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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
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
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
What is diffraction of light? Write the condition for minima in single slit diffraction.
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Dual Nature of Radiation and Matter
Write Einstein's photoelectric equation and explain each term.
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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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Atoms
State the postulates of Bohr's model of the hydrogen atom.
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Atoms
Calculate the energy of the photon emitted when an electron in a hydrogen atom jumps from n = 3 to n = 2. (Use En = -13.6/n^2 eV)
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Nuclei
The mass defect of a helium nucleus is 0.0304 u. Calculate its binding energy. (1 u = 931 MeV)
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Nuclei
State the radioactive decay law. What fraction of a radioactive sample remains after 3 half-lives?
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Semiconductor Electronics
Explain how a p-n junction diode works as a half-wave rectifier.
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Semiconductor Electronics
Explain the working of a full-wave rectifier using two diodes.
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Semiconductor Electronics
Write the truth tables of the OR, AND and NOT logic gates for inputs A and B.
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