Question

An object is placed 20cm in front of a concave mirror whose focal length is 15cm. Determine; the image distance and calculate the magnification of the mirror?

Answer 1

Given :-

• Object distance, u = - 20 cm ( Object Distance is always -ve )
• Focal Length, f = - 15 cm ( Focal length of concave mirror is always negative. )

To Find :-

• Image distance, v.
• Magnification, m.

Solution :-

We are given object distance, focal length, so we can find the image distance using the mirror formula.

⇒ 1/f = 1/v + 1/u

⇒ -1/15 = 1/v - 1/20

⇒ 1/v = 1/20 - 1/15

⇒ 1/v = (3 - 4)/60

⇒ 1/v = -1/60

⇒ v = -60 cm

So, We have image distance and object distance, Let us find the magnification as requested in the question,

⇒ Magnification, m = - v / u

⇒ m = -(-60/-20)

⇒ m = -(3)

⇒ m = -3

Hence,

• Image distance, v = - 60 cm
• Magnification, m = -3

Answer 2

Explanation:

Given :-

• Height of the object, h = 20cm.

• Object distance, u = 20cm

• Focal length, f = 15cm

To Find :-

• The height of the image

Solution :-

using mirror formula

$\bigstar$ A mirror formula may be defined as the formula which gives the relationship between the distance of image (v), distance of object (u), and the focal length (f) of the mirror. It may be written as:

$</p><p>{ \boxed {\bf \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}}</p><p></p><p>$

$\\ \\ \sf : \implies \: \: \: \: \:\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$

$\\ \\ \sf : \implies \: \: \: \: \:\dfrac{1}{v} = \dfrac{1}{ - 15} - \dfrac{1}{ - 20}</p><p>$

$\\ \\ \sf : \implies \: \: \: \: \:\dfrac{1}{v} = \dfrac{1}{ - 15} + \dfrac{1}{ 20}$

$\\ \\ \sf : \implies \: \: \: \: \:\dfrac{1}{v} = \dfrac{ - 4 + 3}{ 60}$

$\\ \\ \sf : \implies \: \: \: \: \:\dfrac{1}{v} = \dfrac{ - 1}{ 60}$

$\\ \\ \sf : \implies \: \: \: \: \: v= -60cm$

$\\\\\sf : \implies \: \: \: \: \: m = \frac{ - v}{u} \\ \\ \\ \sf : \implies \: \: \: \: \: m = \: \cancel{ \frac{ - ( - 60)}{ - 20} } \\ \\ \\ \sf : \implies \: \: \: \: \: m = - 3$