Tag: problems on mirror and magnification formula

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Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A flim projector magnifies a flim of area $100 $ square centimeter on screen. If linear magnification is $4$ then area of magnified image on screen will be-

  1. $1600 sq. cm$
  2. $800 sq. cm$
  3. $400 sq. cm$
  4. $200 sq. cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As linear magnification, $M=4$

Hence, a real magnification ${ m } _{ r }={ m }^{ 2 }$
${ \left( 4 \right)  }^{ 2 }=16$
Surface area of film image on screen $=16\times 100=1600$ ${ cm }^{ 2 }$.

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A short linear object of length $b$ lies along the axis of a concave mirror of focal length $f$ at a distance u from the pole of the mirror. The size of the image is approximately equal to :

  1. $b\left (\dfrac {u-f}{f}\right )^{\dfrac {1}{2}}$
  2. $b\left (\dfrac {b}{u-f}\right )^{\dfrac {1}{2}}$
  3. $b\left (\dfrac {u-f}{f}\right )$
  4. $b\left (\dfrac {f}{u-f}\right )^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From mirror formula,


$\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \longrightarrow (1)$

Differentiating, we get  


$\Rightarrow -{ \upsilon  }^{ -2 }dv-{ u }^{ -2 }du=0$

or $\left| d\upsilon  \right| =\left| \cfrac { { \upsilon  }^{ 2 } }{ { u }^{ 2 } }  \right| du \ \longrightarrow (2)$         

Here $\left| dv \right| =$size of image,

$\left| du \right| =$size of object $\left( =b \right) $

From the equation $1$, we write

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

Squaring both sides, we get

$\cfrac { { \upsilon  }^{ 2 } }{ { v }^{ 2 } } ={ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$

Substituting in equation $2$ we get

Size of the image  $dv=b{ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

Magnification for erect and invented image is

  1. $+ve$ and $-ve$ respectively
  2. $-ve$ and $+ve$ respectively
  3. $+ve$
  4. $-ve$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnification is the ratio of height of image and object .

And by convention , height of image formed below principal axis is taken negative and above is taken positive.

Hence, for erect image, $m=+ve$ and for inverted image $m=-ve$.

Answer-(A).

Multiple choice physics ray optics and optical instruments problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A convex lens of 2 D power is joined with a concave mirror of 1 D power. Equivalent power of instrument will be

  1. $-3D$
  2. $+3D$
  3. $-5D$
  4. $+5D$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

Multiple choice physics ray optics and optical instruments problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

If image is real and inverted, mangification will be:

  1. two

  2. negative

  3. one

  4. zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

The object being always erect, the object size is always positive.
So, magnification for a real image, being a ratio of image size and object size is considered negative.