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ISC Class 12 Physics Solved Question Paper 2010

Short Answer Type

31.

(i) What is a continuous emission spectrum? Name one source of light which produces such a spectrum.

(ii) Explain in brief why dark lines are observed in the solar spectrum.

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32. An equiconvex lens of glass, having focal length of 10 cm is split into two identical Plano convex lenses each having focal length fj as shown below.

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33. An illuminated point object O is kept 20 cm from a thin convex lens L₁ of focal length 15 cm as shown below. A thin diverging lens L₂ of focal length 25 cm is kept co-axial with the first lens and 35 cm from it, as shown in Figure 10. Find the position of the final image formed by this combination of lenses.

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34.

(i) What are coherent sources?

(ii) In young’s double slit experiment, what is the path differences between the two light waves forming 5^th bright band (fringe) on the screen?

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35. State one similarity and one difference between interference of light and diffraction of light.

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36. A ray EF of monochromatic light is incident on the refracting surface AB of a regular glass prism (refractive index = 1.5) at an angle of incidence of i = 55° (Figure 11). If it emerges through the adjacent face AC, calculate the right angle of emergence ‘e’.

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37.

(i) In case of polarized light, what is means by the plane of polarization?

(ii) Find refractive index of glass if its polarising angle is 60°.

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38.

(i)Explain the statement: “Angular magnification of a compound microscope in normal use is 30”.

(ii) State how the resolving power of an astronomical telescope can be increased.

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39.
In Millikan’s oil drop experiment, the two plates are 2 cm apart. When a potential differences of 2355 V is applied between them, an oil drop of radius 1μm is found to remain suspended. Calculate the number of excess electrons on the drop,

[Density of oil = 900 kg/m ³=Density of air may be ignored]

Under the equillibrium,

qE= mg

where

q= change on the oil drop,

E= Applied electric field between the places

$straight q space straight x space straight V over straight l space equals space mg straight q space equals space mgl over straight v where comma space straight V space equals space applied space potential straight g space equals space acceleration space due space to space gravity straight q space equals space fraction numerator begin display style bevelled fraction numerator straight D over denominator Vol. end fraction end style space straight x space straight g space straight x space straight l over denominator straight V end fraction straight D space equals space 900 space kg divided by straight m cubed space semicolon space straight g space equals space 9.8 space straight m divided by straight s squared semicolon space straight l space equals space 2 space cm space equals space 0.2 space straight m straight V space equals 2355 space straight V Using space these space values space in space the space above space formula comma space straight q space equals fraction numerator 900 over denominator begin display style bevelled 4 over 3 end style space straight x space straight Z space straight x space straight r cubed end fraction space straight x space fraction numerator 9.8 space straight x space 0.2 over denominator 2355 end fraction space space space equals space 0.178 space straight x space 10 to the power of 18 space space space space equals space 178 space straight x space 10 to the power of 14 space end exponent straight C straight q space equals space ne No. space of space excess space electron space equals space straight q over straight e space equals space fraction numerator 178 space straight x space 10 to the power of 14 space end exponent over denominator 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 111.25 space straight x space 10 to the power of 33 space electrons$

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40. Figure 12 below shows a simple X ray tube. P₁ and P₂ are power supplies which generate 6 V and 40,000 V respectively. Show how you will connect these power supplies to the X ray tube so that it starts producing X rays.