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 difference between interference and diffraction explaining wave phenomena diffraction

In Young's experiment, the source is red light of wavelength $7\times 10^{-7}m$. When a thin glass plate of refractive index $1.5$ at this wavelength is put is the path of one of the interfering beam, the central bright fringe shifts by $10^{-3} m$ to the position previously occupied by the $5^{th}$ bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength $5\times 10^{-7}m$ the central fringe shifts to a position initially occupied by the $6^{th}$ bright fringe due to red light. Find the refractive index of glass for the green light. Also estimate the change in fringe width due to the change in wavelength.

  1. $7\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
  2. $8\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
  3. $9\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
  4. $17\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The shift x = (mu - 1) * t * D / d. For red light, x = 5 * beta_red = 5 * lambda_red * D / d. Given x = 10^-3 m, lambda_red = 7 * 10^-7 m, we find t. The calculations for thickness, refractive index, and change in fringe width are complex but consistent with option A.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In Young's experiment two coherent sources are placed $0.90\ mm$ apart and fringe are observed one metre away. If it produces second dark fringes at a distance of $1\ mm$ from central fringe., the wavelength of monochromatic light is used would be 

  1. $60\ \times 10^{-4}\ cm$
  2. $10\ \times 10^{-4}\ cm$
  3. $10\ \times 10^{-5}\ cm$
  4. $6\ \times 10^{-5}\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Formula,

$x=(2n-1)\dfrac{\lambda D}{2d}$

$\Rightarrow \lambda =\dfrac{2xd}{(2n-1)D}$

$=\dfrac{2\times 10^{-3}\times 0.9\times 10^{-3}}{(2\times 2-1)\times 1}$

$=6 \times 10^{-5}cm$
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

The yellow light source in Young's double- slit experiment is replaced by a monochromatic red light source of same intensity. Then the fringe width of the interference pattern in comparison with that of the previous pattern will 

  1. increase

  2. decrease

  3. remain unchanged

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

Fringe width beta = lambda * D / d. Since the wavelength of red light is longer than that of yellow light, the fringe width increases.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In a Young's double slit experiment, $d = 1\ mm,$ $\lambda = 6000 \overset {0}{A}$ & $D = 1\ m.$ The slits produce same intensity on the screen. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is:

  1. $0.45\ mm$
  2. $0.40\ mm$
  3. $0.30\ mm$
  4. $0.20\ mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity I at a point with phase difference phi is I = I_max * cos^2(phi/2). For I = 0.75 * I_max, cos^2(phi/2) = 3/4, so phi/2 = pi/6 or 5pi/6. The path difference is delta = (lambda/2pi) * phi, and y = (D/d) * delta. Calculating the positions for both phase values gives the distance between them as 0.40 mm.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In a young's bouble slit experiment ,$d=1mm,\lambda  = 6000\mathop A\limits^0 $ and $D=1m$(where d,$\lambda$ and D have unit meaning). Each of slit individually produces same intensity on the screen. The minimum distance between two points  on the screen having $75$% intensity of the maximum intensity is:

  1. $0.45mm$
  2. $0.40mm$
  3. $0.30mm$
  4. $0.20mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

 In Young's experiment , the separation between 5th maxima and 3rd minima is how many times as that of fringr width?

  1. 5 time

  2. 3 times

  3. 2 times

  4. 2.5times
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For interference pattern, in double slit experiment maxima will be found at a distance y from the central maximum,
$y=n\lambda D/d$
$D=$distance between slit and screen
$d=$ separation of the slits
For $n=5$,
$y=5\lambda D/d$
For minima
$y=(2n+1)\lambda D/2d$
For $n=3$,
$y=7\lambda D/d$
Difference between maxima and minima$=7\lambda D/d-5\lambda D/d=2\lambda D/d$
Now, fridge width is given by $x=\lambda D/d$
Thus, the given difference is $2$ times the fringe width.
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Monochromatic light of wavelength 580 mm incident on a slit of width 0.30 mm. The screen 2 m from the slit. the width of the center maximum is 

  1. $3.35\times 10^{-3}m$
  2. $2.25\times 10^{-3}m$
  3. $6.20\times 10^{-3}m$
  4. $7.7\times 10^{-3}m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$d=580nm=580\times {10}^{-9}m$

$d=0.3mm=0.0003m$
width of the cental maxima $=\cfrac{2\lambda D}{d}$
$=\cfrac{2\times 580\times {10}^{-9}\times 2}{0.0003}$
$=0.0077m$

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Monochromatic green light of wavelength 5 $\times 10^{-7}$ m illuminates a pair of slits 1 mm apart. The separation of bright lines in the interference pattern formed on a screen 2 m away is :

  1. 0.25 mm

  2. 0.1 mm

  3. 1.0 mm

  4. 0.01 mm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


$\lambda=5\times 10^{-7}m$


$d=1mm=0.001m$

$D=2m$

The separation between two bright fringe is called fringe width.

$B=\dfrac{D\lambda}{d}$

$B=\dfrac{2\times 5\times 10^{-7}}{10^{-3}}$

$B=1mm$

The correct option is C.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

To make the central fringe at the centre O, a mica sheet of refractive index $1.5$ is introduced. Choose the correct statements (s).

  1. The thickness of sheet is $2\left( {\sqrt 2 - 1} \right)d$ in front of ${S _1}$
  2. The thickness of sheet is $\left( {\sqrt 2 - 1} \right)d$ in front of ${S _2}$
  3. The thickness of sheet is $2\sqrt 2 d$ in front of ${S _1}$
  4. The thickness of sheet is $\left( {2\sqrt 2 - 1} \right)d$ in front of ${S _1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Monochromatic green light of wavelength $5x{ 10 }^{ -7 }$ m illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2m away is 

  1. 0.25 mm

  2. 0.1 mm

  3. 1.0 mm

  4. 0.01 MM
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The fringe width beta is given by lambda * D / d. Here, lambda = 5 * 10^-7 m, D = 2 m, d = 10^-3 m. beta = (5 * 10^-7 * 2) / 10^-3 = 10^-3 m = 1.0 mm.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Each of the four pairs of light waves arrives at a certain point on a screen. The waves have the same wavelength. At the arrival point, their amplitudes and phase differences are :
$2 \mathrm { a } _ { 0 } , 6 \mathrm { a } _ { 0 } $ and $\pi$ rad
$3 \mathrm { a } _ { 0 } , 5 \mathrm { a } _ { 0 } $ and $\pi$ rad
$9 \mathrm { a } _ { 0 } , 7 \mathrm { a } _ { 0 } $ and $3\pi$ rad
$2 \mathrm { a } _ { 0 } , 2\mathrm { a } _ { 0 } $ and $0$
The pair/s which has greatest intensity is /are :

  1. $I$
  2. $II$
  3. $II, III$
  4. $I, IV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Intensity I is proportional to the square of the resultant amplitude. Resultant amplitude R = sqrt(A1^2 + A2^2 + 2*A1*A2*cos(phi)). For I and IV, the phase differences are pi and 0 respectively, leading to high resultant amplitudes.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In Young's double slit. experiment, distance between two sources is 0.1 mm. The distance of screen from the sources is 20 cm. Wavelength of light used is 5460 k Then angular position of first dark fringe is 

  1. $0.20 ^ { \circ }$
  2. $0.32 ^ { \circ }$
  3. $0.08 ^ { \circ }$
  4. $0.16 ^ { \circ }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The path difference for the first dark fringe is lambda/2. The angular position theta is given by sin(theta) = path_diff / d = (lambda/2) / d. With lambda = 5460 * 10^-10 m and d = 10^-4 m, sin(theta) = 2.73 * 10^-3. theta is approximately 0.16 degrees.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

The waves of $600 \mu m$ wave length are incident normally on a slit of $1.2\ mm$ width. The value of diffraction angle corresponding to the first minima will be (in radian):

  1. $\dfrac{\pi}{2}$
  2. $\dfrac{\pi}{6}$
  3. $\dfrac{\pi}{5}$
  4. $\dfrac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

$\lambda=600\mu m$
$d=1.2mm$
condition for minima 
$dsin\theta=m\lambda$
for $m=1$
$sin\theta=\dfrac{m\lambda}{d}=\dfrac{\lambda}{d}$

$sin\theta=\dfrac{600\times 10^{-6}}{1.2\times 10^{-3}}=0.5$

$\theta=sin^{-1}(0.5)=\dfrac{\pi}{6}$
The correct option is B.