Tag: derivation of formula for curved mirrors

Questions Related to derivation of formula for curved mirrors

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)

  1. True

  2. False


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) 
  1. $f=\dfrac{u}{2}$

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

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

  4. None of these


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:

  1. equal to 1

  2. less than 1

  3. more than 1

  4. zero


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:

  1. real and inverted

  2. virtual and erect

  3. real

  4. none of these


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 :

  1. 3 cm

  2. 6 cm

  3. 12 cm

  4. 24 cm


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.

  1. 4

  2. -2

  3. -4

  4. +2


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 $4 cm$:

  1. $3$

  2. $6$

  3. $9$

  4. $4$


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?

  1. $10 cm$

  2. $-5 cm$

  3. $-10 cm$

  4. $2.5 cm$


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:

  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


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

  1. 20 cm

  2. 8 cm

  3. 16 cm

  4. 24 cm


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, $\displaystyle u=-20cm$