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The questions your board exam loves to ask
800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.
PhysicsClass 123 marksmedium
Ray Optics and Optical Instruments
The refractive index of glass is 1.5. Calculate the critical angle for the glass-air interface.
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The critical angle C is given by sin C = 1/n, where n is the refractive index of the denser medium (glass). So sin C = 1/1.5 = 0.667, giving C = sin inverse (0.667) = 41.8 degrees (approximately 42 degrees). For angles of incidence greater than this, total internal reflection occurs.
sin C = 1/n
Marking-scheme points
- ✓sin C = 1/n = 1/1.5 = 0.667
- ✓C = sin inverse (0.667)
- ✓Critical angle = about 41.8 degrees
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Ray Optics and Optical Instruments
Write the lens maker's formula and explain the meaning of each term.
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The lens maker's formula is 1/f = (n - 1)(1/R1 - 1/R2), where f is the focal length of the lens, n is the refractive index of the lens material with respect to the surrounding medium, and R1 and R2 are the radii of curvature of the two surfaces of the lens (taken with proper sign convention). It shows that the focal length depends on the material and the shape (curvature) of the lens.
1/f = (n - 1)(1/R1 - 1/R2)
Marking-scheme points
- ✓1/f = (n - 1)(1/R1 - 1/R2)
- ✓n = refractive index of lens material relative to surroundings
- ✓R1, R2 = radii of curvature of the two surfaces (with sign convention)
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Ray Optics and Optical Instruments
An object is placed 30 cm from a convex lens of focal length 20 cm. Find the position and nature of the image.
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Using the sign convention, u = -30 cm and f = +20 cm. From the lens formula 1/v - 1/u = 1/f, we get 1/v = 1/f + 1/u = 1/20 + 1/(-30) = 1/20 - 1/30 = (3 - 2)/60 = 1/60, so v = +60 cm. The magnification m = v/u = 60/(-30) = -2. The image is real (v positive for a lens), inverted (m negative) and magnified twice, formed 60 cm on the other side of the lens.
1/v - 1/u = 1/f
Marking-scheme points
- ✓u = -30 cm, f = +20 cm; 1/v = 1/f + 1/u
- ✓v = +60 cm; m = v/u = -2
- ✓Image is real, inverted and magnified 2 times
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Ray Optics and Optical Instruments
Define the power of a lens. State its SI unit and the formula for the power of two thin lenses in contact.
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The power of a lens is a measure of its ability to converge or diverge light and is defined as the reciprocal of its focal length in metres: P = 1/f (f in metres). Its SI unit is the dioptre (D). A converging (convex) lens has positive power and a diverging (concave) lens has negative power. For two thin lenses in contact, the total power is the sum: P = P1 + P2.
P = 1/f; P = P1 + P2
Marking-scheme points
- ✓P = 1/f (f in metres); SI unit dioptre (D)
- ✓Convex lens: positive power; concave lens: negative power
- ✓Lenses in contact: P = P1 + P2
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Ray Optics and Optical Instruments
Write the prism formula relating refractive index to the angle of the prism and the angle of minimum deviation.
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When a ray passes through a prism, it is deviated; the deviation is least at the angle of minimum deviation Dm, when the ray passes symmetrically through the prism. The refractive index of the prism material is given by n = sin((A + Dm)/2)/sin(A/2), where A is the refracting angle of the prism and Dm is the angle of minimum deviation. This formula is used to determine the refractive index of a transparent material.
n = sin((A + Dm)/2)/sin(A/2)
Marking-scheme points
- ✓Minimum deviation Dm occurs for symmetric passage
- ✓n = sin((A + Dm)/2)/sin(A/2)
- ✓A = angle of prism; used to find refractive index
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Ray Optics and Optical Instruments
Write the expression for the magnifying power of an astronomical telescope in normal adjustment and state the required focal lengths.
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An astronomical telescope has an objective lens of large focal length fo and an eyepiece of small focal length fe. In normal adjustment (final image at infinity), the magnifying power is M = fo/fe, and the length of the telescope is fo + fe. For high magnification, the objective should have a large focal length (and large aperture) and the eyepiece a small focal length. The final image is inverted.
M = fo/fe
Marking-scheme points
- ✓Magnifying power (normal adjustment): M = fo/fe
- ✓Length of telescope = fo + fe
- ✓Objective: large fo; eyepiece: small fe; image inverted
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Wave Optics
State Huygens' principle of secondary wavelets.
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Huygens' principle states that: (1) every point on a given wavefront acts as a source of new disturbance called secondary wavelets, which spread out in all directions with the speed of the wave; and (2) the new wavefront at a later instant is the forward envelope (tangential surface) of all these secondary wavelets. This principle is used to explain the laws of reflection and refraction and the propagation of light as a wave.
Marking-scheme points
- ✓Every point on a wavefront is a source of secondary wavelets
- ✓Wavelets travel with the speed of the wave
- ✓New wavefront = forward envelope of the secondary wavelets
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Wave Optics
Write the expression for the fringe width in Young's double slit experiment and explain the terms.
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In Young's double slit experiment, two coherent sources produce alternate bright and dark fringes on a screen. The fringe width (the distance between two consecutive bright or dark fringes) is beta = lambda D/d, where lambda is the wavelength of light used, D is the distance between the slits and the screen, and d is the separation between the two slits. All fringes are of equal width, and the width increases with wavelength and D but decreases as d increases.
beta = lambda D/d
Marking-scheme points
- ✓Fringe width beta = lambda D/d
- ✓lambda = wavelength, D = slit-to-screen distance, d = slit separation
- ✓Fringes are equally spaced; beta increases with lambda and D
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Wave Optics
In Young's double slit experiment, light of wavelength 600 nm is used with a slit separation of 1 mm and a screen 1 m away. Calculate the fringe width.
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Given lambda = 600 nm = 600 x 10^-9 m, d = 1 mm = 1 x 10^-3 m, D = 1 m. Fringe width beta = lambda D/d = (600 x 10^-9 x 1)/(1 x 10^-3) = 600 x 10^-6 = 6 x 10^-4 m = 0.6 mm.
beta = lambda D/d
Marking-scheme points
- ✓Convert units: lambda = 6e-7 m, d = 1e-3 m
- ✓beta = lambda D/d = (6e-7 x 1)/1e-3
- ✓beta = 6 x 10^-4 m = 0.6 mm
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Wave Optics
State the conditions for constructive and destructive interference in terms of path difference.
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For constructive interference (bright fringe), the path difference between the two interfering waves must be an integral multiple of the wavelength: path difference = n lambda, where n = 0, 1, 2, ... For destructive interference (dark fringe), the path difference must be an odd multiple of half the wavelength: path difference = (2n - 1) lambda/2. Equivalently, the phase difference is 2n pi for constructive and (2n - 1) pi for destructive interference.
constructive: n lambda; destructive: (2n-1) lambda/2
Marking-scheme points
- ✓Constructive: path difference = n lambda
- ✓Destructive: path difference = (2n - 1) lambda/2
- ✓Phase difference 2n pi (bright) or (2n-1) pi (dark)
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Wave Optics
What are coherent sources? State the conditions for obtaining sustained interference of light.
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Coherent sources are two sources of light that emit waves of the same frequency (or wavelength) and have a constant phase difference between them. Conditions for sustained (steady) interference: (1) the two sources must be coherent; (2) they must have the same frequency and nearly equal amplitudes; and (3) they must be narrow and close together, and the light should preferably be monochromatic. In practice, coherent sources are obtained from a single source (for example, using two slits).
Marking-scheme points
- ✓Coherent sources: same frequency and constant phase difference
- ✓Need equal frequency and nearly equal amplitude
- ✓Obtained from a single source (e.g. two slits)
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Wave Optics
What is diffraction of light? Write the condition for minima in single slit diffraction.
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Diffraction is the bending of light around the edges of an obstacle or aperture and its spreading into the geometrical shadow region. In diffraction at a single slit of width a, a central bright maximum is flanked by alternate dark and bright fringes. The condition for the dark fringes (minima) is a sin theta = n lambda, where n = 1, 2, 3, ... and theta is the angle of diffraction. The central maximum is the brightest and widest.
a sin theta = n lambda
Marking-scheme points
- ✓Diffraction: bending/spreading of light around obstacles/apertures
- ✓Single slit minima: a sin theta = n lambda (n = 1, 2, 3...)
- ✓Central maximum is the brightest and widest
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Wave Optics
State two differences between interference and diffraction of light.
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(1) Interference is due to the superposition of waves from two (or more) different coherent sources, whereas diffraction is due to the superposition of secondary wavelets coming from different parts of the same wavefront. (2) In interference all bright fringes are of equal intensity and equal width, whereas in diffraction the central maximum is the brightest and the intensity of the secondary maxima decreases rapidly on either side.
Marking-scheme points
- ✓Interference: two coherent sources; diffraction: parts of the same wavefront
- ✓Interference fringes: equal width and intensity
- ✓Diffraction: central maximum brightest, others decrease
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Wave Optics
What is polarisation of light? State Brewster's law.
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Polarisation is the phenomenon of restricting the vibrations of the electric field of a light wave to a single plane perpendicular to the direction of propagation; it shows that light is a transverse wave. Brewster's law states that when unpolarised light is incident on a transparent surface at a particular angle called the polarising angle (theta_p), the reflected light is completely plane-polarised, and the refractive index of the medium is n = tan(theta_p).
n = tan(theta_p)
Marking-scheme points
- ✓Polarisation restricts vibrations to one plane (light is transverse)
- ✓At the polarising angle, reflected light is fully plane-polarised
- ✓Brewster's law: n = tan(theta_p)
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Dual Nature of Radiation and Matter
What is the photoelectric effect?
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The photoelectric effect is the phenomenon of emission of electrons (called photoelectrons) from the surface of a metal when light of suitable frequency (usually ultraviolet or visible for some metals) falls on it. The emitted electrons carry kinetic energy. The effect occurs only when the frequency of the incident light is greater than a certain minimum value called the threshold frequency, and it provided evidence for the particle (photon) nature of light.
Marking-scheme points
- ✓Emission of electrons from a metal when light falls on it
- ✓Occurs only above the threshold frequency
- ✓Evidence for the particle (photon) nature of light
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Dual Nature of Radiation and Matter
Write Einstein's photoelectric equation and explain each term.
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Einstein's photoelectric equation is h f = W0 + KE(max), where h f is the energy of the incident photon (h is Planck's constant and f the frequency of light), W0 is the work function (the minimum energy needed to eject an electron from the metal surface), and KE(max) is the maximum kinetic energy of the emitted photoelectron. It expresses conservation of energy: the photon's energy is used partly to free the electron and the rest appears as its kinetic energy. Thus KE(max) = h f - W0.
h f = W0 + KE(max)
Marking-scheme points
- ✓h f = W0 + KE(max)
- ✓h f = photon energy; W0 = work function
- ✓KE(max) = h f - W0 (energy conservation)
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Dual Nature of Radiation and Matter
Define work function and threshold frequency.
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The work function (W0) of a metal is the minimum energy required to just remove an electron from the surface of the metal without giving it any kinetic energy. The threshold frequency (f0) is the minimum frequency of the incident light below which no photoelectric emission takes place, however intense the light may be. They are related by W0 = h f0, where h is Planck's constant.
W0 = h f0
Marking-scheme points
- ✓Work function W0 = minimum energy to free an electron
- ✓Threshold frequency f0 = minimum frequency for emission
- ✓Relation: W0 = h f0
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Dual Nature of Radiation and Matter
Light of wavelength 400 nm is incident on a metal of work function 2 eV. Calculate the maximum kinetic energy of the emitted photoelectrons. (Use hc = 1240 eV nm)
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The energy of the incident photon E = hc/lambda = 1240/400 = 3.1 eV. Using Einstein's equation, the maximum kinetic energy KE(max) = E - W0 = 3.1 - 2.0 = 1.1 eV. Since the photon energy (3.1 eV) is greater than the work function (2 eV), emission occurs and the photoelectrons have a maximum kinetic energy of 1.1 eV.
KE(max) = hc/lambda - W0
Marking-scheme points
- ✓Photon energy E = hc/lambda = 1240/400 = 3.1 eV
- ✓KE(max) = E - W0 = 3.1 - 2.0
- ✓KE(max) = 1.1 eV
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Dual Nature of Radiation and Matter
State any two laws of the photoelectric effect.
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(1) For a given metal, photoelectric emission occurs only if the frequency of the incident light is greater than a certain minimum value (threshold frequency), whatever the intensity. (2) The maximum kinetic energy of the emitted photoelectrons depends on the frequency of the incident light and the nature of the metal, but is independent of the intensity of the light. (3) The number of photoelectrons emitted per second (photoelectric current) is directly proportional to the intensity of the incident light. (4) The emission is instantaneous, with no measurable time lag.
Marking-scheme points
- ✓Emission only above the threshold frequency
- ✓Max KE depends on frequency, not on intensity
- ✓Number of photoelectrons is proportional to intensity; emission is instantaneous
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Dual Nature of Radiation and Matter
What is the de Broglie hypothesis? Write the expression for the de Broglie wavelength.
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The de Broglie hypothesis states that every moving particle has a wave associated with it, called a matter wave. The de Broglie wavelength is lambda = h/p = h/(m v), where h is Planck's constant, p is the momentum, m is the mass and v the velocity of the particle. In terms of kinetic energy, lambda = h/sqrt(2 m KE). This shows the dual (wave-particle) nature of matter; the wavelength is significant only for very small particles like electrons.
lambda = h/(m v)
Marking-scheme points
- ✓Every moving particle has an associated matter wave
- ✓lambda = h/p = h/(m v)
- ✓In terms of KE: lambda = h/sqrt(2 m KE)
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