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.

_____________________________

★Formulas used :-

Mirror formulas

1/v + 1/u = 1/f

Magnification, m = - v/u

Answer 2

★GIVEN★

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

★To Find★

• Position of image.
• Its nature.
• Its size.

★SOLUTION★

Radius of Curvature (R) = 30 cm.

We know that,

Focal length (f) is half of Radius of Curvature (R).

So,

$\large\boxed{\bf{f=\dfrac{R}{2}}}$

$\large\implies{\sf{f=\dfrac{30}{2}}}$

$\large\implies{\sf{f=\dfrac{\cancel{30}}{\cancel{2}}}}$

$\large\implies{\sf{Focal\:Length=15\:cm.}}$

Now,

Now, We have to find the dist. of image (v).

By Mirror Formula,

$\large{\green{\underline{\boxed{\bf{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}}}}$

where,

• Dist. of object (u) = -20 cm. (convex mirror)
• Focal Length (f) = +15 cm. (convex mirror)
• Dist. of image (v)

Putting the values,

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

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

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

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

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

$\large\implies{\sf{v=8.57\:cm}}$

$\large\therefore\boxed{\bf{Distance\:of\:image(v)=8.6\:cm.}}$

★Size of Image★

From the relation,

$\large{\green{\underline{\boxed{\bf{\dfrac{I}{O}=\dfrac{v}{-u}}}}}}$

where,

• v = Image distance = 8.6 cm.
• u = Object distance = -20 cm.
• I = Length of image
• O = Length of image = 5 cm.

Putting the values,

$\large\implies{\sf{\dfrac{I}{5}=\dfrac{8.6}{-20}}}$

By cross multiplying,

$\large\implies{\sf{I\times(-20)=5\times8.6}}$

$\large\implies{\sf{I=\dfrac{43}{-20}}}$

$\large\implies{\sf{I=-2.15}}$

$\large\therefore\boxed{\bf{I=-2.2\:cm.}}$

Now,

$\large\boxed{\bf{Magnification=\dfrac{v}{-u}}}$

$\large\implies{\bf{Magnification=\dfrac{8.6}{-20}}}$

$\large\therefore\boxed{\bf{Magnification=-0.4}}$

This indicates that the image is diminished because the magnification of the image is less than 1.

1)) Position of the image is behind the mirror.

2)) Nature of the image is virtual and erect.

3)) Size of the image is diminished.