NCERT Exemplar Wave Optics - Physics Class 12 - NEET PDF Download

 MULTIPLE CHOICE QUESTIONS 

Q.1. Consider a light beam incident from air to a glass slab at Brewster's angle as shown in Fig.

A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.
(a) For a particular orientation there shall be darkness as observed through the polaoid.
(b) The intensity of light as seen through the polaroid shall be independent of the rotation.
(c) The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the polaroid.
(d) The intensity of light as seen through the polaroid shall go through a minimum for four orientations of the polaroid.

Ans: (c)

Explanation: When unpolarised light is incident at Brewster's angle, the reflected beam is completely plane polarised with its vibration plane perpendicular to the plane of incidence. The transmitted (emergent) beam is not completely polarised - it remains partially polarised because only the reflected component becomes fully polarised. On placing and rotating a polaroid in the path of the emergent ray, the transmitted intensity varies but cannot fall to zero since a portion of the emergent beam remains unpolarised and passes through for every orientation. The intensity therefore shows minima (when the analyser's axis is nearly perpendicular to the dominant polarisation direction) but not complete extinction. These minima occur twice in a full 360° rotation (separated by 180°), so option (c) is correct. Options (a) and (b) are incorrect since complete darkness never occurs and intensity does depend on rotation; option (d) is incorrect because there are two, not four, minima per rotation.

Q.2. Consider sunlight incident on a slit of width 104 Å. The image seen through the slit shall
(a) Be a fine sharp slit white in colour at the center.
(b) A bright slit white at the center diffusing to zero intensities at the edges.

(c) A bright slit white at the center diffusing to regions of different colours.
(d) Only be a diffused slit white in colour.

Ans: (a)

Explanation: The slit width is 10⁴ Å = 10 000 Å. Visible wavelengths lie roughly between 4 000 Å and 8 000 Å, so the slit width is of the same order as visible wavelengths. Diffraction will therefore occur, producing a central maximum in which the contributions of all visible wavelengths overlap. Since all colours overlap at the central maximum, it appears white and relatively sharp compared with the diffracted coloured regions away from centre. Thus option (a) is the best description.

Q.3. Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

Ans: (a)

Explanation: The two reflected rays differ in two ways: (i) one of the reflections (air → glass) produces a phase shift of π, while the other (glass → air) produces no phase shift, and (ii) the ray that penetrates and reflects from the bottom surface travels an extra optical path inside the slab. The extra optical path is 2 n d cos r, where r is the angle of refraction inside the slab. The corresponding phase difference due to path is (2π/λ)·2 n d cos r = (4π n d cos r)/λ. Including the π phase inversion on one reflection, the net phase difference between the two reflected rays is

(4π n d cos r)/λ + π.

This is the expression used to determine constructive or destructive interference between the two reflected waves.

Q.4. In a Young's double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter.

In this case
(a) There shall be alternate interference patterns of red and blue.
(b) There shall be an interference pattern for red distinct from that for blue.
(c) There shall be no interference fringes.
(d) There shall be an interference pattern for red mixing with one for blue.

Ans: (c)

Explanation: For a stable interference pattern the two sources must be coherent - that is, they must maintain a constant phase relationship and have the same frequency (wavelength). Covering one slit with a red filter and the other with a blue filter produces light of different wavelengths and frequencies from the two slits, so the waves are not mutually coherent. Therefore no sustained interference fringes are produced and option (c) is correct.

Q.5. Figure shows a standard two slit arrangement with slits S₁, S₂. P₁, P₂ are the two minima points on either side of P (Fig).

At P₂ on the screen, there is a hole and behind P₂ is a second 2- slit arrangement with slits S₃, S₄ and a second screen behind them.
(a) There would be no interference pattern on the second screen but it would be lighted.
(b) The second screen would be totally dark.

(c) There would be a single bright point on the second screen.
(d) There would be a regular two slit pattern on the second screen.

Ans: (d)

Explanation: The point P₂ corresponds to a minimum on the first screen because the two waves arriving there from S₁ and S₂ are equal in amplitude but opposite in phase. Nevertheless, by Huygens' principle P₂ acts as a secondary source of light. The light emerging through the hole at P₂ illuminates slits S₃ and S₄; since these slits are illuminated by parts of the same wavefront, they act as coherent sources and will produce a regular two-slit interference pattern on the second screen. Thus option (d) is correct.

Q.6. Two source S₁ and S₂ of intensity I₁ and I₂ are placed in front of a screen [Fig.(a)]. The patteren of intensity distribution seen in the central portion is given by Fig.(b).

 In this case which of the following statements are true.
(a) S₁ and S₂ have the same intensities.
(b) S₁ and S₂ have a constant phase difference.
(c) S₁ and S₂ have the same phase.
(d) S₁ and S₂ have the same wavelength.

Ans: (a), (b) and (d)

Explanation: The intensity pattern shows a stable system of equally spaced interference fringes with a dark fringe at the center $(x=0)(x=0)$. A complete central minimum implies that the two waves cancel each other perfectly, which is possible only if the two sources have equal amplitudes, hence equal intensities. The existence of a steady fringe pattern also means that the sources must maintain a constant phase difference (they are coherent). Since the central fringe is dark and not bright, the sources cannot be in the same phase; instead, they must differ in phase by $\pi \backslash p$ (or an odd multiple of it). Moreover, clear and stationary interference fringes can occur only when both sources have the same wavelength (same frequency). Therefore, statements (a), (b), and (d) are correct, while (c) is incorrect.

Q.7. Consider sunlight incident on a pinhole of width 10³A. The image of the pinhole seen on a screen shall be
(a) A sharp white ring.
(b) Different from a geometrical image.
(c) A diffused central spot, white in colour.
(d) Diffused coloured region around a sharp central white spot

Ans: (b, d)

Explanation: The pinhole width is 10³ Å = 1 000 Å, which is much smaller than visible wavelengths (4 000-8 000 Å). Such a small aperture produces strong diffraction; the observed image therefore differs from a simple geometrical projection (so (b) is correct). The diffraction pattern has a central bright region in which many wavelengths overlap (appearing largely white) surrounded by diffracted rings in which different wavelengths disperse, producing coloured regions. Thus option (d) is also correct.

Q.8. Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
(a) The size decreases.
(b) The intensity increases.
(c) The size increases.
(d) The intensity decreases.

Ans: (a, b)

Explanation: The angular width of the central diffraction maximum from a circular aperture is proportional to λ/d (where d is the hole diameter). Therefore increasing the hole size d reduces the angular width of the central maximum - the size decreases, so (a) is correct. Increasing the hole size also allows more light to pass, so the intensity of the diffraction pattern increases, making (b) correct. Options (c) and (d) are therefore incorrect.

Q.9. For light diverging from a point source
(a) The wavefront is spherical.
(b) The intensity decreases in proportion to the distance squared.
(c) The wavefront is parabolic.
(d) The intensity at the wavefront does not depend on the distance.

Ans: (a, b)

Explanation: A point source emits in all directions, so at any instant the disturbance lies on a spherical surface centred on the source; the wavefront is spherical, so (a) is correct. The energy spreads over the surface of a sphere whose area grows as 4πr², so the intensity (power per unit area) falls off as 1/r², giving (b). Options (c) and (d) are incorrect.

VERY SHORT ANSWER TYPE QUESTIONS

Q.10. Is Huygen's principle valid for longitudunal sound waves?
Ans: Yes. Huygens' principle applies to any wave phenomenon. For longitudinal sound waves the wavefront consists of surfaces of constant phase formed by compressions and rarefactions. Each small element of such a wavefront can be regarded as a source of secondary spherical wavelets; their envelope gives the new wavefront. Thus Huygens' construction is valid for longitudinal sound waves as well.

Q.11. Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?

Ans: A point at the focus of the first converging lens emits rays which the first lens makes parallel. These parallel rays entering the second converging lens are brought to its focus, which therefore behaves like a point source. The wavefronts emerging from the final image are spherical, centred on the image point formed by the second lens.

Q.12. What is the shape of the wavefront on earth for sunlight?

Ans: The Sun is effectively at infinity compared with Earth, so rays from the Sun arriving at Earth are nearly parallel. Therefore the wavefronts reaching Earth are effectively plane wavefronts.

Q.13. Why is the diffraction of sound waves more evident in daily experience than that of light wave?

Ans: Diffraction is significant when the size of an obstacle or aperture is comparable with the wavelength. Sound wavelengths in everyday situations range roughly from 15 m down to about 15 mm, so common openings and obstacles are often comparable in size and diffraction is easily noticed. Visible light wavelengths are about 0.4-0.7 μm, far smaller than typical macroscopic objects and apertures, so diffraction of light is usually negligible in daily life and is not easily observed without special small-scale apertures.

Q.14. The human eye has an approximate angular resolution of φ = × 5.8 10^-4 rad and a typical photo printer prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.

Ans: The printer resolution 300 dpi means the distance between centres of two adjacent dots is s = (2.54 cm)/300 = 0.0084667 cm = 8.4667×10^-5 m.

The eye cannot resolve two points separated by an angle smaller than φ. For small angles, φ ≈ s/z, so z ≈ s/φ.

Substitute s = 8.4667×10^-5 m and φ = 5.8×10^-4 rad:

z ≈ (8.4667×10^-5)/ (5.8×10^-4) ≈ 0.146 m ≈ 14.6 cm.

Thus, if the page is held at about 15 cm or farther, individual dots at 300 dpi cannot be resolved by a typical eye, which is less than the usual near-point distance of about 25 cm; so the dots are not seen individually at comfortable reading distances.

Q.15. A polariod (I) is placed in front of a monochromatic source. Another polatiod (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.

Ans: Yes, in general some light will emerge from (II) when a third polaroid (III) is inserted between crossed polaroids (I) and (II).

Explanation: If the transmission axes of I and II are at 90° (crossed), no light is transmitted. Inserting a third polaroid III with its axis at an intermediate angle θ to the axis of I rotates the plane of polarisation of the transmitted light. According to Malus' law, the intensity transmitted through successive polaroids is proportional to cos² of the angle between axes. If the axis of III is neither parallel to I nor to II, some component of the light after III is aligned with II and therefore a nonzero intensity emerges from II. Only when the axis of III is exactly parallel to I or exactly parallel to II will the final transmitted intensity again be zero.

SHORT ANSWER TYPE QUESTIONS

Q.16. Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?
Ans: Yes. Polarisation by reflection occurs whenever the angle of incidence (measured from the normal on the incident side) equals the Brewster angle for that pair of media. Brewster's law in general gives tan i_B = n₂/n₁ (where n₁ is the refractive index of the incident medium and n₂ that of the second medium). If light comes from the denser medium, the same condition applies: when the incidence angle equals the appropriate Brewster angle measured in that medium, the reflected ray is plane polarised. Thus reflection from the higher-index side can produce plane polarised light provided the incidence is at the Brewster angle for the interface.

Important point: Brewster's angle is the special incidence angle at which the reflected light is perfectly plane polarised; it is named after Sir David Brewster.

Q.17. For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Å and electrons accelerated through 100V used as the illuminating substance.

Ans: Resolution limit is proportional to the wavelength of the probe. For light, λ_light = 5 000 Å = 5 000×10^-10 m = 5.0×10^-7 m.

The de Broglie wavelength of electrons accelerated through V volts (non-relativistic) is approximately λ_e (in Å) ≈ 12.27/√V. For V = 100 V, λ_e ≈ 12.27/10 = 1.227 Å = 1.227×10^-10 m.

Ratio of least separations (light : electrons) ≈ λ_light/λ_e = (5 000 Å)/(1.227 Å) ≈ 4.07×10³.

Hence the minimum resolvable separation with light of 5 000 Å is about 4 070 times larger than that with electrons accelerated through 100 V.

Q.18. Consider a two slit interference arrangements (Fig.) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.

Ans:

Explanation: Let the slit separation be d and the distance from the slits to the screen be L. The condition given is L = d/2. Let the first minima be at a transverse distance x = D from the centre O. For small angles the path difference between the two slits at the point P is ≈ d sin θ ≈ d x / L. For the first minima the path difference = λ/2. Therefore

d x / L = λ/2.

Substitute L = d/2:

d x / (d/2) = 2 x = λ/2 ⇒ x = λ/4.

Hence D = λ/4.

SHORT ANSWER TYPE QUESTIONS

Q.19. Figure shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I₀ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

Explanation: 

The amplitudes of the interfering components from both slits are equal. Since each has intensity I/2, their amplitudes are √(I/2) and √(I/2).For the principal maximum (constructive interference), the resultant amplitude is the sum of the two amplitudes. Therefore the intensity of the interfering part becomes( √(I/2) + √(I/2) )²

= ( 2 √(I/2) )²

= 2I.In addition to this, the perpendicular component coming from slit S₂ does not interfere and simply adds its intensity. This non-interfering intensity is I/2.Hence the total intensity at the principal maximum is2I + I/2 = 5I/2.Using I = I₀ / 4, we getI_max = 5I₀ / 8.

Now for first minima:

This is the intensity at first minima with polariser.

Q.20. A small transparent slab containing material of µ =1.5 is placed along AS₂ (Fig.). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.

AC = CO = D, S₁C = S₂C = d << D

Ans:

Explanation: When a thin slab of refractive index μ and thickness L is introduced in the path from one slit, it adds an extra optical path (μ - 1)L. The net path difference at an observation point is therefore the usual geometrical difference plus this extra term. If 2 d sin θ is the path difference without slab, the total path difference becomes Δx = 2 d sin θ + (μ - 1)L.

For the central principal maximum the path difference must be zero (or an integral multiple of λ). For the central maximum (θ = 0) the slab introduces an offset so the central maximum shifts to a small angle θ₀ satisfying 2 d sin θ₀ + (μ - 1)L = 0. For small θ₀, sin θ₀ ≈ θ₀ ≈ y/D (if D is the distance to screen).

For the first minima the required path difference (relative to the shifted central maximum) is ±λ/2, so 2 d sin θ + (μ - 1)L = ±λ/2. Solving these for θ (and then converting to transverse distances on the screen using small-angle approximations y ≈ D θ) gives the distances of the principal maximum and first minima from the original central position O. The algebra follows by expanding sin θ ≈ θ and substituting the geometry of the setup; this yields expressions for the shifted positions of principal maximum and the first minima on either side of it in terms of d, D, μ, L and λ.

In short, when we introduce a thin transparent plate in front of one of the slits in YDSE, the fringe pattern shifts towards the side where the slab is present. 

Q.21. Four identical monochromatic sources A,B,C,D as shown in the (Fig.) produce waves of the same wavelength λ and are coherent. Two receiver R₁ and R₂ are at great but equal distaces from B.

(i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when B is turned off?
(iii) Which of the two receivers picks up the larger signal when D is turned off?

(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?

Ans.

Explanation: (i) At R₁ the contributions from the four sources cancel out (two are out of phase with the other two) and the resultant is zero. At R₂ the phases add more constructively, producing a nonzero resultant. Hence R₂ picks up the larger signal.

(ii) If B is switched off both R₁ and R₂ receive identical nonzero contributions from the remaining sources; the resultant magnitudes at R₁ and R₂ become equal in this arrangement, so both pick up the same signal.

(iii) If D is switched off, the symmetry at R₂ is disturbed less than at R₁, and R₂ receives a larger resultant than R₁. Thus R₂ picks up the larger signal.

(iv) Comparison of signals at R₁ and R₂ can indicate which source has been turned off: a change at R₁ that leaves R₂ unchanged suggests B was switched off; a marked increase at R₂ relative to R₁ suggests D was switched off. Therefore, both receivers together can distinguish whether B or D has been turned off by observing the pattern of changes in received amplitude.

Q.22. The optical properties of a medium are governed by the relative permitivity (ε_r) and relative permeability (µ_r). The refractive index is defined as 

  For ordinary material ε_r > 0 and µ_r > 0 and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with ε_r < 0 and µ_r < 0. Since then such 'metamaterials' have been produced in the laboratories and their optical properties studied.
(i) According to the description above show that if rays of light enter such a medium from air (refractive index =1) at an angle θ in 2^nd quadrant, them the refracted beam is in the 3^rd quadrant.
(ii) Prove that Snell's law holds for such a medium.

Ans:

Explanation: (i) For a medium with ε_r < 0 and µ_r < 0 the refractive index n can be taken negative (n = -√(ε_rµ_r)). When an incident ray from air (n = +1) enters such a medium, the wavefront construction shows that the direction of the transmitted (refracted) ray is on the opposite side of the normal compared with ordinary materials. Geometrically this means an incident ray in the second quadrant is refracted into the third quadrant (the refracted ray is on the same side of the normal as the incident ray measured from the interface but with reversed direction of the wavevector component parallel to the surface). Thus negative refraction occurs and the refracted beam lies in the third quadrant for the geometry stated.

(ii) Snell's law follows from the equality of the tangential components of the wavevector (or equivalently from equal phase across the interface). Writing k₁ sin θ₁ = k₂ sin θ₂ and using k = (2π/λ) = (2π n/λ₀) gives n₁ sin θ₁ = n₂ sin θ₂. This form remains valid even if n₂ is negative; the algebraic sign in n₂ reverses the direction of the refracted wavevector but the relation between sines of angles and refractive indices - Snell's law - still holds. A direct wavefront construction using equal optical path/phase across the interface also leads to the same result, confirming Snell's law for metamaterials with negative refractive index.

Q.23. To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens).
A typically used dielectric film is MgF₂ (n = 1.38). What should the thickness of the film be so that at the center of the visible spectrum (5500 Å) there is maximum transmission.

Ans:

Explanation: For an anti-reflection coating on glass the usual design is a single layer of refractive index n_f ≈ √(n_air n_glass) to minimise reflection and a thickness such that the two reflected rays from the top and bottom of the film interfere destructively. For normal incidence the optical path difference between the two reflected rays is 2 n_f d and for destructive interference this should equal λ/2 (for the wavelength in vacuum λ). Thus the condition is

2 n_f d = λ/2 ⇒ d = λ/(4 n_f).

Substitute λ = 5 500 Å and n_f = 1.38:

d = (5 500 Å)/(4 × 1.38) ≈ (5 500)/(5.52) Å ≈ 996 Å ≈ 1.00×10³ Å.

Therefore a film thickness of about 1 000 Å (100 nm) gives maximum transmission at λ = 5 500 Å for MgF₂ coating at near normal incidence.