(i) Derive an expression for the fringe width ‘y’ in Young�

Subject

Physics

Class

ICSE Class 12

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 Multiple Choice QuestionsShort Answer Type

41. The following equation represents a fusion reaction :

straight H presubscript 1 presuperscript 2 space plus space straight H presubscript 1 presuperscript 3 space rightwards arrow space He presubscript 2 presuperscript 4 space plus space straight n presubscript 0 presuperscript 1 space plus space straight Q space semicolon

space where space straight Q space is space the space enrgy space released

Mass space of space straight H presubscript 1 presuperscript 2 space space atom space equals space 2.014102 space straight u

Mass space of space straight H presubscript 1 presuperscript 3 space atom space equals space 3.016050 space straight u

Mass space of space He presubscript 2 presuperscript 4 space atom space equals space 4.002603 space straight u

Mass space of space space neutron space space straight n presubscript 0 presuperscript 1 space equals space 1.008665 space straight u

Calculate the value of Q.

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42. What is a Zener diode? Explain how it can be used to stabilise the voltage in a circuit.
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43. The output of an OR gate is connected to the input of a NOT gate. Draw the logic circuit of this combination and write its truth table.  
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 Multiple Choice QuestionsLong Answer Type

44.

The current flowing through a conductor is given by I = neAvd. 

(i) Identify each term in the equation.

(ii) Obtain an expression for vd, if the current flowing through the conductor of length I has its ends maintained at a potential difference of V volts.

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45.

(i) Write an expression for Biot-Savart’s law in the vector form. Write an expression for the magnetic field at the centre of a circular coil of radius R, with N turns and carrying a current I. 

(ii) A helium nucleus completes one round of a circle of radius 0.8 m in 2 seconds. Find the magnetic field at the centre of the circle.

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46. Show that straight I subscript rms space equals space fraction numerator straight I subscript straight o over denominator square root of 2 end fraction where Irms is the root mean square value of alternating current and I0 is its peak value.
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47.

(i) Derive an expression for the fringe width ‘y’ in Young’s double slit experiment. Use X as the wavelength of the monochromatic source, D as the distance of the screen from the slits and d as the distance of separation between the slits.

(ii) How does the fringe pattern change if the monochromatic source of light in the above experiment is replaced by a white source of light ?


(i)

Expression for fringe width : Considering a point P at a distance x from C. The path difference between two waves arriving at P.

    = BP -AP 

  space equals space open square brackets D squared space plus space open parentheses x space plus space straight d over 2 close parentheses squared close square brackets to the power of begin inline style bevelled 1 half end style end exponent space minus space open square brackets D squared space plus space open parentheses x space minus space straight d over 2 close parentheses squared close square brackets to the power of begin inline style bevelled 1 half end style end exponent space

equals space D open square brackets 1 space plus space fraction numerator open parentheses x plus begin display style d over 2 end style close parentheses squared over denominator 2 D squared end fraction minus 1 space minus fraction numerator open parentheses x italic minus begin display style d over 2 end style close parentheses squared over denominator 2 D squared end fraction space close square brackets

equals space fraction numerator D over denominator 2 D squared end fraction space open square brackets open parentheses x space plus d over 2 close parentheses squared space minus open parentheses x space minus d over 2 close parentheses squared space close square brackets space equals space fraction numerator x d over denominator D end fraction

T h e r e f o r e comma space f o r space b r i g h t space f r i n g e s space left parenthesis m a x i m u m right parenthesis comma space

fraction numerator x d over denominator D end fraction space equals space eta lambda

eta space equals space space fraction numerator n lambda D over denominator d 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 left parenthesis 1 right parenthesis space

 

where n = 0, 1, ....

for, and dark fringe (minima)

xd over straight D space equals space left parenthesis 2 n minus 1 right parenthesis lambda over 2

x space equals space space left parenthesis 2 n minus 1 right parenthesis space fraction numerator λD over denominator 2 straight d end fraction space space space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis

where comma space straight n space equals space 1 comma 2 comma.... straight n

So fringe width Y is defined as the distance between two consecutive dark fringe or distance between two consecutive bright fringe.



So, 

    Y = xn - xn - 1

 space space space space space equals space left parenthesis 2 straight n space minus 1 right parenthesis straight lambda over 2 space straight D over straight d space minus space open curly brackets 2 space left parenthesis straight n minus 1 right parenthesis space minus space 1 close curly brackets space fraction numerator straight lambda space straight D over denominator 2 straight d end fraction

space space straight y space equals space λD over straight d

straight y space equals space straight x subscript straight n space minus space straight x subscript straight n space minus space 1 space

space space space equals fraction numerator straight n space straight lambda space straight D over denominator straight d end fraction space minus space left parenthesis straight n minus 1 right parenthesis space fraction numerator straight lambda space straight D over denominator straight d end fraction

straight y space equals space fraction numerator straight lambda space straight D over denominator straight d end fraction 

ii)

When we use a source of white light containing light of different colours i.e. (different wavelength X), they at a particular point, constructive interference may be satisfied only for some particular value of X.

Thus, the colour corresponding to this value of X alone shall be visible at that point. Hence the interference fringe will be coloured.

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