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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.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

93 questions · clear filters

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.

Reveal model answer + marking points▾

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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PhysicsClass 123 marksmedium

Wave Optics

Write the expression for the fringe width in Young's double slit experiment and explain the terms.

Reveal model answer + marking points▾

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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PhysicsClass 123 marksmedium

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.

Reveal model answer + marking points▾

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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