Question

an object of size 5cm is placed in front of convex mirror of focal length 20cm at a distance of 20cm in front of its reflective surface. fine the position , nature and size of the image of class 10 ncert
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Answer 1

Given :

In convex mirror,

Height of object : 5 cm.

Focal length : 20 cm.

Object distance : 20 cm.

To find :

The position, nature and the size of the image.

Solution :

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,

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

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

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

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

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

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{2}{20}$

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

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

Thus, the position of the image is 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,

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

where,

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

By substituting all the given values in the formula,

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

$\dashrightarrow\sf \dfrac{h'}{5} = - \dfrac{10}{ - 20}$

$\dashrightarrow\sf \dfrac{h'}{5} = \dfrac{1}{2}$

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

$\dashrightarrow\sf h'= 2.5 \: cm$

Thus, the size of the image is 2.5 cm.

Nature of the image :

• The image is inverted and virtual.
• The image is diminished.

Answer 2

Explanation:

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