Physics

Wave Optics and Diffraction

163 Questions

Wave optics explores light phenomena like interference and diffraction. This hub features problems on Young's double slit experiment, Fresnel biprism, and single slit diffraction patterns. These topics regularly appear in physics sections of competitive exams.

Young's double slitFresnel biprismSingle slit diffractionConstructive interferenceFringe width calculations

Wave Optics and Diffraction Questions

Multiple choice physics wave optics huygens wave theory and wavefront wave propagation (huygens' construction) theories on light wave behaviour

Huygen's concept of secondary wave

  1. allow up to find the focal length of a thick lens

  2. is a geometrical method to find a wavefront

  3. is used to determine the velocity of light

  4. is used to explain polarization
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Huygen proposed a hypothesis for the geometrical construction of the position of a common wavefront at any instant during the propagations of waves in a medium.

Multiple choice physics wave optics huygens wave theory and wavefront wave propagation (huygens' construction) theories on light wave behaviour

The two coherent sources of equal intensity produce maximum intensity of $100$ units at a point. If the intensity of one of the sources is reduced by $50\%$ by reducing its width then the intensity of light at the same point will be

  1. 90

  2. 81

  3. 67

  4. 72.85
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Maximum intensity I_max is proportional to (a1 + a2)^2 = 100 units, meaning a1 + a2 = 10. Reducing the width of one source by 50% reduces its amplitude by a factor of sqrt(0.5). Solving for the new amplitudes and calculating the new maximum intensity via (a1 + a_new)^2 yields approximately 72.85 units.

Multiple choice physics wave optics huygens wave theory and wavefront wave propagation (huygens' construction) theories on light wave behaviour

 For what distance ray optics a good approximation when the aperture is 4 mm wide and the wavelength is 500nm?

  1. $32m$
  2. $69 m$
  3. $16 m$
  4. $8 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The Good approximation distance till ray optics or valid is given by Fresnel distance which is given by
$z _f=\dfrac{a^2}{\lambda}$
$z _f$ is distance till which diffraction effects of light can be neglected.
$a=$ aperature$=4\times 10^{-3}m$
$\lambda =$wavelength$=500\times 10^{-9}$m
$z _f=\dfrac{(4\times 10^{-3})^2}{(500\times 10^{-9})m}$
$=\dfrac{16\times 10^{-6}}{500\times 10^{-9}}$
$=0.032\times 10^3$m
$\therefore z _f=32m$.
Multiple choice physics wave optics huygens wave theory and wavefront wave propagation (huygens' construction) theories on light wave behaviour

At the first minimum adjacent to the central maximum of a single slit diffraction plate the phase difference between the huygens wavelet from the edge of the slit and the wavelet from the midpoint of the slate

  1. $\pi/8$
  2. $\pi/4$
  3. $\pi/2$
  4. $\pi$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At the first minimum, the path difference between the edges of the slit is lambda. The path difference between the edge and the midpoint is lambda/2, which corresponds to a phase difference of pi.

Multiple choice luminous intensity measurements physics

A lamp placed 60 cm from a screen produces the same illumination as a standard 100 W lamp placed 90 cm away on the other side of the screen. The luminous intensity of the first lamp is 

  1. $49.44W$
  2. $44.44W$
  3. $54.44W$
  4. $34.44W$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For same illumination $\dfrac{Power} { r^{2}}$ will be same


=> $\dfrac {100} { 0.9^{2}} = \dfrac {P} {0.6^{2}}$
=> P = $44.44 W$

Multiple choice luminous intensity measurements physics

Two light sources of $8 Cd$ and $12 Cd$ are placed on the same side of the photometer screen at a distance of $40 cm$ from it. Where should a $80 Cd$ source be placed to balance the illuminance?

  1. $40 cm$
  2. $60 cm$
  3. $20 cm$
  4. $80 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Intensity of illumination = $\dfrac {L} {d^{2}}$


Therefore
$I _1+I _2=I$

$\dfrac {8} {0.4^{2}} + \dfrac{12} {0.4^{2}} = \dfrac{80}{d^{2}}$

=>$\dfrac {20} {0.16} = \dfrac{80}{d^{2}}$

=> d = 0.8 m =80 cm

Answer. D) 80 cm

Multiple choice luminous intensity measurements physics

The luminous intensity of a light source is $300 Cd$. The illuminance of a surface lying at a distance of $10$ $m$ from it will be if light falls normally on it 

  1. $30$ lux
  2. $3$ lux
  3. $0.3$ lux
  4. $0.03$ lux
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit  used  when  characterizing illumination conditions of a surface:
$E = \dfrac {I}{R^2}    lux$
where $I$ is the luminous intensity and $R$ is the distance.
Given $I = 300 cd$ and $R = 10 m$
$\implies E = \dfrac {300}{10^2} = 3   lux$

Multiple choice luminous intensity measurements physics

Two lamps of luminous intensity of $8$ $Cd$ and $32$ $Cd$ respectively are lying at a distance of $1.2$ $m$ from each other. Where should a screen be placed between two lamps such that its two faces are equally illuminated due to the two sources?

  1. $10$ $cm$ from $8$ $Cd$ lamp
  2. $10$ $cm$ from $32$ $Cd$ lamp
  3. $40$ $cm$ from $8$ $Cd$ lamp
  4. $40$ $cm$ from $32$ $Cd$ lamp
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let distance between 8 Cd lamp and screen be $ x  m$

Distance between 32 Cd lamp and screen $=$ $ (120-x) m$

Under equal illumination,
$ \dfrac{8}{x^{2}} = \dfrac{32}{(120-x)^{2}}$
=> x $=$ 40 cm

So the distance from 8 Cd lamp is 40 centimeters.

Multiple choice luminous intensity measurements physics

At what distance should a book be placed from a $50 Cd$ bulb so that the illuminance on the book becomes $2lm $ $m^{-2}$

  1. $1$m
  2. $5$m
  3. $10$m
  4. $50$m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit 

used  when  characterizing illumination conditions of a surface:
$E = \dfrac {I}{R^2}    lux$
$\implies R = \sqrt {\dfrac {I}{E}}$
where $I$ is the luminous intensity and $R$ is the distance.
Given $I = 50 cd$ and $E = 2 lm  m^{-2}$
$\implies R = \sqrt {\dfrac {50}{2}} = 5   m$

Multiple choice luminous intensity measurements physics

The luminous intensity of a $100$ W unidirectional bulb is $100$ candela. The total luminous flux emitted from the bulb will be

  1. $100\pi$
  2. $200\pi$
  3. $300\pi$
  4. $400\pi$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have total luminous flux,
$L = 4 \pi I = 4 \pi \times 100 = 400 \pi$

Multiple choice luminous intensity measurements physics

The illuminance on screen distance $3$ m from a $100$ $W$ lamp is $25$ lm/$m^{2}$. Presuming normal incidence, the luminous intensity of the bulb will be 

  1. $100$ $Cd$
  2. $25$ $Cd$
  3. $225$ $Cd$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have illuminance
$I = \dfrac {\phi}{R^2}$
$\implies \phi = I  R^2 = 25 \times 3^2 = 225 \ Cd$

Multiple choice luminous intensity measurements physics

The illuminance of a surface distance $10$ m from a light source is $10$ lux. The luminous intensity of the source for normal incidence will be 

  1. $10^{1} Cd$
  2. $10^{2} Cd$
  3. $10^{3} Cd$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have illuminance
$I = \dfrac {\phi}{R^2}$
$\implies \phi = I  R^2 = 10 \times 10^2 = 10^3  Cd$

Multiple choice luminous intensity measurements physics

The intensity produced by a long cylindrical light source at a small distance $r$ from the source is proportional to

  1. $\displaystyle \dfrac{1}{r^2}$
  2. $\displaystyle \dfrac{1}{r^3}$
  3. $\displaystyle \dfrac{1}{r}$

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

At a distance $r$ from a line source of power$P$ and length $l$, the intensity will be


$ I=\dfrac { P }{ S } =\dfrac { P }{ 2\pi rl } \Rightarrow I\propto \dfrac { 1 }{ r } $

Multiple choice luminous intensity measurements physics

Two light sources with equal luminous intensity are lying at a distance of 1.2 m from each other. Where should a screen be placed between them such that illuminance on one of its faces is four times that on another face?

  1. 0.2 m

  2. 0.4 m

  3. 0.8 m

  4. 1.6 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$E _2 = 4 E _1$. If x is distance from 1st source,
then, $\displaystyle \frac{I}{(1.2 - x)^2} = \frac{4 I}{x^2} $ or $\displaystyle \frac{1}{1.2 - x} = \frac{2}{x}$
$3x = 2.4, x = 0.8 m$