An object of length 5 cm is placed on the principal axis of a concave mirror (f = -30 cm) at 50 cm from its pole. Find the object as parallel to the principal axis.

Steps to Solve

1. Identify the Given Data

We are given:

• Object distance, ( u = -50 ) cm (object distances are considered negative in mirror formulas)
• Focal length, ( f = -30 ) cm

2. Use the Mirror Formula

The mirror formula relates object distance ( u ), image distance ( v ), and focal length ( f ):
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$

Substituting the known values: $$ \frac{1}{-30} = \frac{1}{v} + \frac{1}{-50} $$

3. Rearrange and Solve for ( v )

Rearranging gives: $$ \frac{1}{v} = \frac{1}{-30} + \frac{1}{50} $$

Finding a common denominator (150) and simplifying: $$ \frac{1}{v} = \frac{-5 + 3}{150} $$ $$ \frac{1}{v} = \frac{-2}{150} $$

Thus, $$ v = -75 \text{ cm} $$

4. Determine the Characteristics of the Image

The image distance ( v ) is negative, indicating the image is formed on the same side as the object (virtual image).

To find the height of the image, we can use the magnification formula: $$ m = -\frac{v}{u} = \frac{h'}{h} $$

where ( h' ) is the image height, ( h = 5 ) cm is the object height. Substituting the known values: $$ m = -\frac{-75}{-50} = \frac{75}{50} = 1.5 $$

Then, substituting back to find ( h' ): $$ 1.5 = \frac{h'}{5} $$ $$ h' = 1.5 \times 5 = 7.5 \text{ cm} $$

So, the image height is ( 7.5 ) cm.