Tag: magic with mirrors

Questions Related to magic with mirrors

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 $\displaystyle { s }^{ -1 }$

Show answer
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 $\theta$.
$\therefore 100-\theta=\theta$
$\Rightarrow \theta=50^o$

Virtual image is always erect

evs magic with mirrors image of an extended object formed by a plane mirror reflection at plane surface introduction to light and mirror

  1. True

  2. False

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Correct Option: A
Explanation:

As we can observe from all the virtual images formed by plane mirror or curved mirror,virtual images are always erect.

Formation of erect image is a characteristic of which of the following mirrors ?

evs magic with mirrors image of an extended object formed by a plane mirror reflection at plane surface introduction to light and mirror

  1. Plane Mirror

  2. Concave mirror

  3. Convex mirror

  4. All the above

Show answer
Correct Option: D
Explanation:
$ \bf{Convex\ mirrors} $ always produce a virtual, erect and diminished image and the decrease in size of image depends on the position of the object in front of the mirror.

$ \bf{Plane\ Mirrors} $ always produce a virtual and erect image having the same size as that of the object irrespective of the position of object.

$ \bf{Concave\ mirrors} $ produce real and virtual, erect and inverted, diminished, same-sized and magnified image depending upon the position of the object on the principal axis. The concave mirror forms a virtual and erect image only when the object is placed between the Focus and the pole of the mirror.

Thus, an erect image can be formed by all the three mirrors.

Hence, the correct answer is OPTION D.

Choose the correct option regarding the image formed by a plane mirror

evs magic with mirrors image of an extended object formed by a plane mirror reflection at plane surface introduction to light and mirror

  1. It is real

  2. It is erect

  3. It is inverted

  4. None of the above

Show answer
Correct Option: B
Explanation:

A $ \bf{plane\ mirror} $ always forms an image which is $ \bf{virtual} $ in nature (meaning that light rays meet behind the mirror or that light rays do not actually come from the image), $ \bf{erect} $ and of the $ \bf{same\ shape\ and\ size} $ as that of the object, it is reflecting.

Hence, the correct answer is OPTION B. 

Which of the following is a characteristic of the image formed by a plane mirror ?

evs magic with mirrors image of an extended object formed by a plane mirror reflection at plane surface introduction to light and mirror

  1. The image is twice in size as the object.

  2. The image is nearer from the mirror as the object is in front of it.

  3. The image is real

  4. The image is erect

Show answer
Correct Option: D
Explanation:

A $ \bf{plane\ mirror} $ always forms an image which is $ \bf{virtual} $ in nature (meaning that light rays meet behind the mirror or that light rays do not actually come from the image), $ \bf{erect} $ and of the $ \bf{same\ shape\ and\ size} $ as that of the object, it is reflecting. The image in  plane mirror is formed as far behind the mirror as the object in front of it. 

Hence, the correct answer is OPTION D.