NEET Physics: Electromagnetic Waves and Optics · Chapter 3

Wave Optics

What to remember

Huygens' principle: every point on a wavefront acts as a source of secondary wavelets. The new wavefront is the forward envelope of these wavelets. It proves the laws of reflection and refraction.
Young's double-slit experiment: fringe width β = λD/d. Bright fringes have path difference nλ, dark fringes have (n + ½)λ. The intensity is I = 4I₀ cos²(φ/2).
Single-slit diffraction: minima occur at a sin θ = nλ (n = 1, 2, 3, …). The central maximum has width 2λD/a and is twice as wide as the other maxima.
Polarisation: it proves that light is a transverse wave. Malus' law: I = I₀ cos²θ. Brewster's law: tan iB = n.

Wavefront and Huygens' principle

A wavefront is the locus of all points that vibrate in the same phase at a given instant. A point source in a uniform medium gives spherical wavefronts. A very distant source (such as the Sun or a star) gives plane wavefronts over a small region. A line source gives cylindrical wavefronts. A ray is a line perpendicular to the wavefront, and it shows the direction of travel of the wave.

Huygens' principle. Each point of a wavefront acts as a source of secondary wavelets that spread out in all directions with the speed of the wave. The forward envelope of the wavelets after a time t is the new wavefront.

Reflection of a plane wave at a plane surface can be proved with this construction. The angle of incidence equals the angle of reflection: i = r.

Refraction of a plane wave. Let the speed be v₁ in medium 1 and v₂ in medium 2. The wavefront that enters medium 2 first moves more slowly (if v₂ is less than v₁), and the wavefront bends toward the normal. The construction gives

sin i / sin r = v₁/v₂ = n₂/n₁ (Snell's law)

During refraction the frequency stays constant. The speed and the wavelength change in the same ratio: λ₂/λ₁ = v₂/v₁. A wave of wavelength λ in vacuum has wavelength λ/n in a medium of refractive index n.

Rarer medium and TIR in brief. When a wave goes from a denser medium to a rarer medium (v₂ greater than v₁), it bends away from the normal and r is greater than i. For one angle of incidence, the critical angle ic, the angle of refraction is 90°, and sin ic = v₁/v₂ = n₂/n₁. For i greater than ic, the construction gives no refracted wavefront, and the light is totally reflected.

Quantity	On reflection	On refraction into a denser medium
Frequency	Unchanged	Unchanged
Speed	Unchanged	Decreases
Wavelength	Unchanged	Decreases (λ/n)
Direction	i = r	Bends toward the normal

Coherent and incoherent addition of waves

Two waves of amplitudes a₁ and a₂ with phase difference φ give a resultant of intensity

I = I₁ + I₂ + 2√(I₁I₂) cos φ

Coherent sources keep a constant phase difference with time. The cosine term is steady, so a stable interference pattern is seen. The maximum intensity (φ = 0, 2π, …) is Imax = (√I₁ + √I₂)². The minimum intensity (φ = π, 3π, …) is Imin = (√I₁ − √I₂)².
Incoherent sources have a phase difference that changes randomly and quickly. The average of cos φ is zero, so the intensities add: I = I₁ + I₂. No fixed fringes are seen. Two separate bulbs are incoherent.

For two coherent sources of equal intensity I₀: Imax = 4I₀ and Imin = 0. For incoherent sources of equal intensity I₀, the intensity is 2I₀ everywhere.

Intensity ratio and amplitude ratio are related by I ∝ a². So Imax/Imin = ((a₁ + a₂)/(a₁ − a₂))².

Interference does not create or destroy energy. It redistributes energy: the average intensity of the pattern is still I₁ + I₂.

Interference: Young's double-slit experiment

Light from a single narrow slit falls on two close slits S₁ and S₂ (separation d). The two beams are coherent because they come from the same source. A screen is at distance D from the slits (D much greater than d). For a point P at distance y from the central point O of the screen, the path difference is

Δ = S₂P − S₁P ≈ d sin θ ≈ yd/D

The phase difference is φ = (2π/λ)Δ.

Bright fringes (constructive): Δ = nλ, so y = nλD/d, with n = 0, ±1, ±2, …
Dark fringes (destructive): Δ = (n + ½)λ, so y = (n + ½)λD/d.
Fringe width (the distance between two consecutive bright or two consecutive dark fringes): β = λD/d. The fringes are equally spaced.
Angular fringe width: θ = λ/d.
Intensity: with each slit giving intensity I₀ at the screen, I = 4I₀ cos²(φ/2). It is 4I₀ at the bright fringes and zero at the dark fringes. The average is 2I₀.

The central fringe (n = 0, Δ = 0) is bright.

Effect of changing quantities (β = λD/d):

Change	Effect on fringe width β
Slit separation d increased	β decreases (β ∝ 1/d)
Screen distance D increased	β increases (β ∝ D)
Wavelength λ increased (red instead of blue)	β increases (β ∝ λ)
Apparatus placed in a medium of index n	β becomes β/n, because λ becomes λ/n
Source made white	Central fringe white; coloured fringes on each side

If one slit is covered, the interference fringes vanish and the screen shows a single-slit pattern.

Diffraction at a single slit

When light passes through a slit whose width a is comparable to the wavelength, it bends into the geometrical shadow and gives a pattern of a broad central bright band with weaker bright bands on both sides.

Consider a slit of width a and a screen at distance D. At angle θ from the centre:

Minima (dark): a sin θ = nλ, with n = ±1, ±2, ±3, …. For small angles, the position is y = nλD/a.
Secondary maxima: they lie roughly midway between minima, near a sin θ = (n + ½)λ, and they are much weaker than the central maximum.
Central maximum: it extends from the first minimum on one side to the first minimum on the other. Angular half-width = λ/a, and the linear width is

2λD/a

Each secondary maximum has width λD/a, which is half the width of the central maximum.
A narrower slit gives a wider pattern (width ∝ 1/a). A longer wavelength also gives a wider pattern.

The path-difference reasoning: at the first minimum, the slit can be divided into two halves. Each point in one half has a partner in the other half with a path difference of λ/2, so the contributions cancel in pairs.

Feature	Interference (double slit)	Diffraction (single slit)
Origin	Superposition of waves from two coherent sources	Superposition of wavelets from different parts of one slit
Fringe width	All equal: λD/d	Central is 2λD/a; others λD/a
Intensity of bright fringes	Almost equal	Falls rapidly from the centre
Dark fringes	Intensity zero (equal sources)	Intensity zero at minima
Condition for dark	Δ = (n + ½)λ	a sin θ = nλ

Polarisation

Interference and diffraction show that light is a wave. Polarisation shows that it is a transverse wave. Sound waves in air are longitudinal and cannot be polarised.

In ordinary (unpolarised) light the electric field vibrates in all directions perpendicular to the direction of travel. In plane polarised light it vibrates in only one direction.

Polaroid. A Polaroid is a sheet that allows only the electric field component parallel to its pass-axis to pass. Light passing through an ideal Polaroid becomes plane polarised, and the intensity of unpolarised light is reduced to half: I = I₀/2.

Malus' law. When plane polarised light of intensity I₀ falls on a second Polaroid (the analyser) whose pass-axis makes angle θ with the polarisation direction, the transmitted intensity is

I = I₀ cos²θ

At θ = 0 the whole intensity passes. At θ = 90° (crossed Polaroids) the intensity is zero. The transmitted amplitude is E₀ cos θ.

θ	0°	30°	45°	60°	90°
I/I₀	1	3/4	1/2	1/4	0

Polarisation by reflection. Light reflected from a transparent surface is partially polarised. At one angle of incidence, the Brewster angle iB, the reflected light is completely plane polarised, with its electric field perpendicular to the plane of incidence. At this angle the reflected and refracted rays are at right angles (iB + r = 90°). Brewster's law:

tan iB = n

where n is the refractive index of the reflecting medium with respect to the medium from which the light comes.

Uses of Polaroids: sunglasses and window glasses to reduce glare, camera filters, three-dimensional (3D) cinema glasses, liquid crystal displays and for controlling the intensity of light.

Formula summary

Topic	Formula	Symbols and units
Refraction by Huygens' principle	sin i / sin r = v₁/v₂ = n₂/n₁	v in m s⁻¹
Wavelength in a medium	λ′ = λ/n	f unchanged
Superposition	I = I₁ + I₂ + 2√(I₁I₂) cos φ	φ in rad
Phase and path difference	φ = 2πΔ/λ	Δ in m
Young's fringe width	β = λD/d	D screen distance, d slit separation
Bright / dark positions	y = nλD/d; y = (n + ½)λD/d	y from the central fringe
Intensity	I = 4I₀ cos²(φ/2)	I₀ from one slit
Single-slit minima	a sin θ = nλ	a slit width
Central maximum width	2λD/a
Malus' law	I = I₀ cos²θ	θ angle between axes
Brewster's law	tan iB = n	iB + r = 90°

Worked examples

Example 1 (fringes). In a double-slit experiment d = 0.5 mm, D = 1 m and λ = 500 nm. Find the fringe width and the positions of the 4th bright fringe and the 3rd dark fringe (counted from the centre; n = 2).

β = λD/d = 500 × 10⁻⁹ × 1 / (0.5 × 10⁻³) = 1 × 10⁻³ m = 1 mm.

4th bright: y = 4β = 4 mm. 3rd dark: y = 2.5β = 2.5 mm.

Example 2 (intensity). Each slit alone gives intensity I₀ at a point. At that point the path difference is λ/4. Find the intensity.

φ = (2π/λ)(λ/4) = π/2 (90°). I = 4I₀ cos²(45°) = 4I₀ × ½ = 2I₀.

Example 3 (single slit). A slit of width 0.1 mm is illuminated by light of wavelength 500 nm. The screen is 1 m away. Find the width of the central maximum.

Width = 2λD/a = 2 × 500 × 10⁻⁹ × 1 / (10⁻⁴) = 1 × 10⁻² m = 10 mm. The first minimum is 5 mm from the centre.

Example 4 (Malus' law). Unpolarised light of intensity I₀ passes through two Polaroids whose axes make 45°. Find the final intensity.

After the first: I₀/2. After the second: (I₀/2) cos²45° = (I₀/2)(1/2) = I₀/4.

Example 5 (Brewster angle). Find the angle of polarisation for glass of refractive index 1.5, and the angle of refraction at that angle.

tan iB = 1.5, so iB = 56.3°. The refraction angle is r = 90° − 56.3° = 33.7°.

NEET traps

In interference all fringes have equal width. In single-slit diffraction the central maximum is twice as wide as the others.
The condition a sin θ = nλ in single-slit diffraction gives dark fringes. In double-slit interference, d sin θ = nλ gives bright fringes.
Fringe width β = λD/d is the distance between two consecutive bright fringes. It is not the distance from the centre to the first dark fringe, which is β/2.
Wavelength changes in a medium, but frequency does not. In a medium of index n, β reduces to β/n.
Two independent bulbs do not give interference, because they are incoherent. Their intensities simply add.
For unpolarised light, the first Polaroid transmits half the intensity without using Malus' law. Malus' law applies only to polarised light.
Polarisation shows that light is transverse. Interference and diffraction only show that light is a wave.
At the Brewster angle the reflected light is fully polarised, but the refracted light is only partly polarised.
Brewster's law is tan iB = n, not sin iB = n. The reflected and refracted rays are at 90° to each other.
Covering one slit of a double-slit set-up does not remove light from the screen. It removes the fringes.

One-liners

A wavefront is a surface of constant phase; a ray is perpendicular to it.
A distant source gives plane wavefronts; a point source gives spherical ones.
Huygens' principle proves i = r and Snell's law.
Frequency never changes on refraction.
Coherent sources have a constant phase difference.
Fringe width β = λD/d.
Bright fringes: path difference nλ. Dark fringes: (n + ½)λ.
I = 4I₀ cos²(φ/2) for equal slits.
Incoherent sources add intensities directly.
Single-slit minima: a sin θ = nλ.
Central maximum of a single slit has width 2λD/a.
A narrower slit spreads the diffraction pattern more.
Polarisation proves that light is transverse.
Malus' law: I = I₀ cos²θ.
Brewster's law: tan iB = n.

Practice questions

Match List I with List II. List I: a. Minima of a single slit b. Width of the central maximum c. Angular half-width of the central maximum d. Width of a secondary maximum List II: i. λ/a ii. a sin θ = nλ iii. 2λD/a iv. λD/a
1. a-iii, b-ii, c-i, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-ii, b-iii, c-iv, d-i
4. a-i, b-ii, c-iii, d-iv
Answer
B. a-ii, b-iii, c-i, d-iv
a-ii, b-iii, c-i, d-iv.
A Young's double-slit set-up in air gives a fringe width of 3 mm. The whole set-up is immersed in water of refractive index 4/3. The new fringe width is
1. 2.25 mm
2. 3 mm
3. 4 mm
4. 1.5 mm
Answer
A. 2.25 mm
In the medium λ′ = λ/n, so β′ = β/n = 3/(4/3) = 2.25 mm.
Two coherent beams of intensities I₀ and 4I₀ meet with a phase difference of 120°. The resultant intensity is
1. 9I₀
2. 5I₀
3. 7I₀
4. 3I₀
Answer
D. 3I₀
I = I₁ + I₂ + 2√(I₁I₂) cos φ = I₀ + 4I₀ + 2(2I₀)(−1/2) = 5I₀ − 2I₀ = 3I₀.
Light of wavelength 600 nm is incident on a single slit of width 1.2 μm. The angle of the first minimum is
1. 90°
2. 30°
3. 45°
4. 60°
Answer
B. 30°
sin θ = λ/a = 0.6 / 1.2 = 0.5, so θ = 30°.
Which of the following statements are correct? 1. The intensity in a double-slit pattern is I = 4I₀ cos²(φ/2). 2. The total energy in the interference pattern is more than the sum of the energies of the two beams. 3. Interference redistributes energy from dark fringes to bright fringes.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Energy is conserved. Interference only redistributes it.
Wavefronts of light from a very distant star, over a small region near the Earth, are
1. plane
2. cylindrical
3. conical
4. spherical
Answer
A. plane
A very distant source gives a nearly plane wavefront over a small area.
Plane polarised light of intensity I₀ falls on a Polaroid whose pass-axis makes 60° with the plane of polarisation. The transmitted intensity is
1. I₀/√2
2. 3I₀/4
3. I₀/2
4. I₀/4
Answer
D. I₀/4
I = I₀ cos²60° = I₀ × (1/2)² = I₀/4.
In single-slit diffraction with slit width a and wavelength λ, the first minimum occurs at an angle θ given by
1. sin θ = 2λ/a
2. sin θ = λ/a
3. sin θ = 3λ/(2a)
4. sin θ = λ/(2a)
Answer
B. sin θ = λ/a
Minima are at a sin θ = nλ. For the first minimum n = 1.
Assertion (A): A narrow slit gives a wider diffraction pattern than a wide slit. Reason (R): The first minimum is at sin θ = λ/a, which becomes larger when a is smaller.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
A smaller a means a larger θ, so a wider pattern.
Match List I with List II. List I: a. Distant source b. Point source c. Secondary wavelets d. Sunglasses List II: i. Spherical wavefront ii. Plane wavefront iii. Huygens' principle iv. Polaroid
1. a-ii, b-i, c-iv, d-iii
2. a-iii, b-i, c-ii, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-i, b-ii, c-iii, d-iv
Answer
C. a-ii, b-i, c-iii, d-iv
a-ii, b-i, c-iii, d-iv.
Match List I with List II. List I: a. Polaroid axis at 0° to plane polarised light b. Polaroid axis at 30° c. Polaroid axis at 45° d. Polaroid axis at 90° List II: i. 1/2 of the intensity passes ii. No light passes iii. All the intensity passes iv. 3/4 of the intensity passes
1. a-i, b-ii, c-iii, d-iv
2. a-iv, b-iii, c-i, d-ii
3. a-iii, b-i, c-iv, d-ii
4. a-iii, b-iv, c-i, d-ii
Answer
D. a-iii, b-iv, c-i, d-ii
cos²θ is 1, 3/4, 1/2 and 0: a-iii, b-iv, c-i, d-ii.
Two coherent sources have intensities in the ratio 9 : 1. The ratio of the maximum to the minimum intensity in the interference pattern is
1. 10 : 8
2. 9 : 1
3. 4 : 1
4. 3 : 1
Answer
C. 4 : 1
Amplitude ratio is 3 : 1. Imax/Imin = ((3 + 1)/(3 − 1))² = 4. Check: Imax = (√9 + √1)² = 16 and Imin = (√9 − √1)² = 4, ratio 4.
Unpolarised light of intensity I₀ falls on Polaroid P₁. A second Polaroid P₂ is at 30° to P₁, and a third Polaroid P₃ is at 90° to P₁. The final intensity is
1. I₀/4
2. 3I₀/16
3. I₀/8
4. 3I₀/32
Answer
D. 3I₀/32
After P₁: I₀/2. After P₂: (I₀/2) cos²30° = 3I₀/8. P₃ is at 60° to P₂: (3I₀/8) cos²60° = 3I₀/32.
A single slit of width 0.2 mm is illuminated by light of wavelength 500 nm. The screen is 2 m away. The width of the central maximum is
1. 2.5 mm
2. 10 mm
3. 20 mm
4. 5 mm
Answer
B. 10 mm
Width = 2λD/a = 2 × 500 × 10⁻⁹ × 2 / (0.2 × 10⁻³) = 10⁻² m = 10 mm. The value 5 mm is the half-width.
In a double-slit experiment λ = 500 nm, d = 1 mm and D = 2 m. The distance between the 3rd bright fringe and the dark fringe of path difference 3λ/2, both on the same side of the centre, is
1. 1.5 mm
2. 1 mm
3. 0.5 mm
4. 3 mm
Answer
A. 1.5 mm
β = λD/d = 1 mm. The 3rd bright fringe is at 3β = 3 mm. A path difference 3λ/2 gives y = 1.5β = 1.5 mm. The separation is 1.5 mm.
Light travelling in air strikes a transparent liquid with angle of incidence 60° and the angle of refraction is 30° (c = 3 × 10⁸ m s⁻¹). The speed of light in the liquid is
1. √3 × 10⁸ m s⁻¹
2. 2 × 10⁸ m s⁻¹
3. 3√3 × 10⁸ m s⁻¹
4. 1.5 × 10⁸ m s⁻¹
Answer
A. √3 × 10⁸ m s⁻¹
n = sin 60°/sin 30° = √3. v = c/n = 3 × 10⁸/√3 = √3 × 10⁸ ≈ 1.73 × 10⁸ m s⁻¹.
Unpolarised light of intensity I₀ passes through two ideal Polaroids with their axes at 60° to each other. The final intensity is
1. I₀/2
2. I₀/4
3. 3I₀/8
4. I₀/8
Answer
D. I₀/8
After the first Polaroid I₀/2. After the second (I₀/2) cos²60° = (I₀/2)(1/4) = I₀/8.
When the width of a single slit is halved (the same light and screen), the width of the central maximum of the diffraction pattern
1. remains the same
2. doubles
3. becomes four times
4. halves
Answer
B. doubles
Width = 2λD/a, which is inversely proportional to a.
Light of wavelength 600 nm in air enters glass of refractive index 1.5. Its wavelength in the glass is
1. 400 nm
2. 300 nm
3. 900 nm
4. 600 nm
Answer
A. 400 nm
λ′ = λ/n = 600/1.5 = 400 nm. The frequency does not change.
Light of wavelength 500 nm falls on two slits 0.1 mm apart. The angular fringe width λ/d is
1. 2 × 10² rad
2. 5 × 10⁻³ rad
3. 5 × 10⁻⁴ rad
4. 5 × 10⁻⁶ rad
Answer
B. 5 × 10⁻³ rad
θ = λ/d = 500 × 10⁻⁹ / 10⁻⁴ = 5 × 10⁻³ rad.
Assertion (A): Light waves can be polarised. Reason (R): Light waves are transverse.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Only transverse waves can be polarised.
Given below are two statements: Statement I: When a double-slit set-up is immersed in a liquid of refractive index 1.5, the fringe width increases. Statement II: The wavelength of light in a medium of refractive index n is λ/n.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
β becomes β/n, so the fringe width decreases.
In a double-slit experiment each slit alone produces intensity I₀ at a point, and the phase difference between the two waves at that point is 60°. The resultant intensity is
1. I₀
2. 2I₀
3. 3I₀
4. 4I₀
Answer
C. 3I₀
I = 4I₀ cos²(φ/2) = 4I₀ cos²30° = 4I₀ × 3/4 = 3I₀.
Which of the following statements are correct? 1. Polarisation shows that light waves are longitudinal. 2. An ideal Polaroid halves the intensity of unpolarised light. 3. At the Brewster angle the reflected light is completely plane polarised.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
Polarisation shows that light is transverse.
In a double-slit set-up d = 0.3 mm, D = 1 m and λ = 600 nm. At a point 3 mm from the central fringe the screen will show
1. a fringe of intensity 2I₀ exactly
2. a bright fringe
3. the central maximum
4. a dark fringe
Answer
D. a dark fringe
Δ = yd/D = 3 × 10⁻³ × 0.3 × 10⁻³ / 1 = 9 × 10⁻⁷ m = 900 nm = 1.5λ. This is (n + ½)λ with n = 1, so it is dark.
Given below are two statements: Statement I: Huygens' principle can be used to derive the laws of reflection and refraction. Statement II: A ray of light is perpendicular to the wavefront.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are standard results of the wavefront model.
Given below are two statements: Statement I: In single-slit diffraction, the central maximum is twice as wide as each secondary maximum. Statement II: The width of the central maximum is 2λD/a.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Central width is 2λD/a, and each secondary maximum is λD/a wide.
Given below are two statements: Statement I: Sound waves in air can be polarised. Statement II: The intensity of ordinary light is reduced to half after passing through an ideal Polaroid.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Sound waves are longitudinal and cannot be polarised. Unpolarised light loses half its intensity.
Match List I with List II. List I: a. Fringe width in a double-slit experiment b. Intensity in a double-slit experiment c. Malus' law d. Brewster's law List II: i. I = I₀ cos²θ ii. tan iB = n iii. λD/d iv. 4I₀ cos²(φ/2)
1. a-i, b-ii, c-iii, d-iv
2. a-iv, b-iii, c-ii, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-iii, b-i, c-iv, d-ii
Answer
C. a-iii, b-iv, c-i, d-ii
a-iii, b-iv, c-i, d-ii.
Given below are two statements: Statement I: The frequency of light does not change when it passes from air into glass. Statement II: The wavelength of light increases when it enters glass from air.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
The wavelength decreases to λ/n; the frequency is unchanged.
Match List I with List II. List I: a. Slit separation d doubled b. Screen distance D doubled c. Wavelength tripled d. Whole set-up in water (n = 4/3) List II: i. 3β ii. β/2 iii. 3β/4 iv. 2β
1. a-ii, b-i, c-iv, d-iii
2. a-ii, b-iv, c-i, d-iii
3. a-iv, b-ii, c-i, d-iii
4. a-i, b-ii, c-iii, d-iv
Answer
B. a-ii, b-iv, c-i, d-iii
β = λD/d: β/2 (a-ii), 2β (b-iv), 3β (c-i), β/n = 3β/4 (d-iii).
Light from air falls on a glass plate at the polarising (Brewster) angle, and the glass has refractive index √3. The angle of refraction in the glass is
1. 45°
2. 90°
3. 60°
4. 30°
Answer
D. 30°
tan iB = √3 gives iB = 60°. At the Brewster angle iB + r = 90°, so r = 30°. Check with Snell's law: sin 60°/sin 30° = √3.
Which of the following statements are correct? 1. The fringe width in a double-slit experiment is λD/d. 2. Bright fringes occur where the path difference is nλ. 3. The fringe width increases when the slit separation is increased.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
β ∝ 1/d, so a larger d reduces the fringe width.
Assertion (A): In single-slit diffraction the central maximum becomes narrower when the slit is made narrower. Reason (R): The first minimum occurs at a sin θ = λ.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A narrower slit gives a wider central maximum, because the width is 2λD/a. R is true.
Assertion (A): Sound waves in air cannot be polarised. Reason (R): Sound waves in air are transverse.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Sound in air is longitudinal, so it cannot be polarised. Reason R is false.
Two independent sodium lamps are placed in front of two slits. No stable interference fringes appear mainly because
1. sodium light is not a wave
2. the lamps are incoherent sources, so their phase difference changes randomly
3. the slits are too wide to diffract light
4. the two lamps have different speeds of light
Answer
B. the lamps are incoherent sources, so their phase difference changes randomly
Independent lamps do not keep a constant phase difference. Their intensities simply add.
Which of the following statements are correct? 1. Minima in single-slit diffraction occur at a sin θ = nλ. 2. The central maximum becomes narrower when the slit width increases. 3. Diffraction is more noticeable when the slit width is comparable to the wavelength.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
D. 1, 2 and 3
All three follow from the single-slit relations.
Given below are two statements: Statement I: Two independent sodium lamps can produce a stable interference pattern. Statement II: Coherent sources must have a constant phase difference with time.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Independent lamps are incoherent. Statement II is the definition of coherence.
In a double-slit experiment the fringe width is 3 mm. The slit separation is doubled and the screen distance is halved. The new fringe width is
1. 1.5 mm
2. 6 mm
3. 0.75 mm
4. 3 mm
Answer
C. 0.75 mm
β ∝ D/d. The new β = 3 × (1/2)/2 = 0.75 mm. Halving only D or doubling only d gives 1.5 mm.
Assertion (A): In Young's experiment the fringe width is larger for red light than for blue light. Reason (R): The central fringe in a double-slit experiment is bright.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
β ∝ λ and red has the longer wavelength. The brightness of the central fringe does not explain this.
Assertion (A): No stable interference pattern is observed from two separate bulbs. Reason (R): The phase difference between the light from independent bulbs changes randomly with time.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Independent bulbs are incoherent, so the interference term averages to zero.
The Brewster angle for a transparent medium of refractive index √3 (in air) is
1. 30°
2. 60°
3. 45°
4. 90°
Answer
B. 60°
tan iB = n = √3, so iB = 60°.
In a Young's double-slit experiment the slit separation is 0.2 mm, the screen is 1 m from the slits and the wavelength is 600 nm. The fringe width is
1. 0.3 mm
2. 12 mm
3. 3 mm
4. 1.2 mm
Answer
C. 3 mm
β = λD/d = 600 × 10⁻⁹ × 1 / (0.2 × 10⁻³) = 3 × 10⁻³ m = 3 mm.

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1.The band of the color component of light beam is called its ________.
2.The brain interprets the frequency of an emitted sound called
3.Since sound wave is a disturbance, which moves in the medium by setting the neighbouring particles into motion, they are categorised in which wave category?
4.Sound travels in a medium as a ________.
5.The smell of hot food reaches a distance because:
6.The speed of a wave is given by which of the following formulas?
7.The speed of sound (v) is calculated by multiplying which two wave properties?
8.This figure exhibits compression and rarefaction in a slinky (spring-shaped toy). What is a dark band called?
9.The intensity of the sound is determined by its ______.
10.The presence of _____ between the two surfaces in contact causes friction.

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