Tag: sign convention for spherical mirrors

R=1.5, u=$-10$

To form a image only two rays are needed .The total amount of light released by the object is not allowed to pass through the lens, intensity of image will decrease

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.

$\dfrac {\text {Area of the image}}{\text {Area of the object}}=$ Aerial magnification $=m^2$
Where m is the linear magnification
$\therefore m=\dfrac {f}{f+u}=\dfrac {0.5}{0.5 - 2000} \approx -\dfrac {1}{4000}$
Now, area of the image is equal to size of the camera film, therefore
Area of object$=\dfrac {\text {Area of the image}}{m^2}$
$=18\times 18\times 4000\times 4000$
$=72000 cm\times 72000 cm$
Note that m has no unit, so area of object is in same units as area of image.

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).

$\dfrac { 1 }{ f } =\left( n-1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right) $
$\Rightarrow \dfrac { 1 }{ f } =\left( \dfrac { 1.5 }{ 1 } -1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right)$ when the lens is placed in air and 
$\dfrac { 1 }{ xf } =\left( \dfrac { 1.5 }{ y } -1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right)$ when the lens is places in the liquid.
where $y=R.l.$ of the liquid
solving we get, $y=\dfrac {3}{2+1/x}$
Hence $(B)$ is correct.

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.