Question

At what distance from the space a person should place a concave mirror if a focal length 0.4m so that magnification is 5 times for a virtual image.

Answer 1

Answer :-

→ The object should be placed 160 cm from the mirror .

Step-by-step explanation :-

We have,

→ focal length = 0.4 m = 40 cm .

→ and , magnification = 5 .

Now,

→ magnification = -v/u .

→ -v/u = 5 .

$\therefore$ so, v = - 5u .

Now, using mirror formula ,

$\sf \rightarrow \frac{1}{v} + \frac{1}{u} = \frac{1}{f} . \\ \\ \sf \implies \frac{1}{ - 5u} + \frac{1}{u} = \frac{1}{40} . \\ \\ \sf \implies \frac{ - 1 + 5}{5u} = \frac{1}{40} . \\ \\ \sf \implies \frac{4}{5u} = \frac{1}{40} . \\ \\ \sf \implies40 \times 4 = 5u. \\ \\ \sf \implies u = \frac{160}{5} . \\ \\ \tt \therefore u = 32 \: cm.$

So, v = -5u = -5 × 32 cm.

= - 160 cm.

Hence, it is solved .

Answer 2

ʜᴏʟᴀ!! ✨

ʜᴇʀᴇ ɪs ʏᴏᴜʀ ᴀɴsᴡᴇʀ :-✨

sᴛᴇᴘ ʙʏ sᴛᴇᴘ ᴇxᴘʟᴀɴᴀᴛɪᴏɴ :-✨

ʜᴇʀᴇ, 

ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ = 0.4ᴍ
 ᴄᴏɴᴠᴇʀᴛɪɴɢ ɪɴᴛᴏ cm, ᴡᴇ ɢᴇᴛ, 
ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ =40ᴄᴍ

ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ = 5

$we \: know \: that \\ \\ magnification = \frac{ - v}{u} \\ \\ \frac{ - v}{u} = 5 \\ \\ so \\ v = - 5u$

Using mirror formula,

We get,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \frac{1}{ - 5u} + \frac{1}{u} = \frac{1}{40} \\ \\ \frac{ - 1 + 5}{5u} = \frac{1}{40} \\ \\ \frac{4}{5u} = \frac{1}{40} \\ \\ 40 \times 4 = 5u \times 1 \\ \\ u = \frac{160}{5} \\ \\ so \: u = 32cm$

Now,
$v = - 5u \\ \\ v = - 5 \times 32 \\ \\ v = - 160cm$

Hence, the object should be placed 160cm from the mirror.

↪Thanks! ↩