Tag: reflection of light at curved surfaces

Questions Related to reflection of light at curved surfaces

Magnification produced is +$\dfrac { 1 }{ 3 }$, then what kind of mirror it is?

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

  1. concave mirror

  2. convex mirror

  3. opaque mirror

  4. plane mirror

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Correct Option: B
Explanation:

Since the magnification produced is positive and less than 1,the mirror is a convex mirror.

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)

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

  1. $\dfrac {u}{u-f}$

  2. $\dfrac {u f}{u-f}$

  3. $\dfrac {f}{u+f}$

  4. None of these

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Correct Option: C
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)

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?

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

  1. 3 cm

  2. 4 cm

  3. 12 cm

  4. 36 cm

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Correct Option: D
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 $f=36 cm$

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

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

  1. $ 3 f/2 $

  2. $2 f/3 $

  3. $3f$

  4. Depends on whether the image is real or virtual.

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Correct Option: A
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$u$.

The magnification of plane mirror is always - 

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

  1. $ <1 $

  2. $ > 1 $

  3. $ = 1 $

  4. Zero

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Correct Option: C
Explanation:

The size of the image is equal to size of the object for plane mirror.

So, magnification is equal to $1$ .

 

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-

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

  1. $1600 sq. cm$

  2. $800 sq. cm$

  3. $400 sq. cm$

  4. $200 sq. cm$

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Correct Option: A
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 $=16\times 100=1600$ ${ cm }^{ 2 }$.

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 :

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

  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. $b\left (\dfrac {f}{u-f}\right )^2$

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Correct Option: D
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  $dv=b{ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$

Magnification for erect and invented image is

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

  1. $+ve$ and $-ve$ respectively

  2. $-ve$ and $+ve$ respectively

  3. $+ve$

  4. $-ve$

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Correct Option: A
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).