Ray Optics and Mirrors

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 $7.5\ cm$

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.

Given 1/f = 1/u + 1/v, the uncertainty is df/f^2 = du/u^2 + dv/v^2. Plugging in the values: df = f^2 * (du/u^2 + dv/v^2) = 40^2 * (3/50^2 + 5/200^2) = 1600 * (3/2500 + 5/40000) = 1600 * (0.0012 + 0.000125) = 1600 * 0.001325 = 2.12 mm.

Concave shaving mirror  $f = 0.2m$

$m = \left(\dfrac{f}{f + u}\right)$

$-2 = \dfrac{20}{20 - x _2}$

$-10 x _2 = 10$
$x _2 = 20 \,cm$

$m = 2 = \dfrac{20}{20 - x _1}$

$20 - x _1 = 10$
$x _1 = 10$

$\dfrac{x _1}{x _2} = \dfrac{10}{20} = \left(\dfrac{1}{2}\right)$

From lens formula
$\dfrac{{1}{v}-\dfrac{1}u}=\dfrac{1}{f}$
$\dfrac{u}{v}-1=\dfrac{u}{f}$
$\dfrac{u}{v}=\dfrac{u+f}{f}$
$\dfrac{v}{u}=\dfrac{f}{u+f}$
For a convex lens f = 0.12m and m=3 then find u
So ,  $3=\dfrac{0.12}{0.12+u}$
$0.36+3u = 0.12$
$u=-0.08 m$

$\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)

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).

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 $(C)$