Question

Describe the size nature and position of image farmed when an object is placed at a distance of 15cm from a concave mirror of focal length 10cm​

Answer 1

Explanation:

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Answer 2

Given :

In concave mirror,

Object distance = - 15 cm

Focal length = - 10 cm

To find :

The position of the image and it's nature.

Solution :

» 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 15} = \dfrac{1}{ - 10}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{ - 10}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 10} + \dfrac{1}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 3 + 2}{30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 1}{30}$

$\dashrightarrow\sf v = - 30$

Thus, the position of image is -30 cm.

Nature of image :

• The image is formed in front of the concave mirror, it's nature will be real and inverted.