Question

2. An object of 2cm in height placed at 4 cm form concave mirror of focal length 2 cm.
What is nature size and position of image?​

Answer 1

Answer:

Height of the Object=2cm

Distance of the object =-4cm

Focal Length ;-2cm

Centre of curvature of the mirror is 2*focal=-4

We can see that the object is placed at Centre of curvature.

So,the image is formed at C.

And its size is same as object =4cm

Explanation:

Answer 2

Given :

In concave mirror,

Height of the object = 2 cm

Object distance = - 4 cm

Focal length = - 2 cm

To find :

The nature, size and position of the image.

Solution :

we know that ,

» 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 4} = \dfrac{1}{ - 2}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{ 4} = \dfrac{1}{ - 2}$

$\dashrightarrow\sf \dfrac{1}{v} = - \dfrac{1}{ 2} + \dfrac{1}{ 4}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 2 + 1}{ 4}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ -1}{ 4}$

$\dashrightarrow\sf v = - 4 \: cm$

Thus, the position of image is -4 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,

$\dashrightarrow\bf m = - \dfrac{v}{u}$

$\dashrightarrow\sf \dfrac{h'}{2} = - \dfrac{( - 4)}{ - 4}$

$\dashrightarrow\sf h' = - 1 \times 2$

$\dashrightarrow\sf h' = - 2 \: cm$

Thus, the size of the image is -2 cm

Nature of the image :

• The image is real, inverted and same size as object.