Question

An object 5cm in length is placed at a distance of 20cm in front of a convex mirror of curvature 80cm find the position of the image, it's nature and size​

Answer 1

Given :

In convex mirror,

Object height = 5 cm

Object distance = 20 cm

Radius of curvature = 80 cm

To find :

The position of the image, it's nature and size.

Solution :

we know that,

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,

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

$\dashrightarrow \sf f = \dfrac{80}{2}$

$\dashrightarrow \sf f = 40 \: cm$

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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 20} = \dfrac{1}{40}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{ 20} = \dfrac{1}{40}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{40} + \dfrac{1}{ 20}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1 + 2}{40}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{3}{40}$

$\dashrightarrow\sf v= \dfrac{40}{3}$

$\dashrightarrow\sf v= 13.33 \: cm$

Thus, the position of image is 13.3 cm

» The Magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and is also equal to the ratio of height of the image to the height of the object .i.e.,

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

where,

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

By substituting all the given values in the formula,

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

$\dashrightarrow\sf - \dfrac{ \dfrac{40}{3} }{ - 20} = \dfrac{h'}{5}$

$\dashrightarrow\sf \dfrac{2}{3} = \dfrac{h'}{5}$

$\dashrightarrow\sf h' = \dfrac{2 \times 5}{3}$

$\dashrightarrow\sf h' = \dfrac{10}{3}$

$\dashrightarrow\sf h' = 3.33 \: cm$

Thus, the size of the image is 3.33 cm

Nature of the image :

• The image is formed behind the convex mirror. So, the image is virtual and erect.
• The image is diminished.