Question

an object of 5cm in length is placed a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm .find position of image .it nature size​

Answer 1

ɢɪᴠᴇɴ :-

Focal length, f = R/2 = 30/2 = 15 cm

Object Distance, u = - 20 cm

Height of object, h₁ = 5 cm

ᴛᴏ ꜰɪɴᴅ :-

Image distance, v

Height of image, h₂

Nature of the image

ꜱᴏʟᴜᴛɪᴏɴ :-

we know,

$\boxed{\sf \pink {\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}$

Where,

v = Image distance

u = Object Distance

f = Focal length

Putting all the values, we get

${\sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$

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

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

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

${\sf \dfrac{1}{v}= \dfrac{60}{7} }$

⇒v = 8.57 cm

Hence, the Image distance is  8.57 cm.

Now, Magnification,

$\boxed{\sf \pink {m = \dfrac{- v}{u}}}$

$\tt m = \dfrac{- v}{u}$

$\tt m =\dfrac{-8.57}{20}$

⇒ m = 0.4

Hence, height of image is 2.175 cm.

The image will be virtual and erect.

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Formulas used :-

Mirror formulas

• 1/v + 1/u = 1/f
• Magnification, m = - v/u

Answer 2

$\huge\boxed{\fcolorbox{purple}{ink}{Answer}}$

height of image is 2.175 cm.

The image will be virtual and erect.