Sign convention for mirrors - class-X

Magnification m = -v/u = 1/3 for a convex mirror. Using the mirror formula 1/v + 1/u = 1/f, we substitute v = -u/3. This gives -3/u + 1/u = 1/f, leading to -2/u = 1/f, so u = -2f. The distance is 2f.

Positive and negative sign depend on the assumption of sign conversion.
either side we can consider positive or negative.
Hence Option D.

we now,$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
multiplying by u in above eq.
$\dfrac{u}{f}=\dfrac{u}{v}+\dfrac{u}{u}$
$\dfrac{u}{f}=\dfrac{u}{v}+1$
$\dfrac{u}{f}-1=\dfrac{u}{v}$
$\dfrac{u}{v}=\dfrac{u-f}{f}$
$\dfrac{v}{u}=\dfrac{f}{u-f}  ,  As, m=\dfrac{v}{u}$
$m=\dfrac{f}{u-f}$
hence,option C is correct.

The sum of the reciprocals of object distance and image distance is equal to the reciprocal of the focal length of a mirror.

The relation between $u,,v;and;f$ for a mirror is given by $f=\displaystyle\frac{u\times v}{u+v}$

The relation between $u,,v;and;f$ for a mirror is given by $f=\displaystyle\frac{u\times v}{u+v}$

Magnification is a ratio of lengths, hence it has no units

magnification ratio is given as:

Since focal length of concave mirror given is 20cm l. Object is at 40cm distance which means it is at centre of curvature. hence, image will be formed of same size and at centre focus only but real and inverted.

Linear magnification is given as Image size/ Object size

According to the sign convention followed for mirrors and for lenses, the distances measured above the principal axis are taken as positive and the distances measured below the principal axis are taken as negative.

Hence, the correct answer is OPTION B.

The magnification is the ratio of height of the image to height of the object. The image is smaller than the object, the magnification will be less than 1. If the image if upside down (inverted) then magnification will be negative.

Since the image is erect , so the magnification ratio must be positive .

$m=+\dfrac{v}{u}=\dfrac{+f}{u-f}$
$3=\dfrac{+f}{+2+f}\Rightarrow f=3cm$.

Given,

Focal length, $f=50\,cm$

Magnification, $m=2$

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

$ v=2u $

From mirror formula,

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

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

$ u=\dfrac{3f}{2}=\dfrac{3\times 50}{2}=75\ cm $

Object Is placed at $75\,cm$ from the mirror.

For a convex mirror, f is positive. With object distance u = -f, the mirror formula 1/v + 1/u = 1/f becomes 1/v - 1/f = 1/f. Thus, 1/v = 2/f, so v = f/2.

Given that,

Focal length $=f$

We know that, the magnification is

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

Now, magnification is double and image is real

  $ -2=\dfrac{-v}{u} $

 $ v=2u $

Now, using formula of mirror

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

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

 $ \dfrac{1}{f}=\dfrac{3}{2u} $

 $ u=\dfrac{3f}{2} $

Hence, the distance is $\dfrac{3f}{2}$ 

Given that,

Distance $u=-10\,cm$

Height $h=2.5\,cm$

Radius of curvature $R=30\,cm$

We know that,

  $ -f=\dfrac{R}{2} $

 $ -f=\dfrac{30}{2} $

 $ f=-15\,cm $

Now, using mirror’s formula

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

 $ -\dfrac{1}{15}=\dfrac{1}{v}-\dfrac{1}{10} $

 $ \dfrac{1}{v}=-\dfrac{1}{15}+\dfrac{1}{10} $

 $ v=30\,cm $

Now, the magnification is

  $ m=\dfrac{-v}{u} $

 $ m=\dfrac{30}{10} $

 $ m=3 $

We know that,

  $ m=\dfrac{h'}{h} $

 $ 3=\dfrac{h'}{2.5} $

 $ h'=7.5\,cm $

Hence, the size of the image is $7.5\ cm$

By definition, rays incident parallel to the principal axis of a concave mirror reflect through the principal focus (F).

$\begin{array}{l} In, y-direction \ { V _{ o } }={ V _{ i } } \ \Rightarrow { V _{ i } }=4\widehat { j }  \ In, z-direction \ { V _{ i } }={ V _{ o } } \ \Rightarrow { V _{ i } }=5\widehat { k }  \ In, x-direction \ { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ I } }-{ V _{ M } }={ V _{ M } }-{ V _{ o } } \ \Rightarrow { V _{ I } }=2{ V _{ M } }-{ V _{ o } } \ =2\left[ { \widehat { i }  } \right] -3\left[ { \widehat { i }  } \right]  \ =-\widehat { i }  \ \therefore \overrightarrow { { V _{ I } } } =-\widehat { i } +4\widehat { j } +5\widehat { k }  \ \therefore \overrightarrow { { V _{ iM } } } =-2\widehat { i } +5\widehat { j } +4\widehat { k }  \ \therefore \left| { \overrightarrow { { V _{ iM } } }  } \right| =\sqrt { 45 }  \end{array}$

The rod of length f/3 is placed along the axis. Let the ends be at u1 and u2. The image ends are at v1 and v2. Magnification m = dv/du. Using 1/v + 1/u = 1/f, differentiating gives dv/v^2 = -du/u^2, so m = -v^2/u^2. Given the geometry, the calculation leads to m = -1.5.

Concave shaving mirror  $f = 0.2m$

According to new Cartesian sign convention, object distance for any lens or mirror is measured as negative. This is because, the object distance is measured against the direction of incident light.

The focal length of a concave mirror, with a real focus, is always positive as is measured against the direction of incident light.

According to cartesian system focal length of a concave mirror is negative whereas radius of curvature of convex mirror is positive.

$\dfrac { 1 }{ v } -\dfrac { 1 }{ 15 } =\dfrac { 1 }{ -10 } $
$\Rightarrow v=-30cm$
$\therefore m=-\dfrac { v }{ u } =-\dfrac { -30 }{ 15 } =2$
So, image size will be $4 cm$ (real)

Distance of object from the mirror, $u=-30$ 

The magnification equation relates the ratio of the image distance(v) and object distance(u) and also to the ratio of the image height ($h _i$) and object height ($h _o$).

R=1.5, u=$-10$

Focal length of concave mirror  $f = -10 \ cm$
Object distance   $u = -15 \ cm$
Using     $\dfrac{1}{v}+\dfrac{1}{u} = \dfrac{1}{f}$
$\therefore$   $\dfrac{1}{v}+\dfrac{1}{-15} = \dfrac{1}{-10}$
$\implies   \ v = -30 \ cm$
Since $v$ is negative, so image formed is real.
Magnification  $m = \dfrac{-v}{u}$
$\therefore$  $m = -\dfrac{(-30)}{-15} = -2$
Since $m$ is greater than $1$, so magnified image is formed.
Correct answer is option C.

1- It is true that all the distances are measured from the pole of a mirror parallel to the principal axis.So, Alpesh is right.

Here, the object lies along the axis. The two ends of the object should
be treated as two point objects and the difference between the
corresponding image distances gives the length of the image. When one
end of the image touches the rod, this end must be at 2 f. In this
situation the other end of the rod can be towards the left or right of 2
f. Since the image of the rod is elongated, the other end of the rod
must lie between f and 2 f, the image (when the object lies between f
and 2f, the image is formed more far away behind 2 f).
So, object
distance for closer end of the rod is 2f-f/3 and that of the farther end
is 2 f. The difference between the corresponding image distance is
found to be $\dfrac {f}{2},$ i.e. length of the image is $\dfrac
{f}{2}$.
magnification$=\dfrac {\text {length of image}}{\text {length of object}}=\dfrac {f/2}{f/3}=\dfrac {3}{2}$
Note that the image is elongated and the only option which is greater than 1 is (c).

Distance of the Candle from the Concave Mirror (u) = $40 cm$.(negative)

Now, as per as the Question,

$Image Height = 4 × Object (or Candle) Height$

 $Image Height/Candle Height = 4$

 Magnification = $4$

[ Magnification = Image Height/Object Height]

Now, Magnification = $-v/u$

 $4 = -v/u$

 $-v = 4u$

 $v = -4u$

 $v = -4 \times 40$

 $v = -160 cm.$

Now, Image Distance(v) = - $160 cm.$

Using the Mirror's Formula,

 On Multiplying both sides by $160$ ,

We get,

⇒  $160/f = -1 - 4$

⇒ $160/f = -5$

⇒ $f = 160/-5$

⇒ $f = -32 cm$.

∴ Focal length of the Concave Mirror is 32 cm.

Now, For the Radius of the Curvature,

Using the Formula,

 $ Focal Length = Radius Of Curvature/2$

∴ $Radius of Curvature = Focal Length \times 2$

∴ $R = F \times 2$

∴ $R = 32 \times 2$

∴ $R = 64 cm.$

Hence, the Radius of the Curvature of the Concave mirror of Focal Length 3 cm is 64 cm.