Question

An object is placed at a distance of 15cm from a converging lens of focal length 20cm .find the distance of image and its nature ?




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Answer 1

Given :-

• An object is placed at a distance of 15cm from a converging lens of focal length 20cm.

To find :-

• Distance of image
• Nature of image

Solution :-

• Object distance (u) = - 15cm

• Focal length (f) = + 20cm

According to lens formula

→ 1/v - 1/u = 1/f

→ 1/v - (-1)/15 = 1/20

→ 1/v + 1/15 = 1/20

→ 1/v = 1/20 - 1/15

→ 1/v = 3 - 4/60

→ 1/v = -1/60

→ v = - 60cm

Hence,

• Distance of image is - 60cm

• Nature of image = Real and inverted

• Magnification = It is ratio of height of image to height of object or ratio of image distance to object distance

Answer 2

GIVEN :-

• Aɴ ᴏʙᴊᴇᴄᴛ ɪs ᴘʟᴀᴄᴇᴅ ᴀᴛ ᴀ ᴅɪsᴛᴀɴᴄᴇ ᴏғ 15ᴄm ғʀᴏᴍ ᴀ ᴄᴏɴᴠᴇʀɢɪɴɢ ʟᴇɴs .

• Tʜᴇ ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ᴄᴏɴᴠᴇʀɢɪɴɢ ʟᴇɴs ɪs 20ᴄm .

TO FIND :-

1. Tʜᴇ ᴅɪsᴛᴀɴᴄᴇ ᴏғ ɪᴍᴀɢᴇ .
2. Tʜᴇ ɴᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ .

SOLUTION :-

Wᴇ ʜᴀᴠᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

☯︎ Lᴇɴ's Fᴏʀᴍᴜʟᴀ :-

$\huge\blue\star$ $\bf\pink{\dfrac{1}{v}\:-\:\dfrac{1}{u}\:=\:\dfrac{1}{f}\:}$

Wʜᴇʀᴇ,

• $\bf\red{v}$ = ᴅɪsᴛᴀɴᴄᴇ ᴏғ ɪᴍᴀɢᴇ .

• $\bf\red{u}$ = ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴏʙᴊᴇᴄᴛ .

• $\bf\red{f}$ = ᴅɪsᴛᴀɴᴄᴇ ᴏғ ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ .

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

• ᴜ = -15 ᴄm

• f = 20 ᴄm

➪ $\bf{\dfrac{1}{v}\:=\:\dfrac{1}{f}\:+\:\dfrac{1}{u}\:}$

➳ $\rm{\dfrac{1}{v}\:=\:\dfrac{1}{20}\:+\:\dfrac{1}{-15}\:}$

➳ $\rm{\dfrac{1}{v}\:=\:\dfrac{1}{20}\:-\:\dfrac{1}{15}\:}$

➳ $\rm{\dfrac{1}{v}\:=\:\dfrac{3\:-\:4}{60}\:}$

➳ $\rm{\dfrac{1}{v}\:=\:-\dfrac{1}{60}\:}$

➳ $\bf\green{v\:=\:-60\:cm}$

$\huge\red\therefore$ [1] Tʜᴇ ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ ɪs "60ᴄm" .

__________________________

☯︎ Tᴏ ғɪɴᴅ ᴛʜᴇ ɴᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ, ᴡᴇ ᴄᴀɴ ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴛʜᴇ ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ .

$\huge\gray\star$ $\bf\purple{Magnification\:(m)\:=\:\dfrac{v}{u}\:}$

➟ m = -60/-15

$\bf\green{Magnification\:(m)\:=\:4\:}$

$\huge\red\therefore$ [2] Tʜᴇ ɪᴍᴀɢᴇ ɪs ᴠɪʀᴛᴜᴀʟ , ᴇʀᴇᴄᴛ ᴀɴᴅ 4 ᴛɪᴍᴇs ᴍᴀɢɴɪғɪᴇᴅ .