Tag: image formation by plane mirror

Questions Related to image formation by plane mirror

A flat mirror revolves at a constant angular velocity making $n=0.4$ revolutions per second. With what velocity (in $ms^{-1}$ ) will a light spot move along a spherical screen with a radius of $15$ metres, if the mirror is at the centre of curvature of the screen?

physics light : reflection and refraction from plane surface introduction to light and mirror magic with mirrors image formation by plane mirror principle of reversibility of path of light refractive index

  1. $37.7$

  2. $60.3$

  3. $68.7$

  4. $75.4$

  5. $90.4$

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Correct Option: D
Explanation:
$\because$ Angular velocity of mirror = $0.4rev/s$
$\Rightarrow 0.4\times 2\pi=0.8\pi\,rad/s$
$\because$ Angular velocity of reflected ray
$\Rightarrow 2\times 0.8\pi=1.6\pi\,rad/s$
Hence, velocity of light spot over the screen
$v=rw=15\times 1.6\pi=75.4m/s$

Light incident on a rotating mirror M is returned to a fixed mirror N placed 22.5 km away from M. The fixed mirror reflects it back to M (along the same path) which in turn reflects the light again along a direction that makes an angle of $\displaystyle { 27 }^{ o }$ with the incident direction. The speed of rotation of the mirror is: 

physics light : reflection and refraction from plane surface introduction to light and mirror magic with mirrors image formation by plane mirror principle of reversibility of path of light refractive index

  1. 250 revolutions $\displaystyle { s }^{ -1 }$

  2. 500 revolutions $\displaystyle { s }^{ -1 }$

  3. 1000 revolutions $\displaystyle { s }^{ -1 }$

  4. 125 revolutions $\displaystyle { s }^{ -1 }$

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Correct Option: A
Explanation:
Suppose the number of revolutions are n. The angle between 2 positions of the rotating mirror  = $\dfrac { 1 }{ 2 } \times 27\quad degrees$.

Since the angle of rotation of mirror is half the angle through which the reflected ray rotates.

The time taken by the mirror in rotating through an angle $\theta $ is given by $t=\dfrac { \theta  }{ 2\pi n } =\dfrac { 13.5\quad degrees }{ 2\times 180\times n } sec.$ -- Eqn 1

This is also the time taken by the light to travel from original point to the fixed mirror and back, thus

$t=\dfrac { 2d }{ c } =\dfrac { 2\times 22500 }{ 3\times { 10 }^{ 8 } } [d=22.5\quad km=22500m\quad and\quad c=3\times { 10 }^{ 8 }m/s]$ -- Eqn 2

From eqns 1 and 2,

$\dfrac { 13.5 }{ 2\times 180\times n } =\dfrac { 2\times 22500 }{ 3\times { 10 }^{ 8 } } $

or n = $\dfrac { 13.5\times 3\times { 10 }^{ 8 } }{ 2\times 180\times 2\times 22500 } =\dfrac { 40.5\times { 10 }^{ 8 } }{ 16200000 } =\dfrac { 40500\times { 10 }^{ 5 } }{ 162\times { 10 }^{ 5 } } =250\quad revolutions/s$.

Hence, the number of revolutions are 250 revolutions/s.

The reflective surface is given by y $=$ 2 sinx. The reflective surface is facing positive x-axis. What is the least values of co ordinate of the point where a ray parallel to positive x axis becomes parallel to positive y axis after reflection 2. 

physics light : reflection and refraction from plane surface introduction to light and mirror magic with mirrors image formation by plane mirror principle of reversibility of path of light refractive index

  1. $\left ( \dfrac{\pi }{3},\sqrt{3} \right )$

  2. $\left ( \dfrac{\pi }{2},\sqrt{2} \right )$

  3. $\left ( \dfrac{\pi }{3},\sqrt{2} \right )$

  4. $\left ( \dfrac{\pi }{4},\sqrt{2} \right )$

Show answer
Correct Option: A
Explanation:

$m (L _1) = 2 cos x _o$


$m(N) = \dfrac {+1}{2 cos x _o}$

$m(N) = \dfrac {+1}{2 cos x _o} = 1$

$cos x _o = \dfrac {1}{2}$

      $x _o = \dfrac {\pi}{3}$

        $y = 2sin (\dfrac {\pi}{3}) = \sqrt {3}$

The angle between the incident and reflected rays is $90^o$. If the plane mirror is rotated by $10^o$ about O in the anti-clockwise direction in the plane perpendicular to the mirror, then the angle between the incident and reflected rays will be _______$^0$.

physics light : reflection and refraction from plane surface introduction to light and mirror magic with mirrors image formation by plane mirror principle of reversibility of path of light refractive index

  1. 70

  2. 100

  3. 90

  4. 110

  5. 80

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Correct Option: A,D
Explanation:

If the plane mirror is rotated by 10 degrees about O in the anti-clockwise direction then the angle between the incident and reflected rays will be reduced by 20 degrees. So the angle between incident and reflected ray will be 70 degrees.
If the plane mirror is rotated by 10 degrees in clockwise direction, then the angle between the incident and reflected rays increases by 20 degrees and becomes 110 degrees.

When a plane mirror is rotated through an angle $\theta$, the reflected ray rotates through an angle $2\theta$. Then the size of the image 

physics light : reflection and refraction from plane surface introduction to light and mirror magic with mirrors image formation by plane mirror principle of reversibility of path of light refractive index

  1. is halved

  2. is doubled

  3. remains unchanged

  4. is quadrupled

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Correct Option: C
Explanation:

When a plane mirror is rotated through an angle $\theta$ , then the reflected ray rotates through an angle 2$\theta$; but the size of the image remain the same.

A ray of light making an angle $10^o$ with the horizontal is incident on a plane mirror making angle $\theta$ with the horizontal. What should be the value of $\theta$ so that the reflected ray goes vertically upward?

physics light : reflection and refraction from plane surface introduction to light and mirror magic with mirrors image formation by plane mirror principle of reversibility of path of light refractive index

  1. $30^o$

  2. $40^o$

  3. $50^o$

  4. $60^o$

Show answer
Correct Option: C
Explanation:
The situation is as shown in figure:
The reflected ray makes an angle, $100-\theta$ with the normal. This is also equal to $\theta$.
$\therefore 100-\theta=\theta$
$\Rightarrow \theta=50^o$

If refraction index of glass with respect to air is $ _{a}{u} _{g} = \dfrac{3}{2}$, the refraction index of air with respect to glass will be $ _{g}{u} _{a} =$

physics light : reflection and refraction from plane surface principle of reversibility of path of light image formation by plane mirror refractive index

  1. ${3}/{2}$

  2. ${2}/{3}$

  3. ${1}/{3}$

  4. ${1}/{2}$

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Correct Option: B
Explanation:
refractive index of glass w.r.t air =${ _{ a }{ \mu  } _{ g }= }\dfrac { 3 }{ 2 } $
then,
refractive index of air w.r.t glass  =$ _{ g }{ \mu  } _{ a }=\dfrac { 1 }{ _{ a }{ \mu  } _{ g } } $

$ _{ g }{ \mu  } _{ a }=\dfrac { 1 }{ _{ a }{ \mu  } _{ g } } =\dfrac { 2 }{ 3 } $
Option B is correct.

The refractive index of water with respect to air is $ _{a}{u} _{w}$ and of glass with respect to air is $ _{a}{u} _{g}$. Express the refractive index of glass with respect to water

physics light : reflection and refraction from plane surface principle of reversibility of path of light image formation by plane mirror refractive index

  1. $\dfrac{ _{a}{u} _{g}}{ _{a}{u} _{w}}$

  2. $\dfrac{ _{g}{u} _{a}}{ _{a}{u} _{w}}$

  3. $\dfrac{ _{a}{u} _{w}}{ _{a}{u} _{g}}$

  4. $\dfrac{ _{a}{u} _{a}}{ _{g}{u} _{g}}$

Show answer
Correct Option: A
Explanation:

Refractive index of water w.r.t air
$ _{a}\mu _{w}=\dfrac{\mu _{w}}{\mu _{a}}$
Refractive index of glass w.r.t air
$ _{a}\mu _{g}=\dfrac{\mu _{g}}{\mu _{a}}$
Hence refractive index of glass w.r.t water
$ _{w}\mu _{g}=\dfrac{\mu _{g}}{\mu _{w}}   =\dfrac{ _{a}\mu _{g}}{ _{a}\mu _{w}}$

At ray of light is incident in medium 1 at an angle of $37^{o}$ and gets refracted in medium 2 at an angle of $53^{o}$. What will be angle of refraction if light is incident in medium 2 at an angle of $53^{o}$.

physics light : reflection and refraction from plane surface principle of reversibility of path of light image formation by plane mirror refractive index

  1. $37^{o}$

  2. $53^{o}$

  3. $36^{o}$

  4. $45^{o}$

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Correct Option: A
Explanation:

Principle of reversibility states that when final path of a ray of light after any number of reflections and refractions is reversed, the ray retraces its entire path or in simple words, light follows exactly the same path if its path of travel is reversed.

Thus angle of refraction of light in medium $1$ will be $37^o$.

A real object is at a distance of 1 m from its virtial image formed by a diverging lens . Determine the focal length of the lens if magnification is  02.6 :- 

physics light : reflection and refraction from plane surface principle of reversibility of path of light image formation by plane mirror refractive index

  1. -3.25m

  2. -205m

  3. -3.75 m

  4. 1.25 m

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Correct Option: B