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!!!
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)
CASE I(For concave mirror):
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,
CASE II(For Convex mirror):
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)