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Derivation of formula for curved mirrors - class-XI

Description: derivation of formula for curved mirrors
Number of Questions: 53
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Tags: physics reflection of light at curved surfaces light light - reflection and refraction ray optics and optical instruments optics reflection of light
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An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is 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. 1 m

  2. 0.5 m

  3. 0.24 m

  4. 2 m

Show answer
Correct Option: C
Explanation:

Here, $m =\dfrac{v}{u} =-4$

$\Rightarrow u = \dfrac{-v}{4}$------------(1)

Also, $|u| + |v| = 1.5$

$\dfrac{v}{4} + v = 1.5$

$\Rightarrow \dfrac{(v+4 v)}{4} = 1.5$

 $\Rightarrow  v = 1.2 m$

So, putting the value of $v$ in equation (1):
 $u= \dfrac{-1.2}{4} = -0.3 m\,\,\,$

$\therefore  f =\dfrac{uv}{u-v}$

$\Rightarrow f=\dfrac{(-0.3 \times 1.2)}{(-0.3 - 1.2)} = 0.24m$
Hence the correct option is $(C)$

A convex mirror used for rear view on an automobile has a radius of curvature of 3.00m. If a bus is located at 5.00m from this mirror, find magnification?

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

  1. +0.23

  2. -0.23

  3. +0.45

  4. -0.45

Show answer
Correct Option: A
Explanation:

$\dfrac{2}{R}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{2}{3}=\dfrac{1}{-5}+\dfrac{1}{v}$
$\dfrac{1}{v}=0.866 \Rightarrow  v = 1.15$ cm
Magnification 
$=-\dfrac{v}{u}=-\dfrac{1.15}{-5}=0.23$

What is known as linear magnification of spherical mirrors?

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

  1. Ratio of size of image to size of object

  2. Ratio of shape of image to size of object

  3. Ratio of size of image to shape of object

  4. None

Show answer
Correct Option: A
Explanation:


Ratio of size of image to size of object is known as linear magnification of spherical mirrors.

Linear magnification (m) is the ratio of height of image to that of the object.

Magnification $(m) =$ ______

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

  1. $\dfrac {v}{u}$

  2. $\dfrac {u}{v}$

  3. $\dfrac {h _{o}}{h _{i}}$

  4. $\dfrac {h _{i}}{h _{o}}$

Show answer
Correct Option: D
Explanation:

Magnification is ratio of the size of the image $h _i$ to the size of the object $h _o$.

$m=\dfrac{h _i}{h _o} = \dfrac{-v}{u}$ where $u$ is object distance and $v$ is image distance.

For magnification in spherical mirrors object height 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. Negative.

  2. Positive.

  3. For real images positive.

  4. For virtual images negative.

Show answer
Correct Option: B
Explanation:

Magnification is the increase in the image size produced by spherical mirrors with respect to the object size. It is the ratio of the height of the image to the height of the object. The height of the object is always positive as the object is always above the principal axis. 

Ratio of the size of the image to the size of the object is known as:

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

  1. Focal plane

  2. Transformation ratio

  3. Efficiency

  4. None of these

Show answer
Correct Option: D
Explanation:

Ratio of the size of the image to the size of the object is known as magnification. It is given by $m = v/u$

In case of a real and inverted image, the magnification of a mirror 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. Positive

  2. Negative

  3. Zero

  4. Infinity

Show answer
Correct Option: B
Explanation:

We know,
Magnification(M)$=\dfrac{height  \ of\   image({h} _{i})}{height\   of  \ object({h} _{o})}$
Here,image is inverted so the $height \  of \  image({h} _{i})$will be negative.
Hence, the magnification of a mirror is negative.

Which of the following quantity does not have any unit?

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

  1. Velocity of light

  2. Light year

  3. Magnification

  4. Power of a lens

Show answer
Correct Option: C
Explanation:

Magnification does not have any unit as it is the ratio of same quantity.

The expression for the magnification of a spherical mirror in the terms of focal length (f) and the distance of the object from mirror (u) 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. $\frac{-f}{u-f}$

  2. $\frac{f}{u+f}$

  3. $\frac{-f}{u+f}$

  4. $\frac{f}{u-f}$

Show answer
Correct Option: D
Explanation:
Equation of spherical mirror is $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ , where v is the image distance.
solving,
Replacing $v$ with $v=mu$ , 
$\dfrac{1}{f} = \dfrac{1}{u}  (1 + \dfrac{1}{m} ) $
 $u = f (\dfrac{1}{m} +1)$ where m = magnification $= \dfrac{v}{u}$
$u =\dfrac{ f}{m} + f$
$m = \dfrac{f }{ (u - f)}$

A short linear object of length $L$ lies on the axis of a spherical mirror of focal length $f$ at a distance $u$ from the mirror. Its image has an axial length $L$ 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. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 1/2 }$

  2. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 1/2 }$

  3. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 2 }$

  4. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 2 }$

Show answer
Correct Option: D
Explanation:

From mirror formula,       $\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } $


On differentiating, we get


      $\cfrac { -dv }{ { v }^{ 2 } } -\cfrac { du }{ { u }^{ 2 } } =0\\ \therefore dv=-du{ \left( \cfrac { v }{ u }  \right)  }^{ 2 }\\ as\quad \cfrac { v }{ u } =\cfrac { f }{ u-f } \\ \therefore dv=-du{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }\\ { L }^{ \prime  }=L{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }$

If an object is placed at a distance of 20cm from the pole of a concave mirror, the magnification of its real image is 3. If the object is moved away from the mirror by 10cm, then the magnification is -1.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$M= \frac{f}{f-d _0}$ and real image has M negative

$-3= \frac{f}{f-20}$

$-3f+60=f$

$f=15 cm$

$M= \frac{15}{15-30}$

$M= -1$

A convex lens is given, for which the minimum distance between an object and its rel image is $40cm$. An object is placed at a distance of $15cm$ from this lens. The liner magnification of adjustment 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. $\dfrac{5}{3}$

  2. $-2$

  3. $2$

  4. $\dfrac{1}{2}$

Show answer
Correct Option: A
Explanation:

Given object distance $u=15$ cm

Distance between object and real image produced $=40 $cm
Thus image distance $v=40-15=25$ cm
Also we know linear magnification,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$ 

An object is placed in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm, the position and nature of the image formed 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. $+30 cm, virtual \ and \ erect$

  2. $+30 cm, real \ and \ inverted$

  3. $-30 cm, virtual \ and \ erect$

  4. $-30 cm, real \ and \ inverted$

Show answer
Correct Option: D

An object of length $6\ cm$ is placed on the principle axis of a concave mirror of focal length $f$ at a distance of $4\ f$. The length of the image 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. $2\ cm$

  2. $12\ cm$

  3. $4\ cm$

  4. $1.2\ cm$

Show answer
Correct Option: A
Explanation:

Given that,

The object distance $u=-6\,cm$

Now, magnification is

  $ m=\dfrac{I}{O} $

 $ m=\dfrac{f}{f-u} $

 $ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $

 $ I=-2\,cm $

Hence, the length of image is -$2\ cm$

An astronomical telescope has focal lengths $100$ & $10$cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope 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. -15

  2. -14

  3. -17

  4. -19

Show answer
Correct Option: B
Explanation:

Given focal length of eye piece${f} _{e}=10cm\$

focal length of objective${f} _{o}=100 cm\$
Also we know least distance $D=25 cm\$ 
Magnifying power $M=\dfrac{-{f} _{0}}{{f} _{e}}(1+\dfrac{{f} _{e}}{D})\$
$M=-\dfrac{100}{10}(1+\dfrac{10}{25})\$
$M=-14$

In the displacement method, a convex lens is placed in between an object and a screen. If one of the magnification is $3$ and the displacement of the lens between the two positions is $24$cm, then the focal length of the lens 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. $10$ cm

  2. $9$ cm

  3. $6$ cm

  4. $16/3$ cm

Show answer
Correct Option: B
Explanation:

Given magnification $M=\dfrac{v}{u}=3$


Thus $v=3u$, where v and u are the image and object distance respectively.

Also Distance between lenses$=v-u=24$
Thus $u=12 cm$, than $v=36 cm$

From lens formula we have,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{f}=\dfrac{1}{36}+\dfrac{1}{12}$

$\dfrac{1}{f}=\dfrac{4}{36}$

$f=\dfrac{36}{4}$

$f=9 cm$

A concave mirror of focal length $20\ cm$ produces an image twice the height of the object. If the image is real, then the distance of the object from the mirror 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. $20\ cm$

  2. $60\ cm$

  3. $10\ cm$

  4. $30\ cm$

Show answer
Correct Option: D

In a concave mirror an object is placed at a distance x from the focus, and the image is formed at a distance y from the focus. The focal length of the mirror 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. $xy$

  2. $\sqrt{xy} $

  3. $\dfrac{x+y}{2} $

  4. $\sqrt{\dfrac{x}{y} }$

Show answer
Correct Option: B

Sun subtends an angle of $0.5^{o}$ at the pole of a concave mirror of radius of curvature 15 m. The diameter of the image of the sun formed by the mirror 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. $8.55 cm$

  2. $7.55 cm$

  3. $6.55 cm$

  4. $5.55 cm$

Show answer
Correct Option: A

A light ray travelling parallel to the principle axis of a concave mirror strikes the minor at angle of incidence $\theta$. If the radius of curvature of the mirror is $R$, then after reflection, the ray meets the principle axis at distance $d$ from the centre of curvature, then $d$ 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. $\dfrac {R}{2}$

  2. $R\left(1-\dfrac {1}{2\cos \theta}\right)$

  3. $\dfrac {R}{2\cos \theta}$

  4. $\dfrac {R}{2}(1+\cos \theta)$

Show answer
Correct Option: A

The focal length of a concave mirror is f and the distance from the object to the principal focus is p. The ratio of the size of the real image to the size of the object 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. $-\displaystyle \frac{f}{p}$

  2. $\displaystyle \left(\frac{f}{p}\right)^2$

  3. $\displaystyle \left(\frac{f}{p}\right)^{\frac{1}{2}}$

  4. $-\displaystyle \frac{p}{f}$

  5. $-fp$

Show answer
Correct Option: A
Explanation:

Distance of object is $u= -(f+p)$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ gives:
$\dfrac{1}{v}-\dfrac{1}{f+p}= -\dfrac{1}{f}$
or, $\dfrac{1}{v}= -\dfrac{1}{f}+\dfrac{1}{f+p}$
or, $\dfrac{1}{v}= -\dfrac{p}{(f+p)\times f}$
or, $v= -\dfrac{(f+p)\times f}{p}$      (-ve sign indicates image is real) 
   Magnification $=-\dfrac{v}{u}$ 
           $=-\dfrac{(f+p)\times f}{p\times (f+p)}$   
           $=-\dfrac{f}{p}$  (-ve sign indicates inverted)
    So, ratio of size of image to that of object is: $-\dfrac{f}{p}$

A concave spherical mirror has a focal length of 12 cm. if an object is placed 6 cm in front of it, the position of the magic 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. 4 cm behind the mirror

  2. 4 cm in front of the mirror

  3. 12 cm behind the mirror

  4. 12 cm in front of the mirror

Show answer
Correct Option: A

A glass hemisphere of radius R and of material having refractive index 1.5 is silvered on its flat face as shown in figure . a small object of height h is located at distance 2R from the surface of hemisphere as shown in the figure. the final image will form

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

  1. At a distance of R from silvered surface, on the right side

  2. on the object itself

  3. at hemisphere surface

  4. refractive index

Show answer
Correct Option: B

A small piece of wire bent into L shape such that the upright and horizontal portions are of equal length. It is placed with the horizontal portion along the axis of concave mirror of radius of curvature 20 cm. If the bend is 40 cm from the pole of the mirror, then the ratio of the length of the images of the upright and horizontal portions of the wire 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. 1 : 9

  2. 1 : 3

  3. 3 : 1

  4. 2 :1

Show answer
Correct Option: C
Explanation:
$u= -40$ ; $f= -10$

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

$\dfrac{1}{v}-\dfrac{1}{40}=\dfrac{-1}{10}$

$v=\dfrac{-40}{3}$

magnification=$\dfrac{-v}{u}=\dfrac{-1}{3}$

lateral magnification is $\dfrac{f^{2}}{(u-f)^{2}}$

                                 =  $\dfrac{100}{(40-10)^{2}}$

                                 =  $\dfrac{1}{9}$

ratio of up-right portion to lateral portion is $\dfrac{\dfrac{1}{3}}{\dfrac{1}{9}}=3:1$, hence option $C$ is correct 

A small piece of wire bent into an L shape, with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is 10 cm. If the bend is 20 cm from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire 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. 1:2

  2. 3:1

  3. 1:3

  4. 2:1

Show answer
Correct Option: B
Explanation:
Given, $u= -20$ ; $f= -5$

From mirror formula, $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{v}-\dfrac{1}{20}=\dfrac{-1}{5}$

$v=\dfrac{-20}{3}$

magnification=$\dfrac{-v}{u}=\dfrac{-1}{3}$

lateral magnification is $\dfrac{f^{2}}{(u-f)^{2}}$

                                 =  $\dfrac{25}{(20-5)^{2}}$

                                 =  $\dfrac{1}{9}$
Ratio of up-right portion to lateral portion is $\dfrac{\dfrac{1}{3}}{\dfrac{1}{9}}=3:1$

option $B$ is correct 

An object is kept at 15 cm from a convex mirror of focal length 25 cm. What is the magnification?

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

  1. 4/9

  2. 5/8

  3. 9/4

  4. 8/5

Show answer
Correct Option: B
Explanation:

Magnification for a mirror, $m = \dfrac{f}{f-u}$

As per sign convention: $u = -15\ cm$, $f = 25\ cm$
So, $m=\dfrac{25}{25+15}=\dfrac{5}{8}$

The image of an object placed on the principal axis of a concave mirror of focal length 12 cm is formed at a point which is 10 cm more distance from the mirror than the object. The magnification of the 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. 8/3

  2. 2.5

  3. 2

  4. 1.5

Show answer
Correct Option: D
Explanation:
Let the object distance be $u$ then image distance is $u+10$
$u= -u$ ; $v= -(u+10)$ ; $f= -12$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{-1}{u+10}+\dfrac{-1}{u}=\dfrac{-1}{12}$
$\dfrac{2u+10}{u(u+10)}=\dfrac{1}{12}$
$u=20$cm
$v=-30$cm
Magnification is $-\dfrac{v}{u}= -\dfrac{30}{20}= -1.5$

Mark the correct statement(s) w.r.t. a concave spherical 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. for real extended object, it can form a diminished virtual image

  2. for real extended object, it can form a magnified virtual image

  3. for virtual extended object, it can form a diminished real image

  4. for virtual extended object, it can form a magnified real image

Show answer
Correct Option: B,C
Explanation:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

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

magnification is $\dfrac{-v}{u}=\dfrac{f}{f-u}$

if $f>|u|$ (u<0) then a magnified image is formed which is virtual 

if $u>2f$ (u>0) then a diminished image is formed which is real 

option $B,C$ are correct

A beam of light converges towards a point O, behind a convex mirror of focal length 20 cm. Find the magnification and nature of the image when point O is 30 cm behind 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. 2 (virtual, inverted)

  2. 3 (real, inverted)

  3. 3, (virtual, enlarged)

  4. +1 (real, enlarged)

Show answer
Correct Option: A
Explanation:

$u=30$ ; $f=20$ 


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

$\dfrac{1}{v}+\dfrac{1}{30}=\dfrac{1}{20}$

$v= 60$

Image is virtual (v>0) 

Magnification is $-\dfrac{v}{u}= -\dfrac{60}{30}= -2$ (<0) hence it is inverted.

A beam of light converges towards a point O, behind a convex mirror of focal length 20 cm. Find the magnification and nature of the image when point O is 10 cm behind 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. $2$ (Virtual, Inverted)

  2. $3$ (Real, Inverted)

  3. $5$ (Real, Erect)

  4. $2$ (Virtual, Erect)

Show answer
Correct Option: D
Explanation:
$u=10$ ; $f=20$ 
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}+\dfrac{1}{10}=\dfrac{1}{20}$
$v= -20$
Image is real  , Magnification is $-\dfrac{v}{u}= -\dfrac{-20}{10}=2$ ( > 0) Hence, it is erect. 

A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing

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

  1. a concave mirror of suitable focal length

  2. a convex mirror of suitable focal length

  3. a convex lens of focal length less than 0.25 m

  4. a concave lens of suitable focal length

Show answer
Correct Option: C
Explanation:

Image can be formed on the screed if it is real. Real image of reduced size can be formed can be formed by a concave mirror or a convex lens.


The object is beyond $2f$. 

So let $u=2f+x$

And using lens formula we have

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

or

$\dfrac{1}{2f+x}+\dfrac{1}{v}=\dfrac{1}{f}$

or

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{2f+x}$

Solving we get

$v=\dfrac{f(2f+x)}{f+x}$

We have $u+v=1$

or

$2f+x+\dfrac{f(2f+x)}{f+x}=1$

or

$\dfrac{(2f+x)^2}{f+x}<1$

$(2f+x)^2<(f+x)$

This is valid only when $f<0.25m$

An object is placed at a distance $2 f$ from the pole of a convex mirror of focal length $f$. The linear magnification 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. $\displaystyle \frac {1}{3}$

  2. $\displaystyle \frac {2}{3}$

  3. $\displaystyle \frac {3}{4}$

  4. 1

Show answer
Correct Option: A
Explanation:

$\displaystyle \frac {1}{V} - \frac {1}{2f} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac {3}{2f}  \Rightarrow v = \frac{2}{3}f$
$\therefore m = \displaystyle \frac {u}{v} = \frac{2}{3} \frac{f}{2f} = \frac {1}{3}$

The linear magnification for a 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. True

  2. False

Show answer
Correct Option: B
Explanation:

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$.........(1)


multiplyng by u in eq.(1)

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

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

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

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

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

hence, $m=\dfrac{f}{u-f}$

If linear magnification for a spherical mirror is $\dfrac{3}{2}$, then we may write: (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. $f=\dfrac{u}{2}$

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

  3. $f=\dfrac{3u}{5}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Mirror equation is: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Multiplying both sides by $u$, we get:
$\dfrac{u}{f}=\dfrac{u}{v}+1$
Or  $\dfrac{u}{v}=\dfrac{u}{f}-1=\dfrac{u-f}{f}$
Or  $\dfrac{v}{u}=\dfrac{f}{u-f}$
Now, magnification, $m=\dfrac{v}{u}=\dfrac{3}{2}$
$\therefore$ $\dfrac{f}{u-f}=\dfrac{3}{2}$
Solving the above equation we get  $\dfrac{5}{2}f=\dfrac{3}{2}u$
 $\implies f=\dfrac{3u}{5}$

Magnification produced by a convex 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. equal to 1

  2. less than 1

  3. more than 1

  4. zero

Show answer
Correct Option: B
Explanation:

A convex mirror always creates a virtual image which is diminished. So, magnification produced by convex mirror is always less than one.

If magnification is positive, the nature of the 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. real and inverted

  2. virtual and erect

  3. real

  4. none of these

Show answer
Correct Option: B
Explanation:

$\text{Magnification}=\dfrac{\text{Image  size}}{\text{Object  size}}$

According to new Cartesian sign convention, size of height of real and inverted image is considered negative and that of virtual and erect image is considered positive.
The heigh of the object, being erect is considered positive always.
So, for positive magnification, the ratio, mentioned above, will be positive. This implies that the image height will also be positive.
The image will be virtual and erect.

A concave mirror is made from a hollow sphere of radius of curvature 30 cm.  If an object of height 2 cm is placed at 10 cm from the pole of the mirror, determine the size of the image :

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. 6 cm

  3. 12 cm

  4. 24 cm

Show answer
Correct Option: B
Explanation:

Given: $u=-10cm$


$f=-15cm$

$h _{o}=2cm$

To find, $h _{o}$

From mirror formula,

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

$\dfrac{1}{v}-\dfrac{1}{10}=-\dfrac{1}{15}$

$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{15}=\dfrac{5}{150}$

$v=30cm$

$magnification=\dfrac{h _{i}}{h _{o}}=\dfrac{-v}{u}$

$\dfrac{h _{i}}{2}=\dfrac{-30}{-10}$

$h _{i}=6cm$

An object is placed on the principal axis of a concave mirror at a distance of 60 cm.  If the focal length of the concave mirror is 40 cm then determine the magnification of the obtained image.

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

  1. 4

  2. -2

  3. -4

  4. +2

Show answer
Correct Option: B
Explanation:

Given: $u=-60 cm$     $f=-40 cm$
To find : $m$
Solution: From mirror formula
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}-\dfrac{1}{60}=\dfrac{-1}{40}$
$\dfrac{1}{v}=\dfrac{1}{60}-\dfrac{1}{40}$
$v=-120 cm$
Hence magnification is given by
$m=\dfrac{-v}{u}=-\dfrac{(-120)}{(-60)}$
$m=-2$

Calculate the magnification of an object if it is kept at a distance of $3 cm$ from a concave mirror of focal length $4 cm$:

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

  1. $3$

  2. $6$

  3. $9$

  4. $4$

Show answer
Correct Option: D
Explanation:

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

$\dfrac{1}{-3}+\dfrac{1}{v}=\dfrac{1}{-4}$
$\dfrac{1}{v}=\dfrac{1}{12}$
v= 12 cm
Magnification
$=\dfrac{12}{3}=4$

A concave mirror produces $2$ times magnified real image of an object placed at $5 cm$ in front of it. Where is image located?

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

  1. $10 cm$

  2. $-5 cm$

  3. $-10 cm$

  4. $2.5 cm$

Show answer
Correct Option: C
Explanation:

Magnification, $m = \dfrac {-v}{u}$
Given, magnification, $m = -2$

$-2 = \dfrac {-v}{-5 cm}$
Therefore, object distance, $v = -2\times 5 cm = -10 cm$

Magnification produced by plane mirror is $+1$. It means:

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 formed by plane mirror is greater than size of the object

  2. Image formed by plane mirror is erect and of same size as the object

  3. Image formed by plane mirror is smaller than size of object and is virtual

  4. None

Show answer
Correct Option: B
Explanation:

Magnification $= +1$ signifies that the image formed in a plane mirror is of same size as the object. Positive sign in the value of magnification signifies that image formed by a plane mirror is erect.

The radius of curvature of concave mirror is 24 cm and the image is magnified by 1.5 times. The object distance 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. 20 cm

  2. 8 cm

  3. 16 cm

  4. 24 cm

Show answer
Correct Option: A
Explanation:

Given that, $\displaystyle R=-24$cm
$\displaystyle f=-12cm$ and $\displaystyle m=1.5$
By the lens formula,
$\displaystyle \frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } $
$\displaystyle \frac { 1 }{ 1.5u } +\frac { 1 }{ u } =-\frac { 1 }{ 12 } $
$\displaystyle \frac { 2.5 }{ 1.5u } =-\frac { 1 }{ 12 } $
or, $\displaystyle u=-20cm$

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

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. $-ve$

Show answer
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).

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