Question

A concave mirror has a radius of curvature 20 cm. Calculate the distance of an object from the mirror so as to form an image of magnification -2. Also, find the location of the image.

Hint:

For concave mirrors, the focal length is half of the radius of curvature. In this case, removing the silver coating would prevent the mirror from reflecting light, rendering it ineffective for image formation.

Answer 1

(a) Object Distance Calculation:
The relationship between magnification (\( m \)), object distance (\( u \)), and image distance (\( v \)) is given by:\[m = \frac{-v}{u}\]Given \( m = -2 \), we have:\[-2 = \frac{-v}{u}\]This simplifies to:\[v = 2u\]The mirror equation is:\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]For a concave mirror, the focal length (\( f \)) is half the radius of curvature (\( R \)):\[f = \frac{R}{2}\]With \( R = 20 \, \text{cm} \), the focal length is:\[f = \frac{20}{2} = 10 \, \text{cm}\]Substituting \( f \) into the mirror equation:\[\frac{1}{10} = \frac{1}{u} + \frac{1}{2u}\]Simplifying the right side:\[\frac{1}{10} = \frac{3}{2u}\]Solving for \( u \):\[u = 15 \, \text{cm}\]Using \( v = 2u \), we find \( v \):\[v = 2 \times 15 = 30 \, \text{cm}\]Therefore, the object distance is \( u = 15 \, \text{cm} \) and the image distance is \( v = 30 \, \text{cm} \).(b) Consequences of Silver Coating Removal:
Removing the silver coating from the center of a concave mirror eliminates its reflective capability at that point. The silver coating is essential for reflection. Without it, the concave mirror loses its reflective properties and cannot form an image because the reflective surface is necessary for focusing light.Consequently, the mirror will not form any image. Reflection will not occur at the center, rendering the mirror incapable of image formation.