Tag: reflection of light in spherical mirrors

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Questions Related to reflection of light in spherical mirrors

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A dentist wants a small mirror that when placed $2$cm from a tooth, will produce $3\times$ upright image. What kind of mirror must be used and what must its focal length be?

  1. Concave mirror, $3.04$ cm
  2. Concave mirror, $1.5$ cm
  3. Convex mirror, $3.0$ cm
  4. Convex mirror, $1.5$cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$m=+\dfrac{v}{u}=\dfrac{+f}{u-f}$
$3=\dfrac{+f}{+2+f}\Rightarrow f=3cm$.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The focal length of a concave mirror is 50 cm where an object is to be placed so that its image is two times magnified, real and inverted :

  1. 75 cm

  2. 72 cm

  3. 63 cm

  4. 50 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

Focal length, $f=50\,cm$

Magnification, $m=2$

$ m=\dfrac{v}{u} = 2$

$ v=2u $

From mirror formula,

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

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

$ u=\dfrac{3f}{2}=\dfrac{3\times 50}{2}=75\ cm $

Object Is placed at $75\,cm$ from the mirror.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A convex mirror has a focal length $f$.A real object is placed at a distance $f$ in front of it from the pole, produces an image at:

  1. $\infty$
  2. $f$
  3. $\dfrac{f}{2}$
  4. $2f$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a convex mirror, f is positive. With object distance u = -f, the mirror formula 1/v + 1/u = 1/f becomes 1/v - 1/f = 1/f. Thus, 1/v = 2/f, so v = f/2.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The distance between an object its doubly magnified by a concave mirror of focal length $f$ is

  1. $3 f/2$
  2. $2 f/3$
  3. $3\ f$

  4. Depend on whether the image is real or virtual
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Focal length $=f$

We know that, the magnification is

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

Now, magnification is double and image is real

  $ -2=\dfrac{-v}{u} $

 $ v=2u $

Now, using formula of mirror

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

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

 $ \dfrac{1}{f}=\dfrac{3}{2u} $

 $ u=\dfrac{3f}{2} $

Hence, the distance is $\dfrac{3f}{2}$ 

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

An object $2.5\ cm$ high is placed at a distance of $10\ cm$ from a concave mirror of radius of curvature $30\ cm$. The size of the image is:

  1. $9.2\ cm$
  2. $10.5\ cm$
  3. $5.6\ cm$
  4. $7.5\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,

Distance $u=-10\,cm$

Height $h=2.5\,cm$

Radius of curvature $R=30\,cm$

We know that,

  $ -f=\dfrac{R}{2} $

 $ -f=\dfrac{30}{2} $

 $ f=-15\,cm $

Now, using mirror’s formula

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

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

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

 $ v=30\,cm $

Now, the magnification is

  $ m=\dfrac{-v}{u} $

 $ m=\dfrac{30}{10} $

 $ m=3 $

We know that,

  $ m=\dfrac{h'}{h} $

 $ 3=\dfrac{h'}{2.5} $

 $ h'=7.5\,cm $

Hence, the size of the image is $7.5\ cm$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The focal length $f$ of a mirror is given by $\cfrac{1}{f}=\cfrac{1}{u}+\cfrac{1}{v}$, where $u$ and $v$ represent object and image distances, respectively

  1. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}$
  2. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{v}+\cfrac{\Delta v}{u}$
  3. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}-\cfrac{\Delta (u+v)}{u+v}$
  4. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}+\cfrac{\Delta U}{u+v}+\cfrac{\Delta v}{u+v}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The focal length $f$ of a mirror is given by
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$. . . . . . . .(1)
where $v=$ image distance
$u=$ object distance
from equation (1),
$f=\dfrac{uv}{u+v}=uv(u+v)^{-1}$
Taking log both sides we get
$logf=logu+log v+log(u+v)^{-1}$
$log f=logu+logv-log(u+v)$
Differentiating with respect to each variable
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta (u+v)}{u+v}$
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta u}{u+v}+\dfrac{\Delta v}{u+v}$
The correct option is D.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

When a ray of light parallel to the principle axis is incident on a concave mirror$,$ the reflected ray

  1. Passes through C

  2. Passes through F

  3. Passes midway between P and F

  4. retraces its path
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By definition, rays incident parallel to the principal axis of a concave mirror reflect through the principal focus (F).

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A mirror faces the negative x-axis. (Normal to its reflecting surface is$- \hat { i } )$  While a particle starts moving such that its image is formed in the mirror. At a certain instant the velocity of the particles is $3 \hat { i } + 4 \hat { j } + 5 \hat { k }$ and that of the mirror is $\hat { 1 } - \hat { j } + \hat { k }$ Choose the correct options.

  1. Magnitude of relative velocity of the image w.r.t mirrror is$\sqrt { 45 }$

  2. Magnitude of relative velocity of image w.r.t object is 4
  3. Magnitude of relative velocity of image w.r.t mirrror is $\sqrt { 45 }$
  4. Absolute velocity of the image w.r.t ground is$\sqrt { 42 }$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\begin{array}{l} In\, y-direction \ { V _{ o } }={ V _{ i } } \ \Rightarrow { V _{ i } }=4\widehat { j }  \ In\, z-direction \ { V _{ i } }={ V _{ o } } \ \Rightarrow { V _{ i } }=5\widehat { k }  \ In\, x-direction \ { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ I } }-{ V _{ M } }={ V _{ M } }-{ V _{ o } } \ \Rightarrow { V _{ I } }=2{ V _{ M } }-{ V _{ o } } \ =2\left[ { \widehat { i }  } \right] -3\left[ { \widehat { i }  } \right]  \ =-\widehat { i }  \ \therefore \overrightarrow { { V _{ I } } } =-\widehat { i } +4\widehat { j } +5\widehat { k }  \ \therefore \overrightarrow { { V _{ iM } } } =-2\widehat { i } +5\widehat { j } +4\widehat { k }  \ \therefore \left| { \overrightarrow { { V _{ iM } } }  } \right| =\sqrt { 45 }  \end{array}$

Hence,
option $(A)$ and $(C)$ are both correct answer.