# What will be the distance at which a linear object must be placed from a concave mirror having a focal length of 0.8 m to obtain an image four times a

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What will be the distance at which a linear object must be placed from a concave mirror having a focal length of 0.8 m to obtain an image four times as long as the object?
1. 0.1 m
2. 0.7 m
3. 0.8 m
4. 0.9 m

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Correct Answer - Option 1 : 0.1 m

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 a focusing mirror/converging mirror.
• The size of the image produced by these mirrors can be larger or smaller than the object, depending upon the distance of the object from the mirror.
• The concave mirror can form both real and virtual images of any object.
• Mirror formula: The expression which shows the relation between object distance (u)image distance (v), and focal length (f) is called the mirror formula.

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

Linear magnification (m):

• It is defined as the ratio of the height of the image (hi) to the height of the object (ho).

$m = \frac{{{h_i}}}{{{h_o}}}$

• The ratio of image distance to the object distance is called linear magnification.

$m = \frac{{image\;distance\;\left( v \right)}}{{object\;distance\;\left( u \right)}} = - \frac{v}{u}$

• positive value of magnification means virtual an erect image.
• negative value of magnification means a real and inverted image.

CALCULATION:

Given - f = 0.8 m, m = 4, let U = x

• The magnification of the image is given by

$\Rightarrow 4 = -\frac{V}{u}$

$\Rightarrow V = -4x$

Substituting the value of V and u  in the equation for focal length

$\Rightarrow \frac{1}{f} = -\frac{1}{x}- \frac{1}{4x} = \frac{-5x}{4x^{2}}$

$\Rightarrow f = - \frac{4x^{2}}{5x} = -\frac{4x}{5}$

$\Rightarrow x = \frac{-5\times f}{4} =- \frac{-5\times 0.8}{4} = -0.1 m$

• Hence, the object must be placed at a distance of 10 cm away from it
• Hence, option 1 is the answer.

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