ ʟᴇɴs

✒Sᴏʟᴜᴛɪᴏɴ :-

Dɪsᴛᴀɴᴄᴇ ᴏғ ᴏʙᴊᴇᴄᴛ (ᴜ) = - 2ᴍ

v = ?

$\sf\implies\dfrac{1}{4}=\dfrac{v}{u}$

$\sf\implies\dfrac{1}{4} = \dfrac{v}{-2}$

$\sf\implies v = -\dfrac{2}{4}$

$\sf\implies\dfrac{-1}{2}$

✒ :-

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

$\sf\implies\dfrac{1}{-0.5}-\dfrac{1}{-2}=\dfrac{1}{f}$

$\sf\implies-\dfrac{1}{0.5}+\dfrac{1}{2}=\dfrac{1}{f}$

$\sf\implies\dfrac{1}{f}=\dfrac{-3}{2}$

$✒Focal \: \: length \: \: of \: \: the \: lens \\ (f) = \sf-\dfrac{2}{3}$