An object is placed in front of a concave mirror of radius of curvature 30 cm. An erect image thrice the size of the object is obtained. The position of the image is?

Steps to Solve

1. Understanding Mirror Properties

For a concave mirror, the radius of curvature ($R$) is given as 30 cm. The focal length ($f$) can be calculated using the formula:

$$ f = \frac{R}{2} $$

So,

$$ f = \frac{30, \text{cm}}{2} = 15, \text{cm} $$

2. Using Magnification

The magnification ($m$) for a mirror is defined as the ratio of the height of the image ($h_i$) to the height of the object ($h_o$). Given that the image is erected and thrice the size of the object, we have:

$$ m = \frac{h_i}{h_o} = 3 $$

3. Relating Object Distance and Image Distance

The magnification is also related to image distance ($v$) and object distance ($u$) by the formula:

$$ m = -\frac{v}{u} $$

Combining this with $m = 3$, we get:

$$ 3 = -\frac{v}{u} $$

Which leads to:

$$ v = -3u $$

4. Applying the Mirror Formula

The mirror formula relates object distance, image distance, and focal length:

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

Substituting $v = -3u$ into the formula gives us:

$$ \frac{1}{15} = \frac{1}{u} - \frac{1}{3u} $$

5. Finding Common Denominator

To solve the equation, find a common denominator:

$$ \frac{1}{15} = \frac{3}{3u} - \frac{1}{3u} $$

This simplifies to:

$$ \frac{1}{15} = \frac{2}{3u} $$

6. Solving for Object Distance

Cross multiplying to solve for $u$ results in:

$$ 2 \cdot 15 = 3u $$

Thus,

$$ 30 = 3u $$

So,

$$ u = 10, \text{cm} $$

7. Finding Image Distance

Now substitute $u$ back to find $v$:

$$ v = -3u = -3 \cdot 10 = -30, \text{cm} $$

This indicates that the image is located 30 cm in front of the mirror.