Question

A 5cm high object is placed in front of a concave mirror with a radius of curvature
of 20 cm. Determine the image height, if the object distance is 5 cm.​

Answer 1

Answer:

I hope help at all

Please like bother and sister

Answer 2

Given :

In concave mirror,

Height of the object : 5 cm

Radius of curvature : 20 cm

Object distance : 5 cm

To find :

The height of the image.

Solution :

1st we have to find focal length we know that,

» For a spherical mirror having small aperture, the principle focus lies exactly mid way between the pole and centre of curvature. So, the focal length of a spherical mirror is equal to the half of its radius of curvature.

if f is the focal length of a mirror and R is its radius of curvature, then f = R/2

by substituting the given values in the formula,

$\dashrightarrow \sf f = \dfrac{R}{2}$

$\dashrightarrow \sf f = \dfrac{20}{2}$

$\dashrightarrow \sf f = 10 \: cm$

Now using mirror formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

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

where,

• v denotes Image distance
• u denotes object distance
• f denotes focal length

By substituting all the given values in the formula,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 5} = \dfrac{1}{ - 10}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{5} = \dfrac{1}{ - 10}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 10} + \dfrac{1}{5}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 1 + 2}{10}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{10}$

$\dashrightarrow\sf v = 10 \: cm$

We know that,

» The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and it is equal to the ratio of image height and object height. that is,

$\boxed{\bf \dfrac{h'}{h} = - \dfrac{v}{u}}$

where,

• h' denotes height of the image
• h denotes height of the object
• v denotes image distance
• u denotes object distance

By substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{h'}{h} = - \dfrac{v}{u}$

$\dashrightarrow\sf \dfrac{h'}{5} = - \dfrac{10}{ - 5}$

$\dashrightarrow\sf \dfrac{h'}{5} = 2$

$\dashrightarrow\sf h' = 5 \times 2$

$\dashrightarrow\sf h' = 10 \: cm$

Thus, the height of the image is 10 cm.