CBSE Class 12 Physics HOTs Wave Optics

Question. The phase difference between any two points of a wavefront is:

(a) 0
(b) π/4
(c) π/2
(d) π
Answer. A

Question. The intensity of light issuing out of two slits in young’s experiment is in ratio 1:4. The intensity of the minimum to the maximum will be in the ratio
(a) 1:2
(b) 1:4
(c) 1:9
(d) none of the above
Answer. C

Question. At the first minimum adjacent to the central maximum of a single -slit diffraction, the phase difference between the Huygens’s wavelet from the edge of the slit and the wavelet from the midpoint of the slit is
(a) π radian
(b) π/8 radian
(c) π/4 radian
(d) π/2 radian
Answer. A

Question. A parallel beam of monochromatic light of wavelength 5000A0 is incident normally on a single narrow slit of width 0.001mm. The light is focussed by a convex lens on a screen placed in focal plane. The first minimum will be formed for the angle of diffraction is equal to
(a) 00
(b) 150
(c) 300
(d) 50
Answer. C

Question. In a Young’s double slit experiment, two slits of equal width are arranged symmetrically w.r.to the sources. The intensity of the central bright fringe is I. If one of the slits is closed, the intensity at the Centre of the screen will be
(a) I
(b) I/4
(c) I/2
(d) 2I
Answer. B

Question. How does the diffraction band of blue light look in comparison with the red light?
(a) No changes
(b) Diffraction pattern becomes narrower
(c) Diffraction pattern becomes broader
(d) Diffraction pattern disappears 
Answer. B

Question. Two coherent sources of light can be obtained from
(a) Two different lamps
(b) Two different lamps and of different colours
(c) Two different lamps of the same colour
(d) None of these
Answer. D

Question. Which of the following phenomenon is not explained by Huygen’s wave theory?
(a) Diffraction
(b) interference
(c) Photoelectric effect
(d) None of these
Answer. C

ASSERTION-REASON TYPE QUESTIONS

For the following questions , two statements are given- one labelled as Assertion (A) and the other labelled as Reason (R) . Select the correct answer to these questions from the codes (a) , (b), (c) and (d) as given below:
(A) Both A and R are true and R is the correct explanation of (a)
(B) Both A and R are true but R is not the correct explanation of (a)
(C) A is true but R is false
(D) A is false and R is also false

Question. Assertion (A) All bright interference bands have the same intensity
Reason (R) Because all bands do not receive same light from two sources.
(a) A
(b) B
(c) C
(d) D
Answer. C

Question. Assertion (A) No diffraction is produced in sound waves near a very small opening.
Reason (R) For diffraction to takes place, the aperture of opening should be of same order as the wavelength of the waves.
(a) A
(b) B
(c) C
(d) D
Answer. A

Question. Assertion (A) Crystalline solids can cause X- rays to diffract.
Reason (R) Interatomic distance in crystalline solids is of the order of 0.1 nm
(a) A
(b) B
(c) C
(d) D
Answer. A

Question. Assertion (A) In YDE, the fringe width for dark fringes is different from that for the white fringes
Reason (R) In YDE, when fringes are formed with a source of white light, then only dark and bright fringes are observed
(a) A
(b) B
(c) C
(d) D
Answer. D

Question. Assertion (A) The phase difference between any two points on a wavefront is zero.
Reason (R) corresponding to a beam of parallel rays of light, the wavefronts are planes parallel to one another.
(a) A
(b) B
(c) C
(d) D
Answer. B

Question. Assertion: According to Huygens’s principle, no backward wave-front is possible.
Reason: Amplitude of secondary wavelet is proportional to (1 + cos θ) where θ is the angle between the ray at the point of consideration and the direction of secondary wavelet.
(a) A
(b) B
(c) C
(d) D
Answer. A

Question. Assertion: No interference pattern is detected when two coherent sources are infinitely close to each other.
Reason: The fringe width is inversely proportional to the distance between the two sources.
(a) A
(b) B
(c) C
(d) D
Answer. A

Question. Assertion : It is necessary to have two waves of equal intensity to study interference pattern.
Reason : There will be an effect on clarity if the waves are of unequal intensity.
(a) A
(b) B
(c) C
(d) D
Answer. D

Question. Assertion: When tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of the shadow of the obstacle.
Reason: Destructive interference occurs at the centre of the shadow.
(a) A
(b) B
(c) C
(d) D
Answer. C

Question. Assertion: When tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of the shadow of the obstacle.
Reason: Destructive interference occurs at the centre of the shadow.
(a) A
(b) B
(c) C
(d) D
Answer. C

CASE BASED QUESTIONS

1.Young’s Double slit experiment

Question. What is the path difference between the two light waves coming from coherent sources, which produces 3rd maxim(a)
a) λ
b) 2λ
c) 3λ
d 0
Answer. C

Question. What is the correct expression for fringe width(β).
(a) λd/D
(b) λ/dD
(c) d/λD
(d) λD/d
Answer. D

Question. what is the phase difference between two interfering waves producing 1st dark fringe.
a) π
b) 2π
c) 3π
d) 4π
Answer. A

Question. The ratio of the widths of two slits in Young’s double slit experiment is 4: 1. Evaluate the ratio of intensities at maxima and minima in the interference pattern.
a) 1:1
b) 1:4
c) 3:1
d) 9:1
Answer. D

Question. In a Young’s double slit experiment, the separation between the slits is 0.1 mm, the wavelength of light used is 600 nm and the interference pattern is observed on a screen 1m away. Find the separation between bright fringes.
(a) 6.6 mm
(b) 6.0 mm
(c) 6 m
(d) 60cm
Answer. B

2.Diffraction
In the single slit diffraction experiment, we can observe the bending of light or diffraction that causes light from a source interfere with itself and produce a distinctive pattern called the diffraction pattern.Diffraction is evident when the sources are small enough that they are relatively the size of the wavelength of light.

Question. To observe diffraction, the size of the obstacle
a) Should be λ/2, where λ is the wavelength of light
b) Should be of the order of wave length of light
c) Has no relation to wavelength of light
d) Should be much larger than the wavelength
Answer. B

Question. When violet light of wavelength 415nm falls on a single slit, it creates a central diffraction peak that is 9.2cm wide on a screen that is 2.55m away. How wide is the slit
a) 3x10⁸m
b) 3.8x10^-7m
c) 2.3x10^-5m
d) 4.2x10^-7m
Answer. C

Question. A single slit 1mm wide is illuminated by 450nm light. What is the width of the central maximum in the diffraction pattern on a screen 5cm away?
a) 300x10^-7m
b) 800x10^-7cm
c) 450x10^-7m
d) 500 x10^-7m
Answer. C

Question. What happen with the diffraction pattern if the whole apparatus is immersed in water
a) Width of central maximum increases
b) Width of central maximum decreases
c) Wavelength of light increases
d) Frequency of light decreases
Answer. B

3. Interference
Interference is based on the superposition principle. According to this principle, at a particular point in the medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves. If two sodium lamps illuminate two pinholes S1 and S2. the intensities will add up and no interference fringes will be observed on the screen. Here the source undergoes abrupt phase change in times of the order of 10-10 seconds

Question. In Young’s double slit experiment, the central bright fringe can be identified
(a) By using white light instead of monochromatic light
(b) As it narrower than other bright fringes
(c) As it wider than other bright fringes
(d) As it has greater intensity than the other bright fringes
Answer. A

Question. In a Young's double-slit experiment, the slit separation is double(d) To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to
(a) 2D
(b) 4D
(c) D/2
(d) D/4
Answer. A

Question. In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain 5th bright fringe at the same point
(a) 500nm
(b) 420nm
(c) 750nm
(d) 630nm
Answer. B

Question. The resultant amplitude of a vibrating particle by the superposition of the two waves y₁=a sin[ωt+π/3] and y₂=a sinωt is
(a) a
(b) 2√2 a
(c) 2a
(d) √3 (a)
Answer. D

VERY SHORT ANSWER TYPE QUESTIONS

Question. How does the angular separation of interference fringes change in young’s experiment, if the distance between the slits is increased?
Answer. Angular separation θ= λ/𝑑 𝑠𝑜 θ= 1/d

Question. State the reason, why two independent sources of light cannot be considered as coherent sources.
Answer. Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms, when they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

Question. How does the fringe width of interference fringes change, when the whole apparatus of Young’s experiment is kept in a liquid of refractive index 1.3?
Answer. β(water)=β(air)/n=β(air)/1.3

Question. A interference pattern is obtained using a red light, what will be effect on interference fringes, if the red light is replaced by blue light?
Answer. Fringe width β is directly proportional to wavelength. The wave length of blue light is less than that of red light, hence if red light is replaced by the blue light, the fringe width decreases

Question. A plane wavefront is incident on a convex lens. What will be the shape of the wave front emerging from the lens?
Answer.  Emerging wavefront will be spherical.

Question. The ratio of intensities of two waves is 1: 25, what is the ratio of their amplitudes?
Answer.  1: 5

Question. How does the fringe width, in young’s double-slit experiment, change when the distanceof separation between the slits and screen is doubled?
Answer. Doubled, Fringe width α D

SHORT ANSWER TYPE QUESTIONS 

Question. Define secondary wavelets and how can we construct new wavefront with them?
Answer. According to the Huygens’ principle, every particle of the medium, situated on the wavefront, acts as a new source of light wave from which new similar waves originate. These waves are called secondary wavelets. 1 The envelop of the secondary wavelets in the forward direction at any instant gives the new wavefront at that instant

Question. In a single slit diffraction experiment, width of the slit is increase(d) How will the (a) size and (b) intensity
of central bright band be affected? Justify your answer
Answer. The width of the central bright band = 2D × λ d where d = width of the slit. (i) As the width of the slit is doubled, the size of the central diffraction band will be half. 1 (ii) Intensity of central bright band is proportional to d2. So, the intensity will get quadruple(d)

Question. (i)State one feature by which the phenomenon of interference can be distinguished from that of diffraction. (ii)A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance between the slits and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the screen is. 15 mm, calculate the width of the slit.
Answer. (i) In interference all the maxima are of equal intensity. In diffraction pattern central fringe is of maximum intensity while intensity of secondary maxima falls rapidly. (ii) Position of nth maximum,
y=(2n+1) λD/2d,
here n=2 y=5λD/2d Substituting and solving, we get, d=80μm

Question. State two conditions required for obtaining coherent sources In Young’s arrangement to produce interference pattern, show that dark and bright fringes appearing on the screen are equally space(d)
Answer. Two conditions for obtaining coherent sources: (i) Two sources should give monochromatic light. (ii)Coherent sources of light should be obtained from a single source by some device.
The fringe width (dark and bright) is given by Hence, it is same for both dark and bright fringes, so they are equally spaced on the screen.

Question. (i) In what way is diffraction from each slit related to the interference pattern in a double slit experiment? (ii) When a tiny circular obstacle is placed in the path of light from a distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain, why.
Answer. (i) Diffraction from each slit is related to interference pattern in a double slit experiment in the following ways: The intensity of minima for diffraction is never zero, while for interference it is generally zero. All bright fringes for diffraction are not of uniform intensity, while for interference, these are of uniform intensity
(ii) Waves from the distant source are diffracted by the edge of the circular obstacle and these diffracted waves interfere constructively at the centre of the obstacle’s shadow producing a bright spot

Question. Write the distinguishing features between a diffraction pattern due to a single slit and the interference fringes produced in Young’s double slit experiment?
Answer.   Interference                                                                            Diffraction
1. Interference is due to superposition                                      Diffraction is due to superposition of the secondary 
of two distinct waves coming from                                            wavelets coming from different
two coherent sources.                                                             parts of the same wavefront.
2. Interference fringes may or may not be of the same width.    Diffraction fringes are not to be of the same width
3. The intensity of minima is generally zero.                              The intensity of minima is never zero.
4. All bright fringes are of uniform intensity.                              All bright fringes are of not uniform intensity.

03 Marks questions

Question. Two monochromatic waves emanating from two coherent sources have the displacements represented by
y₁ = a cos ωt and y₂ = a cos (ωt + ϕ ), where ϕ is the phase difference between the two displacements.
(a)Show that the resultant intensity at a point due to their superposition is given by I = 4I₀ cos2 ϕ/2, where I₀ = a2 . (b) Hence obtain the conditions for constructive and destructive interference.
Answer. (a) Let the two waves be represented by equations y₁=acosωt,y₂=a cos(ωt+φ) From the Principle of superposition of waves we get, y=y₁+y₂=2acos(φ/2) cos (ωt+φ/2) Resultant Amplitude is A =2acos(φ/2) Hence, Resultant Intensity α (amplitude)2 =4a² cos² (φ/2)
(a) For constructive interference: (ii)For destructive interference: Cos (Ф/2) = 0 or Ф/2 = (2n-1)π/2 Ф = (2n-1)π

Question. In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Write the expression for fringe width
Answer. Let the Resultant wave Amplitude be ‘A’ A²=a₁ 2+a₂²+2a₁a₂cosφ
For constructive interference: Cosφ=+1, so, Phase difference is φ=2nπ and path difference is x=nλ For
destructive interference,Cosφ=-1,So φ=(2n-1)π,n=1,2,3….
Hence path difference is x= (2n-1)λ/2 Fringe width β= λD/d

Question.The intensity pattern at the central maxima in Young’s double slit experiment is I0. Find out the intensity at a point where the path difference is λ/4 &λ/3.
Answer. If the path difference =λ/4 Here for the given, first case, the path difference is given as
Δx=λ/4 , Δx=2π/λ× ϕ ⇒ϕ=π/2 Now substituting the value of ϕ=π/2 in the equation I=4I₀cos²ϕ/2
⇒I=4I₀cos²π/4 ⇒I=2I₀ Therefore the intensity at a point on a screen in Young’s double-slit experiment where interfering waves of equal intensity have a path difference of λ/4 is given as 2I₀.
Ii). For λ/3 ⇒ϕ=2π/3 ⇒I=4I₀cos² (2π/6 )⇒ ∴I=I0 Therefore the intensity at a point on a screen in Young’s double-slit experiment where interfering waves of equal intensity have a path difference of λ/3 is given as I0.

Question. A beam of light consisting of two wavelength 800 nm and 600 nm is used to obtain the interference fringes in young’s double slit experiment on a screen held 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum, where the bright fringes of the two wavelengths coincide.
Answer. The two bright fringes will coincide when n₁λ₁=n₂λ₂
⇒ n₁×800×10⁻⁹= n₂×600×10⁻⁹
⇒4n₁=3n₂ n₁≠=0,n₂≠ 0
For y to be minimum and since n are integers,
n₁=3,n₂=4
y=n₁λ₁D/d= 3x800x10^-9 x1.4/0.28x10^-3 ⇒y=1.2×10⁻²m ⇒y=1.2 cm

Question. In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Answer. Distance between the slits, d = 0.28 mm = 0.28 × 10 ^-3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe = 1.2 cm = 1.2 × 10 ^-2 m
For constructive interference, the distance between the two fringes is given by relation: x = nλ D/d where, n = Order of fringes wavelength of the light can be given as:
λ = xd/nD = 1.2x10^-2x0.28x10^-3 /4x1.4 = 6x10^-7 m

Question. In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3. Answer Distance of the screen from the slits, D = 1 m
Answer. Wavelength of light used, λ1 =600 nm Angular width of the fringe in air, θ1 = 0.2°
Angular width of the fringe in water = θ, Refractive index of water, μ = 4/3
Refractive index is related to angular width as: μ = θ1/ θ2
θ2 = 3/4 θ1 = 3/4 x 0.2 = 0.15°.

Question. A slit of width “a” is illuminated by light of wavelength, 650nm.What will be the value of the width ‘a’ when
a) first minimum falls at an angle of diffraction 300?
b) first maximum falls at an angle of diffraction 300?
(c) Why does the intensity of the secondary maximum become less as compared to the central maximum?
Answer.
(i) first minimum at 300 satisfies the condition, d sinθ=λ, d= λ/sin 30 =1300nm
(ii) first maxima at 300 satisfies the condition d sinθ =3λ/2 d= 3λ/2sin 30 =1950nm
(b) As the order increases only 1/nth (where n is an odd number) of the slit, will contribute in producing brightness at a point in diffraction. So the higher order maxima are not so bright as the central maximum

Question. (a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light
Answer. (a) After the introduction of the glass sheet (say, on the second slit),
we have = 50% = 1/ 2
Ratio of the amplitudes = a₁/ a₂ = √1/2
Hence
I max/ I min = (a₁+a₂)²/(a₁-a₂)²
(1+1/√2)² / (1-1/√2)²= (√2+1)²/ (√2-1)² =34.55
b) central fringe remains white. No clear fringe pattern is seen after a few (coloured) fringes on either side of the central fringe

Question. (a) Why cannot the phenomenon of interference be observed by illuminating two pin holes with two sodium lamps?
(b) Two monochromatic waves having displacements y1 = a cos ωt and y2 = a cos (ωt + θ) from two coherent sources interfere to produce an interference pattern. Derive the expression for the resultant intensity and obtain the conditions for constructive and destructive interference.
Answer. (a) The interference phenomenon cannot be observed by using two illuminating pin holes with two sodium lamps because the light emitted from sodium lamps undergoes abrupt phase changes in 1s which will not have any fixed phase relationship as they are incoherent.
(b) The displacement equations for two monochromatic waves are given as
y₁ = acos wt and
y₂ = acos (wt + θ)
Now, net displacement y = y₁ + y₂ = acos wt + acos (wt + θ)
= acos wt + acos wt cos θ – asin wt. sinθ
= acos wt (1 + cosθ) – asin wt. sin θ
put a(1 + cos θ) = Asin φ...(i)
and – asin wt = Acos φ ...(ii)
y = Asin φ. cos wt + Acos φ sin wt
y = Asin (wt + φ)
Now, from equation (i) and (ii), A² (sin² φ+ cos² φ) = [a(1 + cos wt)]² + (– asin wt) ²
A² = a² [1 + cos2 wt + cos² wt+ sin² wt]
A² = 2a² [1 + cos 2wt] A² = 4a² cos² wt/2 This is the required expression.
For constructive interference, I should be maximum I max = 4a² , if wt= 0, ± 2π, ± 4 π...
For destructive interference, I should be minimum I min = 0, if wt = ± π, ± 3 π, ± 5 π....