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Light : reflection and refraction from plane surface - class-X

Description: light : reflection and refraction from plane surface
Number of Questions: 23
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Tags: physics refraction of light at plane surfaces light : reflection and refraction from plane surface
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A plane mirror is in a vertical plane and is rotating about a vertical axis at 100 mm horizontal beam of light is incident on the mirror. The reflected beam will rotate at

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. $100 rpm$

  2. $141 rpm$

  3. $0 rpm$

  4. $200 rpm$

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Correct Option: D

When a mirror is rotated through an angle, the reflected ray moves through double that angle, the instrument based on the above principle 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. Periscope

  2. Odometer

  3. Refractometer

  4. Sextant

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Correct Option: D
Explanation:

A periscope is based on the principle of reflection.

An Odometer is used to measure the distance travelled by a vehicle.
A Refractometer is based on the principle of refraction (Snell'slaw).
A Sextant is used to measure the angle between two visible objects. Baesd on the principle that when a mirror is rotated through an angle the reflected ray moves through double of that angle.

A source of light lies on the angle bisector of two plane mirrors inclined at an angle $\theta$. The values of $\theta$, so that the light reflected from one mirror does not reach the other mirror will be

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. $\theta \geq 120^0$

  2. $\theta \geq 90^0$

  3. $\theta \leq 120^0$

  4. $\theta < 30^0$

Show answer
Correct Option: A

In a Michelson experiment for measuring speed of light, the distance traveled by light between two reflections from the rotating mirror is 4.8 km. The rotating mirror has a shape of a regular octagon. At what minimum angular speed of the mirror (other than zero) the image is formed at the position where a non-rotating mirror forms it?

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. $780$ $rev/s$

  2. $7800$ $rev/s$

  3. $690$ $rev/s$

  4. $7500$ $rev/s$

Show answer
Correct Option: B
Explanation:

Minimum angular speed $ f=\dfrac { c }{ dN }$


$ =\dfrac { 3\times { 10 }^{ 8 } }{ 4.8\times { 10 }^{ 3 }\times 8 }$

$ =7812rev/s\cong 7800rev/s$

A plane mirror which rotates $10^4$ times per minute reflects light on to a stationary mirror $50$ m away. This mirror reflects the light normally so that it strikes the rotating mirror again. The image observed in the rotating mirror is shifted through $2.4$ minutes from the position it occupies. When the rotating mirror is stationary, what is the speed of light?

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. $3\times 10^8$ m/s

  2. $4\times 10^8$ m/s

  3. $5\times 10^8$ m/s

  4. $6\times 10^8$ m/s

Show answer
Correct Option: A

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. $90.4$

Show answer
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

Show answer
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

Show answer
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.

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. ${1}/{2}$

Show answer
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. $45^{o}$

Show answer
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

Show answer
Correct Option: B

When a ray of light enters a medium of refractive index $\mu $ from air. It is observed that the angle of refraction is half the angle of incidence. The angle of incidence is :

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

  1. $2 cos^{-1}(\frac{\mu }{2})$

  2. $ cos^{-1}(\frac{\mu }{2})$

  3. $2 cos^{-1}(\mu )$

  4. $2 sin^{-1}(\frac{\mu }{2})$

Show answer
Correct Option: C

Refractive index of diamond with respect to glass is $1.6$ and absolute refractive index of glass is $1.5$. Find out the absolute refractive index of diamond.

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

  1. $1.06$

  2. $0.93$

  3. $2.4$

  4. $0.75$

Show answer
Correct Option: C
Explanation:
Given,

Refractive index of diamond with respect to glass, ${ _{ g }{ n } _{ d } } = 1.6$,

Absolute refractive index of glass, $n _{g} = 1.5$

Refractive index of diamond with respect to glass $= \dfrac {\text {Absolute refractive index of diamond}}{\text {Absolute refractive index of glass}}$

Therefore, absolute refractive index of diamond $=$ Refractive index of diamond with respect to glass $\times$ Absolute refractive index of glass $= 1.6 \times 1.5 = 2.4$

The refractive index of water is $\dfrac{4}{3}$ and of glass is $\dfrac{3}{2}$. What will be 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. $9.1$

  2. $2.623$

  3. $1.125$

  4. $1$

Show answer
Correct Option: C
Explanation:
refractive index of water =${ _{ a }{ \mu  } _{ w }= }\dfrac { 4 }{ 2 } $
refractive index of glass = ${ _{ a }{ \mu  } _{ g }= }\dfrac { 3 }{ 2 } $

then refractive index of glass w.r.t. water is
$ _{ w }{ \mu  } _{ g }= _{ a }{ \mu  } _{ g }\times \dfrac { 1 }{ _{ a }{ \mu  } _{ w } } $
${ _{ w }{ \mu  } _{ g }= _{ a }{ \mu  } _{ g }\times \dfrac { 1 }{ _{ a }{ \mu  } _{ w } } =\dfrac { 3 }{ 2 } \times \dfrac { 3 }{ 4 } =\dfrac { 9 }{ 8 } =1.125 }$

Option C is correct.


Principle of reversibility holds true for

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

  1. Reflection

  2. Refraction

  3. Reflection and Refraction

  4. None of these

Show answer
Correct Option: C
Explanation:

Path of ray of light is reversible or light traces back it's path while traveling in reverse direction or a ray of light can pass in either direction along the same path.
So
Principle of reversibility holds true for both reflection and refraction.

Principle of reversibility can applied to 

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

  1. Lens system

  2. Mirror system

  3. Lens and mirror system

  4. None of these

Show answer
Correct Option: C
Explanation:

Principle of reversibility can applied to both lens and mirror system,
Because the principle is followed by both-Path of ray of light is reversible or light traces back it's path while traveling in reverse direction or a ray of light can pass in either direction along the same path

The refractive index of gias with respect to air is $3/2$ and that of water with respect to air is $4/3$. What is 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. $9 : 8$

  2. $1 : 2$

  3. $2 : 1$

  4. $8 : 9$

Show answer
Correct Option: A
Explanation:

Given, 


Refractive index of glass=$\dfrac{3}{2}$


Refractive index of water=$\dfrac{4}{3}$

Refractive Index of glass with respect to water= $\dfrac{3/2}{4/3}$
                         =$\dfrac{9}{8}$

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