Use the mirror equation to show that: (a) an object placed betwe

Subject

Physics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

21.

A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33.

(a) Will it behave as a converging or a diverging lens in the two cases?

(b) How will its focal length change in the two media?
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22.

Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, v1 >v2 , of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer. 

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23.

Write briefly any two factors which demonstrate the need for modulating a signal.

Draw a suitable diagram to show amplitude modulation using a sinusoidal signal as the modulating signal. 

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24.

Use the mirror equation to show that:

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.


Mirror equation is given by,

space space space space space space 1 over straight f space equals space 1 over v plus 1 over u

rightwards double arrow space space space 1 over v space equals space 1 over f space minus space 1 over u 

a)

For a concave mirror, f is negative, i.e., f < 0

For a real object i.e., which is on the left side of the mirror, 

For u between f and 2f implies that 1/u lies between 1/f and 1/2f

i.e.,   space space space space space fraction numerator 1 over denominator 2 straight f end fraction greater than fraction numerator 1 over denominator straight u space end fraction greater than 1 over straight f space left parenthesis as space straight u comma straight f space are space negative right parenthesis

rightwards double arrow space 1 over straight f minus fraction numerator 1 over denominator 2 straight f end fraction less than space 1 over straight f minus fraction numerator 1 over denominator straight u space end fraction less than space 0

rightwards double arrow space fraction numerator 1 over denominator 2 straight f end fraction less than space 1 over straight v less than 0

straight i. straight e. comma space space space space 1 over straight v space is space negative. space

Implies, v is negative and greater than 2f. Therefore, image lies beyond 2f and it is real.

b)

Focal length is positive for convex mirror, i.e., f > 0.

For a real object on the left side of the mirror, u is negative.

1 over straight f space equals 1 over v space plus space 1 over u

That is, 

1 over straight v space equals space 1 over straight f space minus space 1 over straight u

Since u is negative and f is positive so, 1/v should also be positive, so v must be positive.

Hence, image is virtual and lies behind the mirror.

c) 

Using the mirror equation, we have

1 over straight v space equals space 1 over straight f space minus space 1 over straight u

For space straight a space concave space mirror comma space straight f space is space negative semicolon space straight f space less than space 0

As space straight u space is space also space negative comma space so space straight f space less than space straight u less than 0

This space implies comma space

1 over straight f space minus space 1 over straight u greater than 0

Then space from space left parenthesis 1 right parenthesis comma space 1 over straight v space greater than thin space 0 space or space straight v space is space positive. space

straight i. straight e. comma space image space is space formed space on space the space right space
hand space side space and space is space virtual. space

Magnification comma space straight m space equals negative straight v over straight u space equals space minus fraction numerator straight f over denominator straight u minus straight f end fraction

As space straight u space is space negative space and space straight f space is space positive comma space

magnification comma space straight m space equals space fraction numerator vertical line straight f vertical line over denominator vertical line straight f vertical line space minus space vertical line straight u vertical line end fraction greater than thin space 1 

That is, the image is enlarged.

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25.

(a) Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.

(b) The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
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26.

You are given a circuit below. Write its truth table. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to. 


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27.

A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.

 

OR

 

A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.

If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is 3.8 × 108 m.

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 Multiple Choice QuestionsLong Answer Type

28.

(a) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.

(b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.

(c) How is the magnetic field inside a given solenoid made strong?

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29.

State the working of a.c. generator with the help of a labelled diagram.

The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity. Deduce the expression for the alternating e.m.f. generated in the coil.

What is the source of energy generation in this device?

OR

(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by straight pi/2 in phase.

(b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.

(i) Write the expression for the instantaneous value of the e.m.f. induced in the wire.

(ii) What is the direction of the e.m.f.?

(iii) Which end of the wire is at the higher potential?

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30.

(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.

(b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity.” Justify this statement.

(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range.

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