Question

an object of height 2 cn is placed in front of a convex lens of focal length 20cm at a distance of 15 cm from it. find position of image. also find the magnification and the nature of the image​

Answer 1

Explanation:

Given:-

• Height of the object = 2cm

• Focal length = 20cm

• Object distance = 15cm

• Type of lens used = Convex

To Find:-

• The Magification and Nature of the image.

Solution:-

For a convex lens:-

• Focal length (f) = +ve = 20cm

• Object distance = -ve = -15cm

Using lens formula:-

$\boxed{ \sf \leadsto \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }$

$\sf \implies \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{ - 15}$

$\sf \implies \dfrac{1}{20} = \dfrac{1}{v} +\dfrac{1}{ 15}$

$\sf \implies \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{ 15}$

$\sf \implies \dfrac{1}{v} = \dfrac{3 - 4}{60}$

$\sf \implies \dfrac{1}{v} = \dfrac{-1}{60}$

$\sf \implies v = -60cm$

$\sf \underline{\:\:\underline{\: Image \: distance \: (v) = -60cm\:}\:\:}$

$\sf Magnification = \dfrac{v}{u}$

$\sf \implies m = \dfrac{60}{-15}$

$\sf \implies m = 4$

Since, the magnification is positive

Nature of the image is:-

• Virtual

• Erect

• Enlarged

Answer 2

彡Given:-

✿Height of the object = 2cm

✿Focal length = 20cm

✿Object distance = 15cm

✿Type of lens used = Convex

彡To Find:-

✿ The Magification and Nature of the image.

Solution:-

For a convex lens:-

• Focal length (f) = +ve = 20cm

• Object distance = -ve = -15cm

Using Formula ⤵

1/f=1/v -1/15

Then, We get:- v⇢-60

ᴛʜᴇɴ, ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ ᴀᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ǫᴜᴇsᴛɪᴏɴ

ᴍ=ᴠ/ᴜ

ᴍ= -60/-15

ᴍ= 4

sɪɴᴄᴇ ᴛʜᴇ ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ ɪs ɢᴇᴛ ᴘᴏsᴛɪᴠᴇ.