Physics

Wave Optics and Diffraction

148 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 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 $\displaystyle \frac{1}{1.2 - x} = \frac{2}{x}$
$3x = 2.4, x = 0.8 m$

Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

In a Fresnel biprism experiment, the two positions of lens give separation between the slits as $16 $cm and$9 $cm, respectively. What is the actual distance of separation?

  1. $12.5 cm$
  2. $12 cm$
  3. $13 cm$
  4. $14 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A Fresnel Biprism is the variation on the Young's Slits experiment. The Fresnel biprism has two thin prisms which are joint at their bases to form an isosceles triangle. A single wavefront impinges on both prisms.

Separations between the slits,

$d _{1}= 16 cm$

and $d _{2}= 9 cm.$

Actual distance of separation (d)  can be computed with the formula as given:

$d = \sqrt {d _{1} \times d _{2}}\\$

$d = \sqrt {16 \times 9}\\$

$d = \sqrt {144} \\$

$d = 12 cm$

Thus actual distance will be $12 cm$.

Option B is correct.

Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

A parallel beam of light of wavelength 6000 $\overset{0}{A}$ gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is

  1. $2 \, \times \, 10^{-3} \, rad$
  2. $3 \, \times \, 10^{-3} \, rad$
  3. $1.8 \, \times \, 10^{-3} \, rad$
  4. $6 \, \times \, 10^{-3} \, rad$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here, $\lambda \, = \, 6000 \, \overset{0}{A} \, = \, 6000 \, \times \, 10^{-10} \, m \, = \, 6 \, \times \, 10^{-7} \, m \, a \, - \, 0.3 \, mm \, = \, 0.3 \, \times \, 10^{-4} \, m \, = \, 3 \, \times \, 10^{-4} \, m$

For first minima, $a sin \theta \, = \, \lambda$ where a is the slit widht 
$sin \, \theta = \, \dfrac{\lambda}{a} \, = \, \dfrac{6 \, \times \, 10^{-7}}{3 \, \times \, 10^{-4}} \, = \, 2 \, \times \, 10^{-3}$

As $sin \theta$ is very small

$\therefore \, \theta \, \cong \, sin \, \theta \, = \, 2 \, \times \, 10^{-3} \, rad$
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

Light of wavelength 600 nm is incident on an aperture of size 2 mm. The distance upto which light can travel such that its spread is less than the size of the aperture is:

  1. 12.13 m

  2. 6.67 m

  3. 3.33 m

  4. 2.19 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


Given:
 Aperture Width, $a =2mm=2\times 10^{-3}mm$
The distance upto which light can travel is Fresnel distance, $Z _F \, = \, \dfrac{a^2}{\lambda} \, = \, \dfrac{(2 \, \times \, 10^{-3})^2}{600 \, \times \, 10^{-9}} \, = \, 6.67 \, m$

Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength $5500 A^0$, the first minimum is observed at angle $30^0$. The first secondary maximum is observed at an angle $\theta$ =

  1. $sin^{-1} [\frac{1}{\sqrt{2}}]$
  2. $sin^{-1} [\frac{1}{4}]$
  3. $sin^{-1} [\frac{3}{4}]$
  4. $sin^{-1} [\frac{\sqrt{3}}{2}]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Slit width = d
$\lambda = 5500 A^0 = 5.5 \times 10^{-7}m, \theta _n = 30^0$
For first secondary minima, $d sin \theta _n = \lambda$
$ d = \dfrac{\lambda}{sin \theta _n} = \dfrac{5.5 \times 10^{-7}}{sin 30^0} = 11 \times 10^{-7}$m
For the first secondary maxima, $d sin \theta _n = \dfrac{3 \lambda}{2}$
i.e. $sin \theta _n = \dfrac{3 \lambda}{2d} = \dfrac{ 3 \times 5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}} = sin \theta _n = \dfrac{3}{4} or \theta _n = sin^{-1}(3/4)$

Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

Two slits of width $a _1$  and $a _2$ are illuminated by light of same wavelength.The first diffraction minima produced by each of them are in directions inclined at angles $\theta _1$ and $\theta _2$. The ratio of $sin\, \theta _1$ to $sin\,\theta _2$ is

  1. $\dfrac{a _1}{a _2}$
  2. $\sqrt{\dfrac{a _1}{a _2}}$
  3. $\sqrt{\dfrac{a _2}{a _1}}$
  4. $\dfrac{a _2}{a _1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

The average path difference between two waves coming from third and fifth Fresnel zones of a wave front at the centre of the screen is :

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

The third and fifth fresnel zone differ in phase by $2\pi $. Thus as $2\pi $ corresponds to path difference of $\lambda $.

Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

In Young's double slit experiment, angular width of fringes is $0.20^o$ for sodium light of wavelength $5890\overset{o}{A}$. If complete system is dipped in water, then angular width of fringes becomes.

  1. $0.15^o$
  2. $0.22^o$
  3. $0.30^o$

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

b'Angular fringe width$=\theta=\displaystyle\frac{\lambda}{d}$
$\Rightarrow \theta\propto \lambda$
$\lambda _w=\displaystyle\frac{\lambda _a}{\mu _w}$
So, $\theta _w=\displaystyle\frac{\theta _{air}}{\mu _w}=\frac{0.20}{\displaystyle\frac{4}{3}}=0.15^o$'

Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

In the diffraction of light of wavelength $\lambda$ through single slit of width d, the angle between the principal maxima and first minima will be:

  1. $\lambda/d$
  2. $\lambda/2d$
  3. $\lambda/4d$
  4. $\pi/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
theory of diffraction we know that minima condition is given by 
$n\lambda=d\sin \theta$
for first minima, $n=1$
$\lambda =d\sin\theta$
As $\theta\approx$ small
$\sin\theta\approx \theta$
$\lambda=d\theta$
$\theta=\dfrac{\lambda}{d}\rightarrow $ Required answer
$y=\dfrac{\lambda D}{d}$
$y=D\theta$
$\dfrac{\lambda D}{d}=D\theta$
$\theta=\dfrac{\lambda}{d}$
Required answer
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

A double slit arrangement produces interference fringes for sodium light ($\lambda=5890 \mathring {A}$) that are $0.40^0$ apart. What is the angular fringe separation if the entire arrangement is immersed in water?

  1. $0.10^o$
  2. $0.20^o$
  3. $0.30^o$
  4. $0.40^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know the distance between two slits $d=\dfrac{\lambda}{\sin\theta}$, where $\lambda$=wavelength and $\theta$=angle from the original beam.

Given wavelength of sodium light ${\lambda} _{1}=5890 \mathring { A } =5890\times{10}^{-10} m$ and ${\theta} _{1}={0.40}^{0}$
Putting the above values in the given equation we get,
$d=\dfrac{5890\times {10}^{-10}}{\sin0.40}=8.43\times {10}^{-5} m$
Also ${\lambda} _{2}=\dfrac{{\lambda} _{1}}{n}$--------(A) 
where n= refractive index of water =1.33 
Thus A becomes,
${\lambda} _{2}=\dfrac{5890\times {10}^{-10}}{1.33}=4.42\times {10}^{-7} m$
Again,
$d=\dfrac{{\lambda} _{2}}{\sin{\theta} _{2}}$
${\theta} _{2}={\sin}^{-1} (\dfrac{4.42\times {10}^{-7}}{1.33})={0.30}^{0}$