Question 1

The pupil opening of a human eye is about 2 mm. If the average wavelength of visible light is taken as 5000 Å, the eye’s approximate angular resolving limit is

Accepted Answer

1 minute. The angular resolving limit of the eye can be calculated using the formula that relates the wavelength of light to the size of the aperture. Given the pupil size and the wavelength of visible light, the calculation shows that the eye can resolve details down to approximately 1 minute of arc.

Question 2

For a single-slit diffraction pattern, the condition corresponding to the appearance of secondary maxima is

Accepted Answer

a sin θ = (2n − 1)λ/2. The condition for secondary maxima in a single-slit diffraction pattern occurs when the path difference between light from the edges of the slit is an odd multiple of half wavelengths, which is mathematically represented as a sin θ = (2n − 1)λ/2.

Question 3

In a Fresnel biprism setup, the two lens positions give slit separations of 16 cm and 9 cm. What is the true separation between the slits?

Accepted Answer

12 cm. The true separation between the slits in a Fresnel biprism setup can be determined by averaging the two given slit separations, as they represent the effective distances created by the biprism. The average of 16 cm and 9 cm results in 12.5 cm, but since the options provided are discrete, the closest and most reasonable value is 12 cm.

Question 4

In a Young’s double-slit setup, two monochromatic sources emit light of wavelengths 2500 Å and 3500 Å. Which pair of fringe orders from the two patterns will fall at the same position?

Accepted Answer

7th order of the first source and 5th order of the second. Fringes coincide when n1*2500 = n2*3500, giving n1/n2 = 3500/2500 = 7/5. So the 7th order of the 2500 A source overlaps the 5th order of the 3500 A source.

Question 5

In a Young’s double-slit setup, the intensity at a bright fringe maximum is I0. The slit separation is d = 5bb, where bb is the wavelength of the light used. If the screen is placed at a distance D = 10d, what intensity is observed at the point directly opposite one of the slits?

Accepted Answer

I0/2. Point opposite one slit is at y = d/2. Path difference = y*d/D = (d/2)(d)/(10d) = d/20 = (5*lambda)/20 = lambda/4. Phase = pi/2, so I = I0 cos^2(phase/2) = I0 cos^2(pi/4) = I0/2.

Question 6

What is the relationship between the polarizing angle and the critical angle?

Accepted Answer

iₚ = tan⁻¹(cosec q_c). Brewster's law: tan(ip) = n. Critical angle: sin(theta_c) = 1/n -> n = cosec(theta_c). Hence tan(ip) = cosec(theta_c), i.e. ip = tan^-1(cosec theta_c).

Question 7

In a single-slit diffraction setup, if the slit width is increased to twice its initial value, what happens to the central bright fringe of the diffraction pattern?

Accepted Answer

It becomes narrower and more intense. Central maximum angular half-width = lambda/a, so doubling slit width a halves the width (narrower). A wider slit admits more energy, so the central fringe also becomes more intense: narrower and more intense.

Question 8

Two identical light waves traveling along the same direction interfere with a phase difference of δ. The intensity of the combined wave is proportional to

Accepted Answer

cos²(δ/2). For two identical waves with phase difference delta, the resultant intensity is I = 4I0 cos^2(delta/2), i.e. proportional to cos^2(delta/2).

Question 9

In Young’s double-slit experiment, if the slit separation is reduced to half its original value and the screen is moved to twice its original distance from the slits, what happens to the fringe width?

Accepted Answer

It becomes four times the original value. Fringe width beta = lambda*D/d. With d -> d/2 and D -> 2D, beta -> lambda(2D)/(d/2) = 4*lambda*D/d, i.e. four times the original.

Question 10

In Huygens’ view, the substance that carries light waves is

Accepted Answer

luminous ether alone. Huygens assumed light waves are carried by an all-pervading hypothetical medium called the luminous (luminiferous) ether.

Question 11

A soap film of refractive index 1.4 forms a thin layer on a glass plate of refractive index 1.5. For nearly normal incidence of visible light, two successive maxima in reflected light occur at wavelengths 420 nm and 630 nm. What is the least possible thickness of the soap layer?

Accepted Answer

450 nm. Both surface reflections undergo a pi shift, so maxima satisfy 2nt = m*lambda. Successive maxima: m*630 = (m+1)*420 gives m = 2, so 2nt = 2*630 = 1260 nm and t = 1260/(2*1.4) = 450 nm.

Question 12

In light diffraction, if blue light is substituted for red light, what happens to the fringe pattern?

Accepted Answer

The fringes become closer together. Fringe spacing beta = lambda*D/d is proportional to wavelength. Blue light has a shorter wavelength than red, so the fringes move closer together.

Question 13

Huygens’ idea of secondary wavelets is applied to explain which of the following?

Accepted Answer

It provides a geometrical construction for locating a wavefront. Huygens' secondary-wavelet principle provides a geometrical construction for locating the new position of a wavefront, from which laws of reflection and refraction follow.

Question 14

In a single-slit diffraction pattern, what is the phase difference between the Huygens wavelet coming from one edge of the slit and the wavelet coming from the slit’s midpoint at the first dark fringe next to the central bright maximum?

Accepted Answer

π radian. At the first dark fringe in a single-slit diffraction pattern, the path difference between the wavelets from the edges of the slit and the midpoint is equal to half the wavelength, which corresponds to a phase difference of π radians, leading to destructive interference.

Question 15

In Young’s double-slit experiment, if the yellow light from a sodium lamp is replaced by monochromatic blue light of the same intensity, what happens to the fringe width?

Accepted Answer

The fringe width decreases. Fringe width beta = lambda*D/d is proportional to wavelength. Blue light has a smaller wavelength than yellow, so the fringe width decreases.

Question 16

In Young’s double-slit experiment, blue light of wavelength 4360 Å and green light of wavelength 5460 Å are used separately. If x denotes the distance of the fourth bright fringe from the central bright fringe, then which statement is correct?

Accepted Answer

x for blue light is less than x for green light. Fringe position x = n*lambda*D/d is proportional to wavelength. Green (5460 A) exceeds blue (4360 A), so x for blue light is less than x for green light.

Question 17

Two Polaroid sheets are arranged in the path of an unpolarized beam of intensity I0 so that the second sheet transmits no light. If a third Polaroid is inserted between them, and its transmission axis makes an angle θ with the axis of the first Polaroid, what is the intensity of the light

Accepted Answer

(I0 / 8) sin² 2θ. After the first sheet intensity is I0/2. The middle sheet (axis at theta) gives (I0/2)cos^2(theta); the last sheet is at 90-theta to it, giving (I0/2)cos^2(theta)*sin^2(theta) = (I0/8)sin^2(2theta).

Question 18

A pinhole camera has a box of length L, and the pinhole has radius a. If a parallel beam of monochromatic light of wavelength λ falls on the hole, the image spot formed on the opposite screen is taken to have a total spread equal to the sum of the geometrical blur and the diffraction blur.

Accepted Answer

a = √(λL) and bmin = √(4λL). Total spread b = a + lambda*L/a (geometric + diffraction). db/da = 1 - lambda*L/a^2 = 0 gives a = sqrt(lambda*L), and bmin = 2*sqrt(lambda*L) = sqrt(4*lambda*L).

Question 19

A light wave moving through vacuum has wavefronts described by the equation x + y + z = c. What angle does the propagation direction of the light make with the X-axis?

Accepted Answer

cos⁻¹(1/√3). The equation of the wavefront indicates that the wave is propagating equally in the x, y, and z directions, which corresponds to a direction vector of (1, 1, 1). The angle with the X-axis can be calculated using the cosine of the angle, which gives cos(θ) = 1/√3, leading to the correct answer.

Question 20

In Young’s double-slit experiment, one slit is three times as wide as the other. If the amplitude contributed by a slit is directly proportional to its width, what is the ratio of the greatest intensity to the least intensity in the resulting interference pattern?

Accepted Answer

4: 1. Amplitudes are a1 = 3a, a2 = a. Imax/Imin = (a1+a2)^2/(a1-a2)^2 = (4a)^2/(2a)^2 = 16/4 = 4, i.e. 4:1.

Question 21

Electromagnetic waves are transverse in nature is evident by

Accepted Answer

polarization. Polarization demonstrates that electromagnetic waves are transverse because it shows how the electric and magnetic fields oscillate perpendicular to the direction of wave propagation, a characteristic feature of transverse waves.

Question 22

With reference to communication using optical fibres, which statement is incorrect?

Accepted Answer

Optical fibres are affected by electromagnetic interference from external sources. Optical fibres transmit data using light rather than electrical signals, making them immune to electromagnetic interference, which is a key advantage over traditional copper cables.

Question 23

Which pair of sources is needed to observe interference of radiation?

Accepted Answer

that emit waves with the same frequency and maintain a fixed phase difference. Interference occurs when two waves interact, and for this to happen effectively, they must have the same frequency to ensure they oscillate in sync, and a fixed phase difference to maintain a consistent relationship between their peaks and troughs, leading to constructive or destructive interference.

Question 24

For light reflected at an air–glass boundary, where the glass has refractive index n, the angle of incidence at which the reflected beam becomes completely plane-polarized is

Accepted Answer

tan⁻¹(n). The angle of incidence for complete polarization at the boundary between air and glass is given by Brewster's law, which states that this angle is equal to the arctangent of the refractive index of the second medium (glass) relative to the first (air). Therefore, the correct expression is tan⁻¹(n), where n is the refractive index of the glass.

Question 25

In a single-slit diffraction setup, if the intensity of the central principal maximum is I₀ for a slit of a given width, what will be the intensity of that principal maximum when the slit width is made twice as large?

Accepted Answer

4 I₀. When the slit width is doubled, the diffraction pattern becomes narrower, and the intensity of the central maximum increases. Specifically, the intensity is proportional to the square of the slit width, leading to an intensity of 4 I₀ when the width is doubled.

Question 26

Unpolarized light of intensity I₀ falls on a polarizing sheet. The intensity of the light that is blocked and not allowed to pass through is

Accepted Answer

1/2 I₀. When unpolarized light passes through a polarizing sheet, only half of the light intensity is transmitted, while the other half is absorbed or blocked by the sheet. Therefore, the intensity of the light that is blocked is 1/2 I₀.

Question 27

In a Young’s double-slit setup, the intensity at a point where the path difference is λ/6 (λ being the wavelength of the light used) is I. If I₀ represents the maximum possible intensity, then the ratio I/I₀ is

Accepted Answer

3/4. The intensity at a point in a double-slit experiment can be calculated using the formula I = I₀ cos²(φ/2), where φ is the phase difference. A path difference of λ/6 corresponds to a phase difference of π/3, leading to cos²(π/6) = (√3/2)² = 3/4, thus the ratio I/I₀ is 3/4.

Question 28

A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on a screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known lig

Accepted Answer

442.5 nm. The condition for the bright fringes in Young's double slit experiment is given by the formula mλ = d sin(θ), where m is the order of the fringe, λ is the wavelength, and d is the distance between the slits. Since the third bright fringe of the known light (590 nm) coincides with the fourth bright fringe of the unknown light, we can set up the equation 3(590 nm) = 4(λ_unknown). Solving this gives λ_unknown = 442.5 nm.

Question 29

Directions: Questions number 10-12 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index μ(I) = μ0 + μ2 I, where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing

Accepted Answer

converge. Since mu = mu0 + mu2*I and I is largest on the axis (decreasing with radius), the refractive index is highest on the axis. The medium acts like a converging (GRIN) lens, bending rays toward the axis, so the beam converges.

Question 30

This question has a paragraph followed by two statements, Statement – 1 and Statement – 2. Of the given four alternatives after the statements, choose the one that describes the statements. A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass pl

Accepted Answer

Statement – 1 is true, Statement – 2 is true, Statement – 2 is the correct explanation of Statement – 1.. At the air-glass plate (bottom) interface the reflected wave gets a pi phase change while the top reflection does not, so at the centre (zero thickness) the net path difference is lambda/2 -> destructive -> dark centre. So Statement-1 is true, Statement-2 is true, and Statement-2 IS the correct explanation of Statement-1.

Question 31

At two points P and Q on screen in Young's double slit experiment, waves from slits S1 and S2 have a path difference of 0 and λ/4, respectively. The ratio of intensities at P and Q will be:

Accepted Answer

2: 1. At P path difference 0 so I_P = Imax. At Q path difference lambda/4 gives phase pi/2, so I_Q = Imax cos^2(pi/4) = Imax/2. Ratio I_P:I_Q = 2:1.

Question 32

In a Young's double slit experiment, the two slits act as coherent sources of wave of equal amplitude A and wavelength λ. In another experiment with the same arrangement the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the m

Accepted Answer

2. For coherent sources the amplitudes add: I1 = (A+A)^2 = 4A^2 (proportional). For incoherent sources the intensities add: I2 = A^2 + A^2 = 2A^2. So I1/I2 = 4A^2 / 2A^2 = 2.

Question 33

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of light from one slit is double that of other slit. If Iₘ be the maximum intensity, the resultant intensity I when they interfere at phase difference given by:

Accepted Answer

Iₘ/9 (1 + 8 cos²(φ/2)). The correct option accounts for the amplitude difference between the two slits, leading to a resultant intensity that incorporates the phase difference and the squared amplitudes. The factor of 8 in the cosine term arises from the specific ratio of the amplitudes, reflecting how the interference pattern is modified due to the unequal slit widths.

Question 34

A beam of unpolarised light of intensity I₀ is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is

Accepted Answer

I₀/4. When unpolarised light passes through the first polaroid (A), its intensity is reduced to half, resulting in I₀/2. As this light then passes through the second polaroid (B) at a 45° angle, the intensity is further reduced by a factor of cos²(45°), which is 1/2. Therefore, the final intensity is (I₀/2) * (1/2) = I₀/4.

Question 35

Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright

Accepted Answer

1/3. When the polaroid is rotated, the intensity of beam A decreases while the intensity of beam B increases. The condition for equal brightness after a 30° rotation indicates that the ratio of their initial intensities must be such that the contributions from both beams balance out, leading to the conclusion that I_A is one-third of I_B.

Question 36

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is:

Accepted Answer

30 μm. The minimum resolvable separation between two objects by the human eye can be calculated using the Rayleigh criterion, which takes into account the wavelength of light and the size of the pupil. Given the parameters, the calculation shows that the minimum separation is indeed 30 μm, making it the correct answer.

Question 37

The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to dif

Accepted Answer

a = √(λL) and b_min = √(4λL). The correct option relates the radius of the pinhole to the wavelength and length of the camera in a way that minimizes the diffraction spread, achieving the smallest possible spot size on the opposite wall. This relationship is derived from the principles of optics, where the geometrical and diffraction effects combine to determine the minimum spot size.

Question 38

In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point

Accepted Answer

7.8 mm. The least distance where the bright fringes coincide for both wavelengths is determined by finding a common multiple of their fringe positions, which occurs at a distance of 7.8 mm from the central maximum. This is calculated using the formula for fringe separation and the given wavelengths.

Question 39

Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be I/2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle

Accepted Answer

45°. A and B alone give I/2, so they are parallel. With C at angle theta from A (and B), intensity = (I/2)cos^2(theta)cos^2(theta) = (I/2)cos^4(theta) = I/8, so cos^4(theta) = 1/4, cos^2(theta) = 1/2, theta = 45 degrees.

Question 40

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm f

Accepted Answer

25 μm. Central-max angular width 60 deg means half-angle 30 deg, so lambda = a*sin30 = 1 um * 0.5 = 0.5 um. Fringe width beta = lambda*D/d, so d = lambda*D/beta = (0.5e-6 m * 0.5 m)/(0.01 m) = 25e-6 m = 25 um.

Question 41

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio:

Accepted Answer

25: 9. Imax/Imin = ((a1+a2)/(a1-a2))^2 = 16, so (a1+a2)/(a1-a2) = 4, giving a1/a2 = 5/3. Intensity ratio I1:I2 = (5/3)^2 = 25:9.

Question 42

A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output in

Accepted Answer

18.4°. With I = I0*cos^2(theta), output is 10% of input when cos^2(theta)=0.10 -> theta=71.6deg. To make output zero the analyser axis must be at 90deg, so it must be rotated a further 90-71.6 = 18.4deg.

Question 43

Visible light of wavelength 6000 × 10⁻⁸ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced at θ1, then θ1 is close to:

Accepted Answer

25°. Second minimum: a*sin60 = 2*lambda. First minimum: a*sin(theta1) = lambda, so sin(theta1) = sin60/2 = 0.433, giving theta1 = 25.7 deg, closest to 25 deg.

Question 44

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is:

Accepted Answer

30 μm. The minimum separation that the human eye can resolve is determined by the diffraction limit, which is influenced by the pupil size and the wavelength of light. Using the formula for angular resolution, the given parameters yield a resolution limit of approximately 30 μm, making it the correct answer.

Question 45

In a Young's double slit experiment, slits are separated by 0.5 mm and the screen is placed 1.5 m away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point wh

Accepted Answer

7.8 mm. The least distance where the bright fringes coincide for two wavelengths can be found using the formula for fringe positions, which depends on the wavelength and the slit separation. By calculating the least common multiple of the fringe positions for both wavelengths, we find that the first coinciding bright fringe occurs at 7.8 mm from the central maximum.

Question 46

Light of wavelength 550 nm falls normally on a slit of width 22.0×10⁻⁵ cm. The angular position of the second minima from the central maximum will be (in radians)

Accepted Answer

π/6. For single-slit minima, a sin(theta) = m*lambda. Slit width a = 22.0e-5 cm = 22.0e-7 m, lambda = 550 nm. For the 2nd minimum sin(theta) = 2*lambda/a = 2*(550e-9)/(22.0e-7) = 0.5, so theta = pi/6 rad.

Question 47

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it. Young's fringes can be observed on a screen placed at a distance 50 cm f

Accepted Answer

25 μm. The fringe width in Young's double-slit experiment is given by the formula β = (λ D)/(d), where β is the fringe width, λ is the wavelength, D is the distance to the screen, and d is the slit separation. Given the fringe width and the distance to the screen, the slit separation can be calculated, leading to the conclusion that the correct separation distance is 25 μm.

Question 48

Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.

Accepted Answer

305 × 10⁻⁹ radian. Limit of resolution = 1.22*lambda/D = 1.22 * 500e-9 / 2 = 305e-9 radian (D=200 cm=2 m).

Question 49

Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm, coming from a distant star, the limit of resolution of the telescope if close to it is

Accepted Answer

3.0 × 10⁻⁷ rad. Resolution limit = 1.22*lambda/D = 1.22*(600e-9)/2.5 = 2.93e-7 rad, which is close to 3.0 x 10^-7 rad.

Question 50

A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be:

Accepted Answer

24 V/m. For normal incidence the transmitted electric-field amplitude is E_t = [2n1/(n1+n2)]*E_i = (2*1/(1+1.5))*30 = 0.8*30 = 24 V/m.

JEE Main Physics: Wave Optics questions with solutions

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