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

In a Fresnel's biprism experiment the two positions of lens separation between the slits as $16cm$ and $9cm$ respectively. The true distance of separation between the slits is 

  1. $10cm$
  2. $12cm$
  3. $14cm$
  4. $15cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the displacement method for a biprism, the distance between the virtual slits d = sqrt(d1 * d2). Here, d = sqrt(16 * 9) = sqrt(144) = 12 cm.

Multiple choice physics superposition of waves difference between interference and diffraction explaining wave phenomena diffraction

In Young's experiment fourth bright fringe produced by light of $5000\overset{o}{A}$ superposes on the fifth bright fringe of an unknown wavelength. The unknown wavelength is _________$\overset{o}{A}$.

  1. $4000$
  2. $6000$
  3. $5000$
  4. $8000$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ \beta = \dfrac{ \lambda \times D}{d} $
$4\beta _1=5\beta _2$
$4\times 5000\dfrac{D}{d}=5\times \lambda\dfrac{D}{d}$
$\lambda =4000\overset{o}{A}$.
Multiple choice physics superposition of waves difference between interference and diffraction explaining wave phenomena diffraction

Two wave-fronts are emitted from coherent sources of path difference between them is $2.1$ micron. Phase difference between the wave-fronts at that point is $7.692\ \pi$. Wavelength of light emitted by sources will be :

  1. $5386\mathring{A}$
  2. $5400\mathring{A}$
  3. $5460\mathring{A}$
  4. $5892\mathring{A}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given:
$\Delta x=2.1 \times 10^{-6}\\\Delta \phi=7.6922$

$\Phi =\dfrac{2\pi \Delta x}{\lambda}$

$\lambda=\dfrac{2\pi\times 2.1\times 10^{-6}}{7.692\pi}$

$\lambda=5460\times 10^{-10}m$
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

A screen is at a distance of $1\ m$ away from the aperture. If light of wavelength $500\ nm$ falls on an aperture, then area of first HPZ and radius of third HPZ are :

  1. $1.57\ mm^2, 1.22\ mm$
  2. $1.22\ mm^2, 1.57\ mm$
  3. $1.65\ mm^2, 2.79\ mm$
  4. $2.63\ mm^2, 0.22\ mm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of the n-th HPZ is A = pi * b * lambda. For n=1, A = pi * 1 * 500 * 10^-9 = 1.57 * 10^-6 m^2 = 1.57 mm^2. Radius of n-th HPZ is r_n = sqrt(n * b * lambda). For n=3, r_3 = sqrt(3 * 1 * 500 * 10^-9) = sqrt(1500 * 10^-9) = 1.22 * 10^-3 m = 1.22 mm.

Multiple choice physics superposition of waves difference between interference and diffraction explaining wave phenomena diffraction

The sodium yellow doublet has wavelengths $5890\mathring{A}$ and $'\lambda' \mathring{A}$ and resolving power of a grating to resolve these lines is $982$, then value of $\lambda$ is :

  1. $5896\mathring{A}$
  2. $5880\mathring{A}$
  3. $5869\mathring{A}$
  4. $5876\mathring{A}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Resolving power R = lambda / delta_lambda = n * N. Given R = 982, lambda = 5890, delta_lambda = 5896 - 5890 = 6. R = 5890 / 6 = 981.6, which is approximately 982.

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

The phenomenon of diffraction can be treated as interference phenomenon if the number of coherent sources is 

  1. one

  2. two

  3. zero

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

The single slit diffraction pattern acts as an envelope for the multiple slit interference patterns.

Diffraction on a single slit is equivalent to interference of light from infinite number of coherent sources contained in slit.

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

In Fresnel's biprism expt., a mica sheet of refractive index 1.5 and thickness 6 $\times$ 10$^{-6}$m is placed in the path of one of interfering beams as a result of which the central fringe gets shifted through 5 fringe widths. The wavelength of light used is

  1. 6000 $\overset{o}{A}$
  2. 8000 $\overset{o}{A}$
  3. 4000 $\overset{o}{A}$
  4. 2000 $\overset{o}{A}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Where n is equivalent number of fringe by which the centre fringe is shifted due to mica sheet
$\displaystyle\lambda = \frac{(\mu - 1)t}{n} = \frac{(1.5 - 1) 6 \times 10^{-6}}{5} = 6 \times 10^{-7} m = 6000 \overset{o}{A}$

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

An aperture of size a is illuminated by a parallel beam of light of wavelength $\lambda$. The distance at which ray optics has a good approximation is

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

An aperture of size a is illuminated by a parallel beam of light of wavelength $\lambda$. The distance at which ray optics has a good approximation is $\dfrac {a^{2}}{\lambda}$. This is the Fresnel distance.

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

Diffraction gratings provide much brighter interference pattern since more light passes through them compared with double slits.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The statement is true.

That is diffraction gratings provide much brighter interference pattern since more light passes through them compared with double slits.
One difference between the interference of many slits (diffraction grating) and double-slit (Young's Experiment) is that a diffraction grating makes a number of principle maxima along with lower intensity maxima in between.  The principal maxima occur on both sides of the central maximum for which a formula similar to Young's formula holds true.

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

In biprism experiment, the distance of 20th bright band from the centre of the interference pattern is 8 mm. The distance of the 30th bright band is

  1. 4 mm

  2. 8 mm

  3. 12 mm

  4. 16 mm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given

$20\cfrac { \lambda D }{ d } =8\ =\cfrac { \lambda D }{ d } =\cfrac { 8 }{ 20 } \ \therefore  { 30 }^{ th }$ bright,
$=30\times \cfrac { \lambda D }{ d } \ =30\times \cfrac { 8 }{ 20 } \quad =12mm$

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

In a Fresnel's bi-prism experiment, the fringe of width $0.05mm$ is observed on a screen at a distance of $1.5m$ from the source . When a convex lens is placed between the source and the screen, for two positions of the lens image of interfering sources are produced on the screen. The separation between the two images being $0.04$ and $0.01mm$, respectively. The wavelength of light used is

  1. $6.67nm$
  2. $0.667nm$
  3. $667nm$
  4. $667A^o$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Distance between slits, $d=\sqrt{d _{1}d _{2}}=0..02mm$

Fringe width, $\beta=\dfrac{D\lambda}{d}$
hence $\lambda=\dfrac{\beta d}{D}=0.667nm$

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

A thin water of $9.9\mu$ m thickness and refractive index $1.5$ is introduced in front of one of the slits in a Young's double slit experiment. By how many fringe width does the fringe pattern shift? (Given wavelength of the wave is $5.5\times 10^{-7}$m)

  1. $11$
  2. $9$
  3. $3$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Shift in fringes N = (mu - 1) * t / lambda. N = (1.5 - 1) * 9.9 * 10^-6 / 5.5 * 10^-7 = 0.5 * 9.9 * 10^-6 / 5.5 * 10^-7 = 0.5 * 18 = 9.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Monochromatic light of wavelength $400\ nm$ and $560\ nm$ are incident simultaneously and normally on double slits apparatus whose slits separation is $0.1\ mm$ and screen distance is $1\ m$. Distance between areas of total darkness will be:

  1. $4\ mm$
  2. $5.6\ mm$
  3. $14\ mm$
  4. $28\ mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For $\lambda =400$mm

fringe width
$\beta =\cfrac { \lambda D }{ d } \ =\cfrac { 400\times 1\times 10\times 1000 }{ 1\times { 10 }^{ 6 } } $
$ =4$mm
For $\lambda =560$mm
$\beta =\cfrac { \lambda D }{ d } =5.6$mm
$LCM$ of both fringe width will give combined fringe width which is also the distance between $2$ minima .
$LCM 4,5,6=28$mm

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In, which of the following the interference is produced by the division of wave front? 

  1. Young's double slit experiment

  2. Fresnel's bi-prism

  3. Lloyd's mirror

  4. Colours of thin film
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In Young's Double Slit Experiment, wavefront from a single sources divides when passed through the two slits. These two wavefronts interfere after emerging out of the slits.