# A concave mirror of focal length 15 cm produces an image that is two times the size of the object. If the image formed is real, the distance between t

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A concave mirror of focal length 15 cm produces an image that is two times the size of the object. If the image formed is real, the distance between the object and the mirror is
1. 22.5 cm
2. 10.8 cm
3. 20.3 cm
4. 7.5 cm

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Correct Answer - Option 1 : 22.5 cm

The correct answer is option 1) i.e. 22.5 cm

CONCEPT:

• Concave mirror: If the inner surface of the spherical mirror is the reflecting surface then it is called a concave mirror. It is also called the converging mirror.
• The nature of the image formed by a concave mirror is real and inverted except when the object is kept between the focus and pole, where the image is virtual and erect.
• The relation between object distance (u) and image distance (v) with focal length (f) is given by the mirror equation or mirror formula

$⇒ \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

• ​Magnification: It is the ratio of the image distance (v) and object distance (u)
• ​Mathematically it is written as

$⇒ m = \frac{-v}{u} = \frac{h'}{h}$

Where h' is the height of the image and h is the height of the object.

CALCULATION:

Given that:

Focal length, f = 15 cm

The image is two times the size of the object.

∴ Magnification, m = -2

Let the image distance from the mirror be v and the object distance from the mirror be u.

Using sign convention, f = -15 cm

$⇒ m =- \frac{v}{u}$

$⇒ -2 = - \frac{v}{u}$

⇒ v = 2u

Using the mirror formula,

$⇒ \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$⇒ \frac{1}{-15} = \frac{1}{2u} + \frac{1}{u}$

$⇒ \frac{1}{-15} = \frac{3}{2u}$

⇒ u = - 22.5 cm

• Therefore, the distance between the object and the mirror is 22.5 cm.