Tag: derivation of formula for curved mirrors

Questions Related to derivation of formula for curved mirrors

The relation between $u, v$ ( u is the object distance and v is the image distance )  and f for mirror is given by:

  1. $\displaystyle f=\frac{uv}{u-v}$

  2. $\displaystyle f=\frac{2u\times v}{u+v}$

  3. $\displaystyle f=\frac{u\times v}{u+v}$

  4. none of these


Correct Option: C
Explanation:

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

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

A point source of light is kept in front of a convex mirror of radius of curvature $40 cm$. The image is formed at $10 cm$ behind the mirror. Calculate the object distance

  1. 30

  2. 20

  3. 50

  4. 40


Correct Option: B
Explanation:

Given: For a convex mirror, $R = -40 cm$.
$v = -10 cm$ (image is virtual).
From mirror formula, we have

$\displaystyle \frac {2}{R}=\frac {1}{u}+\frac {1}{v}$

$\displaystyle \frac {1}{u}=\frac {2}{R}-\frac {1}
{v}=\frac {2v-R}{vR}$

$\displaystyle u=\frac {vR}{2v-R}=\frac {(-10cm) \times (-40cm)}{2 \times (-10cm) - (-40cm)}$

$=+20cm$
Thus, object is placed $20 cm$ in front of the mirror.

Mirror formula is valid for:

  1. Convex mirror

  2. Concave mirror

  3. Both A and B

  4. For lenses and mirrors


Correct Option: C
Explanation:
Mirror formula is:
$\dfrac {1}{v} + \dfrac {1}{u} = \dfrac {1}{f}$
That is; $\dfrac {1}{\text {Object Distance}} + \dfrac {1}{\text {Image Distance}} = \dfrac {1}{\text {Focal length of the mirror}}$
This relationship is applicable for both concave and convex mirrors.

Mirror formula can also be written as:

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

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

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

  4. $\dfrac {1}{2v} + \dfrac {4}{u} = \dfrac {3}{R}$


Correct Option: B
Explanation:

Focal length, $f = \dfrac {R}{2}$, where $R$ = radius of curvature

So, $\dfrac {1}{v} + \dfrac {1}{u} = \dfrac {1}{f}$ can be written as
$\Rightarrow \dfrac {1}{v} + \dfrac {1}{u} = \dfrac {2}{R}$

Mark the incorrect statement regarding mirror formula.

  1. Values of known and unknown parameters can be used with their proper signs

  2. Sign of unknown parameter comes of its own after calculation

  3. Mirror formula is applicable for both concave and convex mirrors

  4. All


Correct Option: A
Explanation:

The values of known parameter should be used with their proper sign convention. No sign should be attached to the unknown parameter during calculation. Its sign will come of its own after calculation.

The relation among u, v and f for a mirror is

  1. $f = uv/(u+v)$

  2. $v = fu/(u+f)$

  3. $u = fv/(f+v)$

  4. All of these


Correct Option: A
Explanation:

According to mirror equation.

$\cfrac{1}{f} = \cfrac{1}{u}+ \cfrac{1}{v}$
$\cfrac{1}{f} = \cfrac{u+v}{uv}$
$f = \cfrac{uv}{u+v}$

Similarly,
$v = \cfrac{fu}{u-f}$
$u = \cfrac{fv}{v-f}$

The relation among $u,v$ and $f$ for a mirror is:

  1. $f=uv(u+v)$

  2. $v=fu(u+f)$

  3. $u=fv(f+v)$

  4. None of these


Correct Option: D
Explanation:

from mirror formula

              $\dfrac { 1 }{ v } +\dfrac { 1 }{ u } =\dfrac { 1 }{ f } $
              $\Rightarrow \quad \boxed { f=\dfrac { uv }{ u+v }  } $
and
          $1+\dfrac { v }{ u } =\dfrac { v }{ f } \Rightarrow \boxed { u=\dfrac { fv }{ f-v }  } $

A object moving at a speed of $5$ m/s towards a concave mirror of focal length $f=1$m is at a distance of $9$m. The average speed of the image is?

  1. $\dfrac{1}{5}$ m/s

  2. $\dfrac{1}{10}$ m/s

  3. $\dfrac{4}{5}$ m/s

  4. $\dfrac{2}{5}$ m/s


Correct Option: C

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

  1. 1 m

  2. 0.5 m

  3. 0.24 m

  4. 2 m


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?

  1. +0.23

  2. -0.23

  3. +0.45

  4. -0.45


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 
$=-\dfrac{v}{u}=-\dfrac{1.15}{-5}=0.23$