Wave Optics: Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. Sketch the refracted wavefront emerging from the convex lens, If a plane wavefront is an incident normally on it.

Ans: The figure is as shown.

Ques. How would you explain the propagation of light on the basis of Huygen’s wave theory?

Ans: To explain the propagation of light we have to draw a wavefront at a later instant when a wavefront at an earlier instant is known. This can be drawn by the use of Huygen’s principle.

Ques. Draw the shape of the reflected wavefront when a plane wavefront is an incident on a concave mirror.

Ans: The reflected wavefront is as shown.

Ques. Name two phenomena that establish the wave nature of light.

Ans: Interference and diffraction of light.

Ques. State the conditions which must be satisfied for two light sources to be coherent.

Ans:(a) Two sources must emit light of the same wavelength (or frequency).

(b) The two light sources must be either in-phase or have a constant phase difference.

Ques. Draw an intensity distribution graph for diffraction due to a single-slit.

Ans: The intensity distribution for a single-slit diffraction pattern is as shown.

Ques. Name one device for producing plane polarised light. Draw the graph showing the variation of intensity of polarised light transmitted by an analyser.

Ans: Nicol prism can be used to produce plane polarised light. The graph is as shown.

Ques. State Huygens’ principle of diffraction of light. (CBSE AI 2011C)

Ans: Huygens principle states that

1. Each point on a wavefront is a source of secondary waves which travel out with the same velocity as the original waves.
2. The new wavefront is given by the forward locus of the secondary wavelets.

Ques. In what way is a plane polarised tight different from an unpolarised light? (CBSE AI 2012C)

Ans: Plane polarized light vibrates 1n only one plane.

Ques. Which of the following waves can be polarised: (i) Heatwaves (ii) Sound waves? Give a reason to support your answer. (CBSE Delhi 2013)

Ans: Heatwaves are transverse In nature.

Ques. Define the term ‘wavefront’. (CBSE AI 2014C)

Ans: It Is defined as the locus of all points In a medium vibrating in the same phase.

Ques. Define the term ‘coherent sources’ which are required to produce interference patterns in Young’s double-slit experiment. (CBSE Delhi 2014C)

Ans: Two sources that are In phase or have a constant phase difference are called coherent sources.

Ques. What change would you expect if the whole of Young’s double-slit apparatus were dipped into the water?

Ans: The wavelength λ, of light In water, is less than that in air. Since the fringe width β is directly proportional to the wavelength of light, therefore, the fringe width will decrease.

---

Short Answer Questions [2 Marks Questions]

Ques. How can one distinguish between an unpolarised and linearly polarised light beam using a polaroid? (CBSE Delhi 2019)

Ans: The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

Ques. What is meant by plane polarised light? What type of waves shows the property of polarisation? Describe a method of producing a beam of plane polarised light?

Ans: 1. The light that has its vibrations restricted to only one plane is called plane polarised light. 2. Transverse waves show the phenomenon of polarization. Light is allowed to pass through a polaroid. 3. The polaroid absorbs those vibrations which are not parallel to its axis and allows only those vibrations to pass which are parallel to its axis.

Ques. Write the Important characteristic features by which the Interference can be distinguished from the observed diffraction pattern. (CBSE AI 2015)

Ans: (a) In the interference pattern the bright fringes are of the same width, whereas in the diffraction pattern they are not of the same width.

(b) In interference all bright fringes are equally bright while in diffraction they are not equally bright.

Ques. State Brewster’s law. The value of Brewster’s angle for the transparent medium is different for the light of different colours. Give reason. (CBSE Delhi 2016)

Answer: When the reflected ray and the refracted ray are perpendicular then μ = tan ip where ip is the polarising angle or Brewster angle.

Brewster’s angle depends upon the refractive index of the two media in contact. The refractive index in turn depends upon the wavelength of light used (different colours) hence Brewster’s angle is different for different colours.

Ques. Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.

Ans: Let lo be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be l = lo cos 2θ, where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P3 will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

l = lo cos² θ cos² (90° – θ) = lo cos² θ sin² θ = (lo / 4) sin² 2θ

Therefore, the transmitted intensity will be maximum when θ = π/4

Ques. One of the slits of Young’s double-slit experiment is covered with a semi-transparent paper so that it transmits lesser light. What will be the effect on the interference pattern?

Ans. There will be an interference pattern whose fringe width is the same as that of the original. But there will be a decrease in the contrast between the maxima and the minima, i.e. the maxima will become less bright and the minima will become brighter.

Ques. A Polaroid (I) is placed In front of a monochromatic source. Another Polaroid (II) is placed in front of this Polaroid (I) and rotated till no light passes. A third Polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II)? Explain. (NCERT Exemplar)

Ans: Only in the special case when the pass axis of (III) is parallel to fill or (II) there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Also Read:

---

Long Answer Questions [3 Marks Questions]

Ques. (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. (All India 2008)

Ans:

(i) In interference all the maxima are of equal intensity.

In a diffraction pattern, the central fringe is of maximum intensity while the intensity of secondary maxima falls rapidly.

(ii) Given, λ = 600 nm = 6 × 10^-7 m,

D = 0.8 m, γ₂ = 15 × 10^-3 m

To calculate: Width of the slit ‘d’

Calculations: γ_{2 = \(\frac{5}{2} \times \frac{\lambda D}{d}\)}

\(\Rightarrow d=\frac{5}{2} \times \frac{6 \times 10^{-7} \times 0.8}{15 \times 10^{-3}}\)

∴ Distance, d = 8 × 10^-5 = 80 μm

Ques. Define the term ‘linearly polarised light. When does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids? (All India 2008)

Ans:

Linearly polarised light is one in which the vibration of light is present in one line only.

Law of Malus: I = I₀ cos²θ, at θ = 0, cos θ = 1,

Intensity is maximum.

Ques. (i) State the principle on which the working of an optical fibre is based.

(ii) What are the necessary conditions for this phenomenon to occur? (All India 2009)

Ans:

(i) Working of an optical fibre is based on the principle of total internal reflection.

(ii) (a) Light should travel from a denser to a rarer medium.

(b) Angle of incidence should be more than critical angle given by i_c = sin^-1 \(\left( \frac{1}{\mu} \right)\)

Ques. (a) Why are coherent sources necessary to produce a sustained interference pattern?

(b) In Young’s double-slit experiment using monochromatic light of wavelength X, the intensity of light at a point on the screen where the path difference is X, is K units. Find out the intensity of light at a point where the path difference is 2λ/3. (Delhi 2012)

Ans:

(a) Coherent sources have a constant phase difference and, therefore, produce a sustained interference pattern.

These sources are needed to ensure that the position of maxima and minima do not change with time.

Ques. Write the distinguishing features between a diffraction pattern due to a single slit and the interference fringes produced in Young’s double-slit experiment?

Ans: Difference between interference and diffraction of light

Ques. Answer the following questions :

(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. (Comptt. All India 2013)

Ans:

(i) Diffraction from each slit is related to an 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 produc¬ing a bright spot.

Ques. (a) Write the conditions under which light sources can be said to be coherent.

(b) Why is it necessary to have coherent sources in order to produce an interference pattern? (All India 2013)

Ans:

(a) Coherent sources of light. The sources of light, which emit continuously light waves of the same wavelength, same frequency and in the same phase are called Coherent sources of light.

The interference pattern is not obtained. This is because the phase difference between the light waves emitted from two different sodium lamps will change continuously.

(b) Conditions for interference. The important conditions for obtaining interference of light are:

• The two sources of light must be coherent. i.e. they should exist continuous waves of the same wavelength or frequency.
• The two sources should be monochromatic.
• The phase difference of waves from two sources should be constant.
• The amplitude of waves from two sources should be equal.
• The coherent sources must be very close to each other.

---

Very Long Answer Questions [5 Marks Questions]

Ques. Define the term wavefront. Using Huygen’s wave theory, verify the law of reflection. Or, Define the term, “refractive index” of a medium. Verify Snell’s law of refraction when a plane wavefront is propagating from a denser to a rarer medium. (CBSE Delhi 2019)

Ans: The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM–. Let us consider the situation when one end A of was front strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),

∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.

Or

The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

sin i = \(\frac{BC}{AC} = \frac{v_1 \tau}{AC}\)

And sin r = \(\frac{AE}{AC} = \frac{v_2 \tau}{AC}\)

Therefore, \(\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = n_{12}\)

Ques. Find an expression for the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted Intensity be maximum? (CBSE Delhi 2015)

Ans: Let l0 be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be

l = l₀ cos² θ

where θ is the angle between the pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

I = I₀ cos² θ cos²\(\left( \frac{\pi}{2} - \theta \right)\) = I₀ cos² θ sin² θ

= \(\frac{I_0}{4} \sin^2 \theta\)

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Ques. Distinguish between unpolarised and linearly polarised light. Describe with the help of a diagram how unpolarised light gets linearly polarised by scattering. (CBSE Delhi 2015)

Ans: The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in a perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

Ques. Answer the following questions:

(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?

Ans: The size reduces by half according to the relation: size = λ/d. Intensity increases fourfold.

(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?

Ans: The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?

Ans: Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot.

(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily? (NCERT)

Ans: For diffraction or bending of waves by obstacles/apertures by a large angle, the size of the latter should be comparable to the wavelength. If the size of the obstacle/aperture is much too large compared to the wavelength, diffraction is by a small angle. Here the size is of the order of a few metres. The wavelength of light is about 5 × 10^-7 m, while sound waves of, say, 1 kHz frequency have a wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot.

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates

---

Check-Out: