Light and Mirrors

A real image is always inverted.

Light changes its direction after reflection. As we move the object away from the light source, the shadow becomes larger as the distance between object and the light source increases. Reflection is possible from rough surface also.

Hubble space telescope is reflecting telescope it is designed with TWO hyperbolic mirrors. A curved primary mirror is the basic optical element that creates an image at the focal plane. And secondary mirror was added to modify the optical characteristics for visual observation.

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 $f=\displaystyle\frac{u\times v}{u+v}$

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 .

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

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

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