Tag: wave optics

Quizzes (12) Questions (312)

Questions Related to wave optics

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

A person wishes to distinguish between two pillars located at a distance of 11 km. What should be the minimum distance between these pillars (resolving power of normal human eye is 1')?

  1. 1 m

  2. 3.2 m

  3. 0.5 m

  4. 5 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resolving power is given by the distance between two objects to be distinguished per unit distance of objects from the object distinguishing them.

Hence,$\theta=\dfrac{d}{D}$ 
Hence,$d=\theta D=\dfrac{1}{60}\times \dfrac{\pi}{180}\times 110000=3.2m$

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Wavelength of light used in an optical instrument are $\lambda _1 = 4000 A^o  and \lambda _2 = 5000 A^0$, then ratio of their respective resolving powers (corresponding to $\lambda _1   \ and  \ \lambda _2$) is

  1. 16:25

  2. 9:1

  3. 4:5

  4. 5:4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resolving power of an optical instrument $\displaystyle \propto \dfrac {1}{\lambda}$


$\displaystyle \dfrac{\text{Resolving  power  at} \lambda _1} {\text{Resolving  power  at} \lambda _2}=\dfrac {\lambda _1}{\lambda _2}$

$\displaystyle \left [\text{Limit of resolution}  \propto \dfrac{1}{\text{resolving  power}}\right]$

$\therefore$  Ratio  of  resolving  power $= \displaystyle \dfrac {5000}{4000} = \dfrac {5}{4} = 5 : 4$

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Resolving power of a telescope increases with

  1. increase in focal length of eye-piece

  2. increase in focal length of objective

  3. increase in aperture of eye piece

  4. increase in aperture of objective
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resolving power $= \displaystyle \dfrac{\lambda}{d \lambda}$ plane transmission granting 


Resolving power for telescope

$= \displaystyle \frac{1}{\text{limit of resolution}} = \dfrac{d}{1.22 \lambda} = \dfrac{d _0}{d _1}$

by increasing the aperture of objective resolving power can be increased.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The limit of resolution of eye is approximately

  1. $1^0$
  2. $1'$
  3. $1 mm$
  4. $1 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The resolution of the human eye is the smallest object our  eye can see. This is limited by the diffraction limit, which is approximated by the angular size ratio of the object's size versus the distance to the object.
The normal pupil size of a human eye is 4 mm, which sets a minimum angular resolution of the eye  and to able to see the small objects we bring them as close to our eyes as possible, but there is a minimum distance for comfortable viewing which is roughly at 25 cm.But quoted figure for the smallest resolvable size is 0.1 mm, showing that the diffraction limit is a crucial factor in visual resolving power.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Aperture of the human eye is 2 mm. Assuming the mean wavelength of light to be 5000 $\overset{o}{A}$, the angular resolution limit of the eye is nearly:

  1. 2 minute

  2. 1 minute

  3. 0.5 minute

  4. 1.5 minute
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the angular limit of resolution of human eye is R then
R = $\displaystyle\frac{1.22\lambda}{a}$ = $\displaystyle\frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}}$ rad
         
      = $\displaystyle\frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}} \times \frac{180}{\pi} \times 60$ minute = 1 minute
  

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The magnifying power of an astronomical telescope is $8$, then the ratio of the focal length of the objective to the focal length of the eyepiece is : (final image is at $\displaystyle \infty $)

  1. $8$
  2. $\displaystyle \frac { 1 }{ 8 } $
  3. $0.45$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

magnifying power of telescope  (M P)$=8$
The magnifying power of an astronomical telescope is $=\cfrac{focal  \ length \ of \ object (-f _0)  }{focal \ length \ of \ eyepiece(f _e)}=8$
hence, the ratio of the focal length of the objective to the focal length of the eyepiece is $8$.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

A photographer changes the aperture of his camera and reduces it to half of the original aperture. The exposure time now should be:

  1. same as before

  2. double than before

  3. four times than before

  4. half than before
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Answer is D.

The amount of light captured while taking a photo is known as the exposure, and it's affected by three things - the shutter speed, the aperture diameter, and the ISO or film speed. 
Aperture is measured using the "f-number", sometimes called the "f-stop", which describes the diameter of the aperture. A lower f-number relates to a wider aperture (one that lets in more light), while a higher f-number means a narrower aperture (less light).
Because of the way f-numbers are calculated, a stop doesn't relate to a doubling or halving of the value, but to a multiplying or dividing by 1.41 (the square root of 2). For example, going from f/2.8 to f/4 is a decrease of 1 stop because 4 = 2.8 * 1.41. Changing from f/16 to f/11 is an increase of 1 stop because 11 = 16 / 1.41.
Hence, when the aperture of the camera reduces to half of the original aperture, the exposure time should be half than before.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Wavelength of light used in an optical instrument  are $\lambda _1 = 4000 \mathring { A } $ and $ \lambda _2 = 5000 \mathring { A} $ then ratio of their respective resolving powers(corresponding to $\lambda _1$ and $ \lambda _2$) is

  1. 16:25

  2. 9:1

  3. 4:5

  4. 5:4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know, 
$ Resolving\ Power\ \propto \dfrac{1}{\lambda} $

Given, 
$ \lambda _1 = 4000 \mathring{A} $
$ \lambda _2 = 5000 \mathring{A} $

$ \Rightarrow \dfrac{RP _{\lambda _1}}{RP _{\lambda _2}} = \dfrac{ \lambda _2}{ \lambda _1} $

$ \Rightarrow \dfrac{RP _{\lambda _1}}{RP _{\lambda _2}} = \dfrac{5000}{4000} = \dfrac{5}{4} $

$ \Rightarrow RP _{\lambda _1} : RP _{\lambda _2} = 5:4 $

Hence, the correct answer is OPTION D.