Question

An object of size 7.0 cm is placed at 27 CM in front of concave mirror focal length 18 cm. At what distance from the mirror should a screen with placed so that ASAP used image can be obtained? Find the size and nature of the image .


please answer it ......☺☺☺❤

Answer 1

Hey mate..


Here is ur ans

Size of the object =7cm


F=-18cm


u=-27cm.


v=?


1/f. =1/v+1/u

.1/-8 =1/v -1/27

.size of the image =m=1/0=-v/u 1/7


=-57/27


=- 14 cm Ans


Image REAL and inverted


Hope it helps u✌❤

Answer 2

sᴏʟᴜᴛɪᴏɴ –

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ɢɪᴠᴇɴ –

Object distance, u = -27 cm
Image distance, v = ?
Focal length, f = -18 cm
Height of object, h1 = 7.0 cm
Height of image, h2 = ?

ʙʏ ᴜsɪɴɢ ᴍɪʀʀᴏʀ ғᴏʀᴍᴜʟᴀ

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{ - 18} = \frac{1}{v} + \frac{1}{ - 27} \\ \frac{1}{ - 18} = \frac{1}{v} - \frac{1}{27} \\ \frac{1}{v} = - \frac{1}{18} + \frac{1}{27} \\ \frac{1}{v} = \frac{ - 3 + 2}{54} \\ \frac{1}{v} = - \frac{ 1}{54} \\ v = 54 \: cm$

ɴᴏᴡ,

the image distance is - 54 cm,
so,
the screen should be placed at a distance of 54 CM in front of the concave mirror. the nature of image obtained on the screen is real and inverted.

ᴀʟsᴏ,

For a mirror ,

$\frac{h2}{h1} = - \frac{v}{u} \\ \frac{h2}{7.0} = - (\frac{ - 54}{ - 27} ) \\ \frac{h2}{7} = - 2 \\ h2 = - 2 \times 7 \\ h2 = - 14$
Thus, the size of image is 14 cm.


ɪᴍᴀɢᴇ ᴅɪsᴛᴀɴᴄᴇ , ᴠ ɪs - 54 ᴄᴍ,
ʜᴇɪɢʜᴛ ᴏғ ɪᴍᴀɢᴇ, ʜ2 ɪs -14 ᴄᴍ.

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THANKS☺️