# A concave mirror (f = 41 cm) produces an image whose distance from the mirror is one-third the object distance. How do you determine the object distance in cm? What is the (positive) image distance in cm?

Jan 19, 2016

Object distance $= 164 c m$
And Image distance $= \frac{164}{3} c m = 54. \dot{6} c m$

#### Explanation:

For a concave mirror we know that there are various locations of object and the image being formed. These are as follows.
Let F be Focal Point, C its Center of Curvature where C=2F

1. Object is placed between F and the mirror
The image formed is behind the mirror, virtual, upright and Magnified.

2. Object is placed at the Focal Point
Image is formed at infinity. it can be real or virtual upright or inverted depending on approach to F. Formed in front of mirror.

3. Object is placed between F and C
Image is formed between infinity and C. It is real, inverted and magnified.

4. Object at C
Image is formed at C. It is real, inverted and of the same size as the object.

5. Object beyond center of curvature
Image is formed between C and F. It is real, inverted and diminished.

6. Object at infinity
Image is formed at F. t is real, inverted and diminished and tends to be zero.

In the problem, it is given that "produces an image whose distance from the mirror is one-third the object distance". Implies that it is under case number 5.

The ray diagram for this case is as under

Given that $\frac{1}{3} {d}_{o b j e c t} = {d}_{i m a g e}$ and $f = 41 c m$ Substituting in
Mirror and Lens equation
$\frac{1}{d} _ \left(o b j e c t\right) + \frac{1}{d} _ \left(i m a g e\right) = \frac{1}{f}$
$\frac{1}{d} _ \left(o b j e c t\right) + \frac{1}{{d}_{o b j e c t} / 3} = \frac{1}{41}$
$\implies \frac{1}{d} _ \left(o b j e c t\right) + \frac{3}{{d}_{o b j e c t}} = \frac{1}{41}$
$\implies \frac{1}{d} _ \left(o b j e c t\right) \left(1 + 3\right) = \frac{1}{41}$
$\implies {d}_{o b j e c t} = 4 \times 41 = 164 c m$
Also ${d}_{i m a g e} = \frac{164}{3} c m$