Mirror and Magnification Formula Problems

Here, $m =\dfrac{v}{u} =-4$

$\Rightarrow u = \dfrac{-v}{4}$------------(1)

Also, $|u| + |v| = 1.5$

$\dfrac{v}{4} + v = 1.5$

$\Rightarrow \dfrac{(v+4 v)}{4} = 1.5$

 $\Rightarrow  v = 1.2 m$

So, putting the value of $v$ in equation (1):
 $u= \dfrac{-1.2}{4} = -0.3 m,,,$

$\therefore  f =\dfrac{uv}{u-v}$

$\Rightarrow f=\dfrac{(-0.3 \times 1.2)}{(-0.3 - 1.2)} = 0.24m$
Hence the correct option is $(C)$

$\dfrac{2}{R}=\dfrac{1}{u}+\dfrac{1}{v}$

Ratio of size of image to size of object is known as linear
magnification of spherical mirrors.

Linear magnification (m) is the ratio of height of image to that of the object.

Magnification is ratio of the size of the image $h _i$ to the size of the object $h _o$.

Magnification is the increase in the image size produced by spherical mirrors with respect to the object size. It is the ratio of the height of the image to the height of the object. The height of the object is always positive as the object is always above the principal axis. 

Ratio of the size of the image to the size of the object is known as magnification. It is given by $m = v/u$

We know,
Magnification(M)$=\dfrac{height  \ of\   image({h} _{i})}{height\   of  \ object({h} _{o})}$
Here,image is inverted so the $height \  of \  image({h} _{i})$will be negative.
Hence, the magnification of a mirror is negative.

Magnification does not have any unit as it is the ratio of same quantity.

From mirror formula,       $\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } $

$M= \frac{f}{f-d _0}$ and real image has M negative

$-3= \frac{f}{f-20}$

$-3f+60=f$

$f=15 cm$

$M= \frac{15}{15-30}$

$M= -1$

Given object distance $u=15$ cm

Given that,

The object distance $u=-6\,cm$

Now, magnification is

  $ m=\dfrac{I}{O} $

 $ m=\dfrac{f}{f-u} $

 $ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $

 $ I=-2\,cm $

Hence, the length of image is -$2\ cm$

Given focal length of eye piece${f} _{e}=10cm\$

Given magnification $M=\dfrac{v}{u}=3$

For a real image, magnification m = -v/u = -2. Thus v = 2u. Using 1/v + 1/u = 1/f with f = -20 cm, we have 1/(2u) + 1/u = 1/-20. This simplifies to 3/(2u) = -1/20, so 2u = -60, u = -30 cm. The distance is 30 cm.

This is a property of concave mirrors where the focal length f is the geometric mean of the distances of the object and image from the focus. Specifically, f^2 = x * y, so f = sqrt(xy).

The radius of curvature R is 15 m, so the focal length f = R/2 = 7.5 m. The sun subtends an angle theta = 0.5 degrees = 0.5 * (pi/180) radians. The diameter of the image d is given by f * theta, which evaluates to (7.5 m) * (0.5 * pi / 180) approximately equal to 0.0654 m or 6.55 cm.

For a ray parallel to the axis striking at angle of incidence theta, the distance from the pole to the intersection point is f/cos(theta) = (R/2)/cos(theta). The distance from the center of curvature (at R) is R - (R/2)/cos(theta). None of the options match this standard derivation perfectly, but R/2 is the paraxial approximation.

Magnification for a mirror, $m = \dfrac{f}{f-u}$

$u=30$ ; $f=20$ 

Image can be formed on the screed if it is real. Real image of reduced size can be formed can be formed by a concave mirror or a convex lens.

$\displaystyle \frac {1}{V} - \frac {1}{2f} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac {3}{2f}  \Rightarrow v = \frac{2}{3}f$
$\therefore m = \displaystyle \frac {u}{v} = \frac{2}{3} \frac{f}{2f} = \frac {1}{3}$

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$.........(1)

Mirror equation is: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

A convex mirror always creates a virtual image which is diminished. So, magnification produced by convex mirror is always less than one.

$\text{Magnification}=\dfrac{\text{Image  size}}{\text{Object  size}}$

Given: $u=-10cm$

Given: $u=-60 cm$     $f=-40 cm$
To find : $m$
Solution: From mirror formula
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}-\dfrac{1}{60}=\dfrac{-1}{40}$
$\dfrac{1}{v}=\dfrac{1}{60}-\dfrac{1}{40}$
$v=-120 cm$
Hence magnification is given by
$m=\dfrac{-v}{u}=-\dfrac{(-120)}{(-60)}$
$m=-2$

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

Magnification, $m = \dfrac {-v}{u}$
Given, magnification, $m = -2$

Magnification $= +1$ signifies that the image formed in a plane mirror is of same size as the object. Positive sign in the value of magnification signifies that image formed by a plane mirror is erect.

Given that, $\displaystyle R=-24$cm
$\displaystyle f=-12cm$ and $\displaystyle m=1.5$
By the lens formula,
$\displaystyle \frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } $
$\displaystyle \frac { 1 }{ 1.5u } +\frac { 1 }{ u } =-\frac { 1 }{ 12 } $
$\displaystyle \frac { 2.5 }{ 1.5u } =-\frac { 1 }{ 12 } $
or, $\displaystyle u=-20cm$

Magnification by a mirror         $m = \dfrac{-v}{u}$

The image produced is virtual and erect and also diminished.

Since the object and the image move in opposite directions, the position of the object should be in between $f$ and $2f$.
So, $f < x < 2f$

Since the magnification produced is positive and less than 1,the mirror is a convex mirror.

General equation for a spherical mirror says that:
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

The magnification is given as,

$m = \dfrac{{ - v}}{u}$

$2 = \dfrac{{ - v}}{u}$

$v =  - 2u$

Ignoring the sign and using mirror formula, we get

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{{1 + 2}}{{2u}} = \dfrac{1}{f}$

$u = \dfrac{{3f}}{2}$

Here, difference between object distance and image distance is also$u$.

The size of the image is equal to size of the object for plane mirror.

So, magnification is equal to $1$ .

As linear magnification, $M=4$

From mirror formula,

Magnification is the ratio of height of image and object .

The power of a lens is P = 1/f. When a lens and a mirror are combined, the equivalent power is P_eq = P_lens + P_mirror. Given P_lens = +2D and P_mirror = +1D, the total power is +3D.

According to new Cartesian sign convention, image size for erect image is considered positive and that of inverted image is considered negative.

if object distance < f then m>1

Convex mirrors are used in automobiles to see the near field of view as convex mirror has a wider field of view and it has higher magnification.