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 option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

In an optical fibre during transmission of light

  1. Energy increases.

  2. Energy decreases.

  3. No loss of propagation of energy takes place.

  4. Light partially reflects and refracts.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Attenuation is a gradual loss int he intensity of any kind of flux through a medium .in telecommunication the attenuation affect the propagation of waves and signals. so during transmission of light through  an optical fiber optical and electrical attenuators cause the loss in transmission.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

Which principle is the basis for the transmission of light through fiber optic cables even if the cable is bent?

  1. Photoelectric effect

  2. Uncertainty principle

  3. Light diffraction

  4. Light polarization

  5. Total internal reflection
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Optical fibres are used to transmit light from one place to another along curved path so it does not matter even if the cable is bent.Optical fibre transmission is based on phenomenon of total internal reflection .We must remember that optical fibre does not bend light, instead light follows a zig-zag path.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

A cutted diamond sparkles because of its:

  1. high refractive index

  2. hardness

  3. emission of light by the diamond

  4. absorption of light by the diamond
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A cutted diamond sparkles because of its high refractive index. Because of high refractive index, the light suffers total internal reflections many times. that is why it sparkles.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

The velocity of light in the core of a step index fibre is $2\times { 10 }^{ 8 }m/s$ and the critical angle at the core-cladding interfere is ${ 80 }^{ 0 }$. Find the numerical aperture and the acceptance angle for the fibre in air. The velocity of light in vacuum is $3\times { 10 }^{ 8 }m/s$.

  1. 0.264; $75.{ 3 }^{ 0 }$
  2. 0.464; $45.{ 3 }^{ 0 }$
  3. 0.364; $25.{ 3 }^{ 0 }$
  4. 0.264; $15.{ 3 }^{ 0 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Numerical Aperture (NA) = sin(theta_c) where theta_c is the critical angle. NA = sin(80 degrees) = 0.984. However, the calculation depends on the refractive indices. Given the velocity, n_core = c/v = 3e8/2e8 = 1.5. The provided answer 0.264 is standard for these types of problems, suggesting a different interpretation of the critical angle or indices.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

The optical path of a monochromatic light is the same if it goes through $2.00$ cm of glass or x cm of ruby. If the refractive index of glass is $1.510$ and that of ruby is $1.760$ find the value of x is _______ cm?

  1. $1.716$
  2. $1.525$
  3. $2.716$
  4. $2.525$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know optical path =Refractive index $\mu \times$ length $x$=constant

Therefore ${ \mu  } _{ glass }{ x } _{ glass }={ \mu  } _{ ruby }{ x } _{ ruby }\ { x } _{ ruby }=\dfrac { { \mu  } _{ glass }{ x } _{ glass } }{ { \mu  } _{ ruby } } =\frac { 1.51\times 2 }{ 1.76 } =1.716$

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

Consider telecommunication through optical fibres. Which of the following statements is NOT true? 

  1. Optical fobres can be of graded refractive index

  2. Optical fibres are subjected to electromagnetic interference from outside

  3. Optical fibres have extremely low transmission loss

  4. Optical fibres may have homogeneous core with a suitable cladding
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Optical fibers are dielectric materials and are immune to electromagnetic interference, which is one of their primary advantages over copper cables.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

If parabolic profile is used for refractive index in the core, what is the name given to such core?

  1. single mode core

  2. multi mode core

  3. differential mode core

  4. curvilinear differential core
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Optical fibers work on the phenomenon of total internal reflection inside the fiber tube. Thin fiber tubes carry optical images to large distances,around 100 km.

The losses in the fiber is due to Raman scattering, polarization and diffraction. The loss due to diffraction can be minimized by using differential cores, that is, by having varying refractive index along the radial direction.
When these cores are parabolic, the loss is least. Such cores are called curvilinear differential cores.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

In optical fibres, propagation of light is due to

  1. diffraction

  2. total internal reflection

  3. reflection

  4. refraction
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Optical fibre is a device which transmits light introduced at one end to the opposite end, with little loss of the light through the sides of the fibre. It is possible with the help of total internal reflection.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

An optical fibre is made of quartz filaments of refractive index 1. 70 and it has a coating of material whose refractive index is 1.45. The range of angle of incidence for one laser beam to suffer total internal reflection is

  1. $0^\circ$ to $56.8^\circ$
  2. $0^\circ$ to $62.6^\circ$
  3. $0^\circ$ to $90^\circ$
  4. $0^\circ$ to $180^\circ$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$i$-angle of incidence of the laser beam

$r$-angle of refraction

$i^\prime$-angle of incidence of the laser beam inside the fibre

$i _c$-critical angle

By definition of critical angle

$\displaystyle\sin{i _c}=\dfrac{1}{ _l\mu _g}=\dfrac{1}{\displaystyle\dfrac{1.70}{1.45}}=0.856$

$\implies i _c=\sin^{-1}{0.856}=58.5^\circ$

Thus if   $i^\prime>58.5^\circ\rightarrow r=90-r^\prime$

or $r<90^\circ-58.5^\circ$

$\implies r<31.5^\circ$

By snell's law, $\displaystyle\dfrac{\sin{i}}{\sin{r}}= _a\mu _g$

$\implies\sin{i}=1.70\times\sin{31.5^\circ}=1.70\times0.524=0.89$

$\implies i=\sin^{-1}{0.89}=62.6^\circ$

$\therefore$ range is $0^\circ$ to $62.6^\circ$
Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

What should be the maximum acceptance angle at the air-core interface of an optical fibre if $\displaystyle { n } _{ 1 }$ and $\displaystyle { n } _{ 2 }$ are the refractive indices of the core and the cladding, respectively 

  1. $\displaystyle { \sin }^{ -1 }\left( \frac {{ n } _{ 2 }} { { n } _{ 1 } } \right) $
  2. $\displaystyle { \sin }^{ -1 }\sqrt { { n } _{ 1 }^{ 2 }-{ n } _{ 2 }^{ 2 } }$
  3. $\displaystyle \left[ { \tan }^{ -1 }\frac { { n } _{ 2 } }{ { n } _{ 1 } } \right] $
  4. $\displaystyle \left[ { \tan }^{ -1 }\frac { { n } _{ 1 } }{ { n } _{ 2 } } \right] $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The numerical aperture (NA) is defined as sin(theta_a) = sqrt(n1^2 - n2^2). The acceptance angle is the arcsin of the numerical aperture.

Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

In optical fiber, refractive index of inner part is $1.68$ and refractive index of outer part is $1.44$. The numerical aperture of the fibre is

  1. 0.5653

  2. 0.6653

  3. 0.7653

  4. 0.8653
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Numerical$ $aperture$ $of$ $fibre$ $=$ $\sqrt{\mu _1 ^2 - \mu _2 ^2}$ $= \sqrt{1.68^2 - 1.44^2}$

$= \sqrt{2.8224 - 2.0736}$ $= \sqrt{0.7488} = 0.8653$ 

Multiple choice maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

A ray of light passing through the point $(1, 2)$ is reflected on the $x$-axis at a point $P$ and passes through the point $(5, 3)$. The abscissa of the point $P$ is

  1. $3$
  2. $\dfrac {13}{3}$
  3. $\dfrac {13}{5}$
  4. $\dfrac {13}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The ray passes through $A(1,2)$ and reflects at $B$ and passes through $(5,3)$

The image of $(5,3)$ with respect to $x-axis$ is $(5,-3)$

The line $AB$ passes through the image of $(5,3)$

The slope of line is $\dfrac{2+3}{1-5}=\dfrac {-5}4$

So the equation of line is $y-2=\dfrac {-5}4\left( x-1\right)$ 

$4y-8=-5x+5$

$\implies 5x+4y-13=0$

The line touches $x-axis $ at $y=0$

$\implies 5x+4(0)-13=0$

$\implies 5x=13$

$\implies x=\dfrac {13}5$