Question

An object of size 50 cm, placed in front of a spherical mirror at 20 cm, forms an image of size 25 cm above the principal axis. Find out the focal length with sign convention and the type of spherical mirror used?​

Answer 1

$\huge\mathfrak{Bonjour!!}$

$\huge\fcolorbox{black}{aqua}{Solution:-}$

✒ ɢɪᴠᴇɴ:

★ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ, ʜ' = +25 ᴄᴍ (ᴘᴏsɪᴛɪᴠᴇ sɪɴᴄᴇ ɪᴍᴀɢᴇ ɪs ғᴏʀᴍᴇᴅ ᴀʙᴏᴠᴇ ᴛʜᴇ ᴘʀɪɴᴄɪᴘᴀʟ ᴀxɪs)

★ ᴏʙᴊᴇᴄᴛ ᴅɪsᴛᴀɴᴄᴇ, ᴜ = -20 ᴄᴍ

★ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ, ʜ = 50 ᴄᴍ

ʟᴇᴛ ᴍ ʙᴇ ᴛʜᴇ ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ, ᴠ ʙᴇ ᴛʜᴇ ɪᴍᴀɢᴇ ᴅɪsᴛᴀɴᴄᴇ ᴀɴᴅ ғ ʙᴇ ᴛʜᴇ ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ.

ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

ᴍ = -ᴠ/ᴜ = ʜ'/ʜ

=) 25/50 = 1/2 = -ᴠ/ᴜ

=) -ᴠ/-20 = 1/2

ᴛʜᴇʀᴇғᴏʀᴇ, ᴛʜᴇ ɪᴍᴀɢᴇ ᴅɪsᴛᴀɴᴄᴇ, ᴠ= 10 (ᴀғᴛᴇʀ ᴀᴘᴘʟʏɪɴɢ sɪɢɴ ᴄᴏɴᴠᴇɴᴛɪᴏɴ ᴠ ɪs ᴘᴏsɪᴛɪᴠᴇ sᴏ ɪᴛ sʜᴏᴡs ᴛʜᴀᴛ ᴛʜᴇ ɪᴍᴀɢᴇ ғᴏʀᴍᴇᴅ ɪs ᴠɪʀᴛᴜᴀʟ)

ɴᴏᴡ,

ᴀᴘᴘʟʏɪɴɢ ᴛʜᴇ ɢɪᴠᴇɴ ᴠᴀʟᴜᴇs ɪɴ ᴛʜᴇ ᴍɪʀʀᴏʀ ғᴏʀᴍᴜʟᴀ, ᴡᴇ ɢᴇᴛ,

1/ᴠ + 1/ᴜ = 1/ғ

1/10 - 1/20 = 1/ғ

1/ғ = 1/20

ᴛʜᴇʀᴇғᴏʀᴇ, ᴛʜᴇ ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ, ғ = +20 ᴄᴍ.

sɪɴᴄᴇ ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ ɪs ᴘᴏsɪᴛɪᴠᴇ ᴀɴᴅ ᴛʜᴇ ɪᴍᴀɢᴇ ғᴏʀᴍᴇᴅ ɪs ᴠɪʀᴛᴜᴀʟ, ᴛʜᴇ ᴍɪʀʀᴏʀ ɪs ᴄᴏɴᴠᴇx ᴍɪʀʀᴏʀ.

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Answer 2

Answer: Hope this helps you.

Explanation: