Subject

Physics

Class

ICSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

31. (i) In an electromagnetic wave, how are electric vector E, magnetic vector B and velocity of propagation of the wave (c) oriented?

ii) How long would gamma radiation take to travel from sun to earth, a distance of 1.5x1011m?
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32.

With the help of a labelled diagram, show that fringe separation β in Young’s double slit experiment is given by:

                               straight beta space equals space λD over straight d
where the terms have their usual meaning.

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33.

(i) What is the difference between polarised light unpolarised light based on the direction of electric vector E?

(ii) What will be the effect on the width of the central bright fringe in the diffraction pattern of a single slit if :

(1)    Monochromatic light of smaller wavelength is used.

(2)    Slit is made narrower.

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34.

At what angle, a ray of light should be incident on the first face AB of a regular glass prism ABC so that the emergent ray grazes the adjacent face AC? See Figure 7 below. (Refractive Index of glass = 1.6)

 

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35. A convex lens ‘L’ and a plane mirror ‘M’ are arranged as shown in Figure 8 below. Position of object pin ‘O’ is adjusted in such a way that the inverted image ‘I’ formed by the lens mirror combination, coincides with the object pin ‘s. Explain how and when this happens.


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36. Starting with an expression for refraction at a single spherical surface obtain an expression for lens maker’s formula. 


Lens makers formula: It relates focal length of a lens of the refractive index of material of lens and radii of curvature of two surfaces of lens.  

Let us consider a conves lens of glass having RI n2 placed in rarer medium of Rln1. P1 and P2 be poles of C1 and C2. R1= (P1C1), R2 = (P2C2). τ= P1P2(thickness of lens).

Let a point object O be placed on principal axis at a distance is u from P1. A ray OA travelling in rarer medium is incident on first surface of lens. It is refracted towards normal at A.

If the second surface of lens were absent, the ray AB would have met the principle axis at I’ which is image of V.

Let the distance of I1 from P1 be V1

straight n subscript 2 over straight v subscript 1 space minus space n subscript 1 over u space equals space fraction numerator n subscript 2 space minus space n subscript 1 over denominator R end fraction space space space space space space space... space left parenthesis 1 right parenthesis

The roy AB travelling in medium n2 is actually incident an second surface of lens and refracted into rarer medium n1 bending away from normal at B. The emergent ray meets principal axis at I which is final real image of O.

For refraction at second surface I acts as vertual image and I is real image, v → distance of I from P2

Distance space of space straight I apostrophe space from space straight P subscript 2 space equals space left parenthesis straight v apostrophe minus straight t right parenthesis

straight n subscript 1 over straight v space minus space fraction numerator straight n subscript 2 over denominator straight v apostrophe minus straight t end fraction space equals fraction numerator straight n subscript 1 space minus space straight n subscript 2 over denominator straight R subscript 2 end fraction

Lens space is space thin space straight t space less than less than space vl
straight n subscript 1 over straight v space minus space fraction numerator straight n subscript 2 over denominator straight v apostrophe end fraction space equals space fraction numerator straight n subscript 1 space minus space straight n subscript 2 over denominator straight R subscript 2 end fraction space space... space left parenthesis 2 right parenthesis

Adding space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space get
straight n subscript 1 over straight v space minus space straight n subscript 1 over straight u space equals space left parenthesis straight n subscript 1 space minus space straight n subscript 2 right parenthesis open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses

straight n subscript 2 over straight n subscript 1 equals straight n space is space the space refractive space index space
of space the space material space of space space the space lens space straight w. straight r. straight t space
surrounding space medium. space

1 over straight v minus 1 over straight u equals left parenthesis straight n minus 1 right parenthesis space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses

When space object space is space at space infinity comma space the space image
space will space be space formed space at space principal space focus

straight u space equals straight a space semicolon space straight v space equals space straight f
because space 1 over straight f minus 1 over infinity space equals space left parenthesis straight n minus 1 right parenthesis

Distance space of space straight I apostrophe space from space straight P subscript 2 space equals space left parenthesis straight v apostrophe minus straight t right parenthesis

straight n subscript 1 over straight v space minus space fraction numerator straight n subscript 2 over denominator straight v apostrophe minus straight t end fraction space equals fraction numerator straight n subscript 1 space minus space straight n subscript 2 over denominator straight R subscript 2 end fraction

Lens space is space thin space straight t space less than less than space vl
straight n subscript 1 over straight v space minus space fraction numerator straight n subscript 2 over denominator straight v apostrophe end fraction space equals space fraction numerator straight n subscript 1 space minus space straight n subscript 2 over denominator straight R subscript 2 end fraction space space space space space... space left parenthesis 2 right parenthesis

Adding space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space get
straight n subscript 1 over straight v space minus space straight n subscript 1 over straight u space equals space left parenthesis straight n subscript 1 space minus space straight n subscript 2 right parenthesis open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses
straight n subscript 2 over straight n subscript 1 equals straight n space is space the space refractive space index space
of space the space material space of space the space lens space straight w. straight r. straight t space
surrounding space medium. space

1 over straight v minus 1 over straight u equals left parenthesis straight n minus 1 right parenthesis space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses

When space object space is space at space infinity comma space the space image space
will space be space formed space at space principal space focus

straight u space equals straight a space semicolon space straight v space equals space straight f

because space 1 over straight f minus 1 over infinity space equals space left parenthesis straight n minus 1 right parenthesis thin space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses

space 1 over straight f space equals space left parenthesis straight n minus 1 right parenthesis open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses space 

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37. Show that the axial chromatic aberration (fr-fv) for a convex lens is equal to the product of its mean focal length (f) and dispersive power (ω ) of its material i.e. Prove : (fr-fv ) = ω f.
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38.

Draw a labelled diagram of an image formed by a compound microscope, with the image at least distance of distinct vision. Write any one expression for its magnifying power.

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39. What is meant by long-sightedness? How can this defect be corrected?
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40.

(i) What is meant by ‘Quantization of charge? 

(ii) In Thomson’s experiment, prove that the ratio of charge to the mass (e/m) of an electron is given by :

straight e over straight m space equals space fraction numerator 1 over denominator 2 V end fraction E squared over B squared
where the terms have their usual meaning.

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