Tag: option c: imaging

Quizzes (13) Questions (113)

Questions Related to option c: imaging

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

Advantages of optical fibres over electrical wires is:

  1. High band width and EM interference

  2. Low band width and EM interference

  3. High band width low transmission capacity and no EM interference

  4. High band width, high data transmission capacity and no EM interference
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Few advantages of optical fibres are that the number of signals carried by optical fibres is much more than that carried by the Cu wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it.

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

(A): Optical fibres are widely used to communication network.
(R) : Optical fibres are small in size, light weight, flexible and there is no scope for interference in them.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)

  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)

  3. (A) is true but (R) is false

  4. (A) is false but (R) is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Optical fibres are widely used to communication network because they are small in size, light weight, flexible and there is no scope for interference in them. They are easy to handle due to it's small size, light weight and flexibility. They can be placed wherever we need. Further in this case, due to the no scope for interference , information cannot be loss or damage.

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 perceive colours resolution of optical instruments lenses option c: imaging physics

The angles of minimum deviations are 53$^{\circ}$ and 51$^{\circ}$ for blue and red colors respectively produced in an equilateral glass prism. The dispersive power is :

  1. $\dfrac{2}{51}$
  2. $\dfrac{1}{26}$
  3. $\dfrac{1}{52}$
  4. $\dfrac{1}{51}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As $\delta _{Y}=\dfrac{\delta _{B}+\delta _{R}}{2}$


So $\delta _{Y}=\dfrac{53^{\circ }+51^{\circ }}{2}$ =52$^{\circ }$

$\therefore dispersive\ power\ =\dfrac{\delta _{B}-\delta _{R}}{\delta _{Y}}$

                                    $=\dfrac{53-51}{52}$

                                    $=\dfrac{2}{52}$

                                    $=\dfrac{1}{26}$

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

If the refractive indices of crown glass for red, yellow and violet colours are respectively ${ \mu  } _{ r },{ \mu  } _{ y },{ \mu  } _{ v }$, then the dispersive power of this glass would be

  1. $\cfrac { { \mu } _{ v }-{ \mu } _{ r } }{ { \mu } _{ y }-1 } $
  2. $\cfrac { { \mu } _{ v }-{ \mu } _{ y } }{ { \mu } _{ r }-1 } $
  3. $\cfrac { { \mu } _{ v }-{ \mu } _{ y } }{ { \mu } _{ y }-{ \mu } _{ r } } $
  4. $\cfrac { { \mu } _{ v }-{ \mu } _{ r } }{ { \mu } _{ y } } -1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Dispersive power of a glass is given by ratio of difference of reflective index of two extreme wavelength to the difference of intermediate wavelength to unity i.e.


$ { Dispersive\quad Power\quad =\quad \dfrac { { \mu  } _{ v }-{ \mu  } _{ r } }{ { \mu  } _{ y }-1 }  } $

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

The dispersion of a medium for wavelength $\lambda$ is D. Then the dispersion for the wavelength $2\lambda$ will be:

  1. $(D/8)$
  2. $(D/4)$
  3. $(D/2)$

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

As we know, Cauchy's Dispersion formula is :
$ \mu $= $A + \dfrac{B}{ \lambda^{^{2}}} $
And dispersion
D= - $ \dfrac {d\mu}{d\lambda}$
Therefore, from the above 2 equations:
D = $ -(-2\lambda ^{3})B $= $ \dfrac {2B}{\lambda^{3}}$
This implies that
D $ \alpha \dfrac{1}{\lambda^{3}}$
Hence,
$ \dfrac {{D}'}{D}$ = $( \dfrac{\lambda}{{\lambda}' })^{3}$
As $ {\lambda}' = 2\lambda$
Therefore,
$ {D}' =D/8$

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

In a prism, the refractive indices of different colours are
$\mu _{V} =$ 1.6;
$\mu _{R} =$ 1.52;
$\mu _{Y} =$ 1.56.
The dispersive power of the prism is :

  1. $\dfrac{1}{56}$
  2. $\dfrac{1}{8}$
  3. $\dfrac{1}{7}$

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

Dispersive power of prism= $w=\dfrac{\mu _{V}-\mu _{R}}{\mu _{Y}-1}$


$w=\dfrac{1.6-1.52}{1.56-1}$

$w=\dfrac{1}{7}$

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

A flint glass prism is of refracting angle 5$^{\circ}$. Its refractive index for C line is 1.790 and for F line is 1.805. The angular dispersion of C and F lines is:

  1. 0.075$^{\circ}$
  2. 0.085$^{\circ}$
  3. 0.095$^{\circ}$
  4. 0.065$^{\circ}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Deviation for C line $= \text{(Refractive index for C line -1) x Refracting angle}$

                                 $= (1.79-1) \times 5$
                                 $=3.95^o$

Deviation for F line $= \text{(Refractive index for F line -1) x Refracting angle}$

                                 $= (1.805-1) \times 5$      
                                 $= 4.025^o$

So the angular dispersion of C and F line is the deviation difference between F and C line.
Angular dispersion $= 4.025 -3.95 = 0.075^o$

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

Dispersive power of the material of a prism is 0.0221. If the deviation produced by it for yellow colour is 38$^{\circ}$, then the angular dispersion between red and violet colours is :

  1. 0.65$^{\circ}$
  2. 0.84$^{\circ}$
  3. 0.48$^{\circ}$
  4. 1.26$^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As Dispersive power $ = \dfrac{Angular  \  dispersion}{mean  \ deviation}$


We take the deviation for yellow color as the mean deviation

$\Longrightarrow$  $\omega=\dfrac{\delta _{V}-\delta _{R}}{\delta _{Y}}$


$\Longrightarrow$  $0.0221=\dfrac{\delta _{V}-\delta _{R}}{38^{\circ }}$

$\Longrightarrow$  $\delta _{\nu}-\delta _{R}=0.84^{\circ }$