Tag: derivation of formula for curved mirrors

Questions Related to derivation of formula for curved mirrors

We want a mirror that will make an object look larger. What combination of image and object distances (from the mirror) will accomplish this?

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

  1. Image Distance $3.0 cm$, Object Distance $3.0 cm$

  2. Image Distance $2.0 cm$, Object Distance $3.0 cm$

  3. Image Distance $3.0 cm$, Object Distance $5.0 cm$

  4. Image Distance $3.0 cm$, Object Distance $2.0 cm$

  5. Image Distance $3.0 cm$, Object Distance $10.0 cm$

Show answer
Correct Option: D
Explanation:

Magnification by a mirror         $m = \dfrac{-v}{u}$


For option D :    $m = \dfrac{- 3.0}{-2.0} = 1.5$             $\implies m>1$
Thus the data given in option D will make a larger image.

The magnification produced by a mirror is $+\dfrac{1}{3}.$ Then the mirror is a ____________.

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. Plane mirror

  4. plano convex mirror

Show answer
Correct Option: B
Explanation:

The image produced is virtual and erect and also diminished.


Hence, the mirror must be $convex$ mirror. Concave mirror also produces virtual image but it is enlarged.But convex mirror always produces diminished virtual image.

Answer-(B)

In an experiment to determine the focal length ($f$) of a concave mirror by the $u-v$ method, a student places the object pin A on the principal axis at a distance $x$ from the pole $P$. The student looks at the pin and its inverted image from a distance keeping the eye in line with $PA$. When the student shifts the eye towards left, the image appears to the right of the object pin. Then:

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

  1. $x< f$

  2. $f< x< 2f$

  3. $x= 2f$

  4. $x> 2f$

Show answer
Correct Option: B
Explanation:

Since the object and the image move in opposite directions, the position of the object should be in between $f$ and $2f$.
So, $f < x < 2f$

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

Show answer
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

Show answer
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

Show answer
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.

Show answer
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

Show answer
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. $200 sq. cm$

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

Show answer
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  $dv=b{ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$