Question

a concave mirror has a radius of curvature of 0.4 m. Find the position and size of the image of an object 0.2 m high placed o.8 m in front of the mirror

Answer 1

We can use mirror formula to determine the position and the height of the image formed by the concave mirror of the object.

(1/v)-(1/u)=(1/f)

Now here we have to determine the value of v and the value of u and f are given.

Here u is the distance of the object from the mirror which is given 0.8 m and f is the focal length of the mirror which is half of the length of the radius of curvature so f is equals to 0.2 m.

So, we can write,
(1/v)=(1/0.2)+(1/0.8)
(1/v)=(5/0.8)
v=(4/25) m =0.16m

to find the height of the image we can use the magnification formula which is given as.

m=(Hi/Ho) = -(v/u)
here m is equals to magnification.
Hi is equals to height of the image.
Ho is equals to height of the object.
v is the distance of the image from the mirror.
u is the distance of the object from the mirror

So, we can write,

Hi= -(0.16/0.8)x(0.2)
= - 1m
negative represents that the image will be formed in a downward direction to Di principal line.

Answer 2

$\huge\mathfrak{Bonjour!!}$

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

✒ ᴀɴsᴡᴇʀ:-

⛄ 0.067 ᴍ ʟᴏɴɢ. (ɪᴍᴀɢᴇ ғᴏʀᴍᴇᴅ- ɪɴᴠᴇʀᴛᴇᴅ ᴀɴᴅ ʀᴇᴀʟ)

✒ ᴇxᴘʟᴀɴᴀᴛɪᴏɴ:-

⛄ ʜᴇʀᴇ,

★ ʀ= - 0.4 ᴍ (ʙʏ ᴄᴀʀᴛᴇsɪᴀɴ sɪɢɴ ᴄᴏɴᴠᴇɴᴛɪᴏɴ)

★ ғ= - 0.2 ᴍ (sɪɴᴄᴇ ғ = ʀ/2)

★ ᴜ= - 0.8 ᴍ

★ ʜ= 0.2 ᴍ

ᴡᴇ ʜᴀᴠᴇ,

1/ғ = 1/ᴜ + 1/ᴠ

=> 1/ᴠ = 1/ғ - 1/ᴜ

= - 1/0.2 + 1/0.8

= - 10/2 + 10/8

= -5 + 5/4

= -20 + 5/4

= -15/4

=> ᴠ= -4/15 = - 0.267 ᴍ.

ʜᴇɴᴄᴇ ᴀ ʀᴇᴀʟ ɪᴍᴀɢᴇ ɪs ғᴏʀᴍᴇᴅ ᴀᴛ 0.267 ᴍ ɪɴ ғʀᴏɴᴛ ᴏғ ᴛʜᴇ ᴍɪʀʀᴏʀ.

ᴀʟsᴏ ᴍ = ʜ'/ʜ = - ᴠ/ᴜ ᴏʀ ʜ' = ʜ × [- ᴠ/ᴜ].

ʜᴇʀᴇ,

★ ʜ= 0.2 ᴍ;

★ ᴠ= (-4)/15 ᴍ;

★ ᴜ= - 0.8 ᴍ.

ᴛʜᴇʀᴇғᴏʀᴇ,

ʜ' = 0.2 × { - [-4/15/ -0.8]}

= -2 × 4/15 × 1/0.8

= - 0.067 ᴍ.

↪ ᴛʜᴇ sɪᴢᴇ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ ɪs 0.067 ᴍ ʟᴏɴɢ. ᴛʜᴇ ᴍɪɴᴜs sɪɢɴ ʜᴇʀᴇ sʜᴏᴡs ᴛʜᴀᴛ ᴛʜᴇ ɪᴍᴀɢᴇ ɪs ғᴏʀᴍᴇᴅ ʙᴇʟᴏᴡ ᴛʜᴇ ᴘʀɪɴᴄɪᴘᴀʟ ᴀxɪs, ɪ.ᴇ., ɪᴛ ɪs ɪɴᴠᴇʀᴛᴇᴅ ᴀɴᴅ ʀᴇᴀʟ.

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