Please help me!!

A mirror of focal length 26.6 cm creates an image with magnification −0.350. What is the image distance?

if know what the answer is... can you explain it to me. thnxs!!!

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- May 7th 2009, 03:53 PMsoojin89mirror equation...hate them!!
Please help me!!

A mirror of focal length 26.6 cm creates an image with magnification −0.350. What is the image distance?

if know what the answer is... can you explain it to me. thnxs!!! - May 9th 2009, 05:33 AMfardeen_gen
You haven't mentioned whether it is a concave or convex mirror. So I am taking cases.

Using the mirror equation,

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

where:

$\displaystyle u$ - object distance

$\displaystyle v$ - image distance

$\displaystyle f$ - focal length of the mirror

(NOTE: The equation is formulated defining the pole as origin and the quantities $\displaystyle u,v\ \mbox{and}\ f$ denote the x-coordinates)

(For concave mirror):**CASE I**

We have,

$\displaystyle f$ = - 26.6 cm (Considering focus towards left of pole as it is a concave mirror)

So using mirror equation,

$\displaystyle \frac{1}{u} + \frac{1}{v} = -\frac{1}{26.6}$

$\displaystyle \frac{u + v}{uv} = -\frac{1}{26.6}$ --------- ■

We know that,

Magnification = $\displaystyle \frac{\mbox{Height of image}}{\mbox{Height of object}}$ = $\displaystyle -\frac{v}{u}$

(The equation is formulated taking proper signs)

Thus,

$\displaystyle - 0.35 = -\frac{v}{u}$

$\displaystyle \implies v = 0.35u$ -------- ●

Using the above relation in ■,

$\displaystyle \frac{u(0.35 + 1)}{0.35u^2} = -\frac{1}{26.6}$

$\displaystyle \implies (1.35\times 26.6)u = -0.35u^2$

$\displaystyle \implies 35.91u + 0.35u^2 = 0$

$\displaystyle \implies u(35.91 + 0.35u) = 0$

$\displaystyle \implies u = 0$ (rejected as image cannot be formed at the pole)

or $\displaystyle u = -\frac{35.91}{0.35}\ \mbox{cm} = - 102.6\ \mbox{cm}$

Using ●,

$\displaystyle v = 0.35\times -102.6 = -35.91\ \mbox{cm}$

Thus, the image is formed at a distance of $\displaystyle \boxed{35.91\ \mbox{cm}}$ towards the left of the concave mirror. (The negative sign implies negative x-axis which is towards the left of the origin i.e the pole)

Similarly,

(For Convex mirror):**CASE II**

We have,

$\displaystyle f$ = + 26.6 cm [Considering focus to the right of the pole]

Repeating the rest of the process,

$\displaystyle v = +35.91\ \mbox{cm}$

Thus, the image is formed at a distance of $\displaystyle \boxed{35.91\ \mbox{cm}}$ towards the right of the convex mirror.

(The positive sign implies positive x-axis which is towards the right of the origin i.e. the pole)