Tag: reflection of light at curved surfaces

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Questions Related to reflection of light at curved surfaces

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

The linear magnification for a spherical mirror is the ratio of the size of the image to the size of the object, and is denoted by m. Then m is equal to (symbols have their usual meanings)

  1. $\dfrac {u}{u-f}$
  2. $\dfrac {u f}{u-f}$
  3. $\dfrac {f}{u+f}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

General equation for a spherical mirror says that:
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

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

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

$\dfrac{v}{u}=\dfrac{f}{u+f}=m$ (magnification)

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 concave mirror forms the real image of an object which is magnified 4 times. The objects is moved 3 cm away, the magnification of the image is 3 times. What is the focal length of the mirror?

  1. 3 cm

  2. 4 cm

  3. 12 cm

  4. 36 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For mirror $u=\frac {f(m-1)}{m}$
In first case, $u=\frac {f(-4-1)}{-4}$
In the second case, $u+3=\frac {f(-3-1)}{-3}$
On solving, we get $f=36 cm$
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

The distance between an object and its doubly magnified image by a concave mirror is: [ Assume $f$ = focal length]

  1. $ 3 f/2 $
  2. $2 f/3 $
  3. $3f$

  4. Depends on whether the image is real or virtual.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The magnification is given as,

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

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

$v =  - 2u$

Ignoring the sign and using mirror formula, we get

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

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

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

$u = \dfrac{{3f}}{2}$

Here, difference between object distance and image distance is also$u$.

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).