# A concave mirror with a radius of curvature of 1.7 m is illuminated by a candle located on the...

## Question:

A concave mirror with a radius of curvature of 1.7 m is illuminated by a candle located on the symmetry axis 3.7 m from the mirror. Where is the image of the candle? Answer in units of m.

## Mirror Equation & Sign Conventions:

Mirror equation is the relation between the focal length, object distance and image distance of a mirror. It is expressed by the equation;

{eq}\dfrac {1}{v}+\dfrac {1}{u} = \dfrac{1}{f} {/eq}

where,

• {eq}v {/eq} is the image distance,
• {eq}u {/eq} is the object distance,
• {eq}f {/eq} is the focal length of the mirror.

Sign conventions:

The image formation by the spherical mirror follows some sign conventions. These are the followings;

• Distances measured in the direction of the incident ray are taken as positive.
• Distances measured in the direction opposite to the incident ray are taken as negative.
• Distances measured above the principal axis are taken as positive.
• Distances measured below the principal axis are taken as negative.

Given:

• Distance of object from the mirror is {eq}u= -3.7\ m {/eq}
• Radius of curvature of the mirror is {eq}R= -1.7\ m {/eq}

Let:

• {eq}v {/eq} is the image distance.

Focal length of the mirror is given by;

{eq}\begin{align} f&=\dfrac{R}{2}\\ &=\dfrac{-1.7\ m }{2 }\\ &= -0.85\ m\\ \end{align} {/eq}

From the Mirror formula,

{eq}\begin{align} \dfrac {1}{v}+\dfrac {1}{u} &= \dfrac {1}{f}\\ \implies \dfrac {1}{v}+\dfrac {1}{-3.7\ m} &= \dfrac {1}{-0.85\ m}\\ \implies v &=-1.1035\ m \\ \end{align} {/eq}

Therefore, the image of the candle is formed at 1.1035 m behind the mirror.