Physics · Science General

Optics and Light Properties

2,019 Questions

Optics and light properties focus on the behavior of light, including reflection, refraction, dispersion, and polarization. This topic also covers the wave nature of light, illumination, and the functioning of optical instruments like microscopes. Questions on these concepts are common in general science sections of SSC, Railways, and State exams.

Reflection and mirrorsRefraction and mediumsLight wave theoriesPolarization and intensity

Optics and Light Properties Questions

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

Which of the following conditions are necessary for total internal reflection to take place at the boundary of two optical media ?
1. Light is passing from optically denser medium to optically rarer medium. 
2. Light is passing from optically rarer medium to optically denser medium. 
3. Angle of incidence is greater than the critical angle. 
4. Angle of incidence is less than the critical angle. 

  1. 1 and 3 only

  2. 2 and 4 only

  3. 3 and 4 only

  4. 1 and 4 only
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total internal reflection is a strange phenomenon that happens when a propagating wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface. If the refractive index is lower on the other side of the boundary and the incident angle is greater than the critical angle, the wave cannot pass through and is entirely reflected. The critical angle is the angle of incidence above which the total internal reflection occurs.
Hence, the statements present in 1 and 3 are correct.

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

A ray of light passing through an equilateral triangular prism gets deviated at least by $30^\circ$. Then, the refractive index of the material of the prism must be 

  1. $\leq \sqrt{2}$
  2. $\geq \sqrt{2}$
  3. $\leq \sqrt{3}$
  4. $\geq \sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Answer is B.

The refractive index of a prism is calculated from the formula, $\mu =\dfrac { sin\frac { A+D }{ 2 }  }{ sin\frac { A }{ 2 }  } \mu =\dfrac { sin\frac { A+D }{ 2 }  }{ sin\frac { A }{ 2 }  } $.
In this case, as it is an equilateral prism, the angle of prism is 60 degrees and the angle of minimum deviation is given as 30 degrees.
So, $\mu =\dfrac { sin\frac { 60+30 }{ 2 }  }{ sin\frac { 60 }{ 2 }  } =\dfrac { sin\quad 45 }{ sin\quad 30 } =\ge \sqrt { 2 } $.
Hence, the refractive index of the material of the prism must be $\ge \sqrt { 2 } $.

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

If the velocity of light in water is $2.25 \times {10}^{10}   cm$ per second and that is glass is $2 \times {10}^{10}   cm$ per second. A slab of this glass is immersed in water, what will be the critical angle of incidence of a ray of light tending to go from glass slab to water ?

  1. $\sin ^{ -1 }{ { 3 }/{ 5 } } $
  2. $\sin ^{ -1 }{ { 8 }/{ 9 } } $
  3. $\sin ^{ -1 }{ { 4 }/{ 5 } } $
  4. $\sin ^{ -1 }{ { 3 }/{ 4 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
given,
velocity of light in water  $ { v } _{ w }=2.25\times { 10 }^{ 10 }cm/s\\$ 
velocity  of light in glass  ${ v } _{g }=2\times { 10 }^{ 10 }cm/s$
refractive index of glass w.r.t water =${ _{ w }{ \mu  } _{ g } }=\dfrac { velocity\quad of\quad light\quad in\quad water }{ velocity\quad of\quad light\quad in\quad glass\quad  } =\dfrac { 2.25\times { 10 }^{ 10 } }{ 2\times { 10 }^{ 10 } } =\dfrac { 9 }{ 8 } $

refractive index of water w.r.t. glass =${ _{ g }{ \mu  } _{ w } }=\dfrac { 1 }{ { _{ w }{ \mu  } _{ g } } } =\dfrac { 8 }{ 9 } $
let the critical angle be $\angle { i } _{ c }$
then $sin{ i } _{ c }=\dfrac { 1 }{ _{ w }{ \mu  } _{ g } } = _{ g }{ \mu  } _{ w }$
$sin{ i } _{ c }=\dfrac { 8 }{ 9 } \\ \angle { i } _{ c }={ sin }^{ -1 }\left( \dfrac { 8 }{ 9 }  \right) $

Option B is correct.

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

In vacuum, to travel distance $d$, light takes time $t$ and in medium to travel $5d$, it takes time $T$. The critical angle of the medium is :

  1. $\sin ^{ -1 }{ \left( \dfrac { 5T }{ t } \right) } $
  2. $\sin ^{ -1 }{ \left( \dfrac { 5t }{ 3T } \right) } $
  3. $\sin ^{ -1 }{ \left( \dfrac { 5t }{ T } \right) } $
  4. $\sin ^{ -1 }{ \left( \dfrac { 3t }{ 5T } \right) } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In vacuum, $c = {d}/{t}$


In medium, $v = \dfrac{5d}{T}$

As refractive index, $\mu = \dfrac{c}{v} = \dfrac{{d}/{t}}{{5d}{T}} = \dfrac{T}{5t}$

Also,      $\sin{C} = \dfrac{1}{\mu}     \therefore  C = \sin ^{ -1 }{ \left[ \dfrac { 5t }{ T }  \right]  } $

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

The index of refraction for diamond is $2.42$. For a diamond in the air (index of refraction $=1.00$), what is the smallest angle that a light ray inside the diamond can make with a normal and completely reflect back inside the diamond (the critical angle)?

  1. $90^{\circ}$
  2. $45^{\circ}$
  3. $68^{\circ}$
  4. $66^{\circ}$
  5. $24^{\circ}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For a light ray incident on the air-diamond surface to completely reflect back of the smallest possible angle is 
$\mu sini _{min}=1$

$\implies i _{min}=sin^{-1}(\dfrac{1}{\mu})$
$=sin^{-1}(\dfrac{1}{2.42})=24^{\circ}$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence 45$^{\circ}$. The ray undergoes total internal reflection. The possible value of refractive index of the medium with respect to air is

  1. 1.245

  2. 1.324

  3. 1.414

  4. 1.524
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\mu _{medium} \times \sin 45^{\circ} =\mu _{air}$


$\dfrac{\mu _{medium}}{\mu _{air}}=\dfrac{1}{\sin 45^{\circ}}
=\sqrt{2}
=1.414$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

 Light takes t$ _{1}$ sec to travel a distance x cm in vacuum and takes t$ _{2}$ sec to travel 10x cm in a medium.  The critical angle corresponding to the media is :                         

  1. $sin^{-1}(10t _{1}/t _{2}$)
  2. $sin^{-1}(t _{2}/10$)
  3. $sin^{-1}(1/t _{1}$)
  4. $sin^{-1}(t _{1}/10t _{2}$)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\vartheta _{vaccum}=\dfrac{d _{vaccum}}{t _{vaccum}}=\dfrac{\chi }{t _{1}}$


$v _{med}=\dfrac{10\chi }{t _{2}}$

$\mu=\dfrac{\chi \times t _{2}}{t _{1}\times 10\chi }=\dfrac{t _{2}}{10t _{1}}$


$\theta _{cric}=sin^{-1}\left ( \dfrac{1}{\mu} \right )=sin^{-1}\left ( \dfrac{10t _{1} }{t _{2}}\right )$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

If the critical angle of the medium is 30$^{\circ}$, the velocity of light in that medium is :
(velocity of light in air  $3 \times 10^8 $ m/s)

  1. 6 x 10$^{8}$m/ s
  2. 3 x 10$^{8}$m /s
  3. 1.5 x 10$^{8}$m/ s
  4. 1 x 10$^{8}$m/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As, $\theta _{cric}=sin^{-1}\left ( \dfrac{1}{\mu} \right )$


$ sin \theta _{cric}=\dfrac{1}{\mu} $

$ \mu= \dfrac{1}{sin 30}=2 $

Also as $ \mu= \dfrac{c}{v _m}=\dfrac{3\times 10^{8}}{v _m } $

$ v _m = 1.5 \times 10^{8} m/s. $

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

A ray of light from a denser medium strikes a rarer medium at an angle of incidence $i$. If the reflected and refracted rays are mutually perpendicular to each other then the critical angle is :

  1. sin$^{-1}$ (tan i)
  2. cos$^{-1}$ (tan i)
  3. cot $^{-1}$ (tan i)
  4. cosec$^{-1}$ (tan i)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$i+r=90$  where i is the angle of incidence, r is the angle of refraction


$r=90-i$

From Snell's Law, $\mu \times sin i= 1 \times sin r$

$\mu \times sin i = sin \left ( 90-i \right )$

$\mu = cot\ i$

$\theta _{cric} =sin^{-1} \left ( \dfrac{1}{\mu} \right )=sin^{-1}\left ( tan\ i \right )$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

The critical angle for a medium with respect to air $45^0$. The refractive index of that medium with respect to air is:

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

$C=45^0$


$^{med}\mu _a=\dfrac {1}{sin C}=\dfrac {1}{sin 45^0}$

$=\dfrac {1}{(1\sqrt 2)}=\sqrt 2$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

A ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$, which is less than the critical angle. Then there will be?

  1. Only a reflected ray and no refracted ray

  2. Only a refracted ray and no reflected ray
  3. A reflected ray & a refracted ray and the $\angle$ between them would be less than $180^o-20^0$
  4. A reflected ray & a refracted ray and the $\angle$ between them would be greater than $180^o-2\theta$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There will be a reflected ray and a refracted ray.since incident angle is less than critical angle, angle between two resultant rays would be between $180^{\circ} - 20^{\circ}$.hence option c

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular brightspot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is

  1. 1.21

  2. 1.30

  3. 1.36

  4. 1.42
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have, 
$ Sin  C = [ 1 + \dfrac {\mu+b}{\mu _l}] = [\dfrac {\mu _1}{2.72}]$


$\implies [ \dfrac {r}{\sqrt {r^2 + h^2}}] = \dfrac {\mu _1}{2.72}$

$\implies \mu _1 = (\dfrac {2.72}{2}) = 1.36$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence $45^{\mathrm{o}}$ and undergoes total internal reflection. lf $\mu$ is the refractive index of medium the possible values of $\mu$ are

  1. $\mu =1.3$
  2. $\mu =1.4$
  3. $\mu =1.5$
  4. $\mu =1.6$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Since, the ray undergoes total internal reflection, 
$ \mu > \dfrac{1}{sin c} $


Now,$ i = 45$

Thus, $sin \  c < sin  \ i $

Thus, $ \mu > \dfrac{1}{sin i} $

OR, $ \mu > \sqrt{2} $

Multiple choice

Why (according to canon) are some lightsaber red?

  1. The crystals are cracked.

  2. The crystals bleed.

  3. The crystals are naturally red.

  4. The crystals are made on mustafar in a red color.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In current Star Wars canon, red lightsabers are created when a dark side user pours their pain and anger into a kyber crystal. This process is referred to as making the crystal bleed.