Subject

Physics

Class

CBSE Class 12

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.

Book Store

Download books and chapters from book store.
Currently only available for.
CBSE

Test Series

Pre-Board Question Papers are now available online for students to practice and excel the exams based on the latest syllabus and marking scheme given by respective boards.

CBSE Physics 2014 Exam Questions

Long Answer Type

31.

(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.

(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10−6 m. The distance between the slit and the screen is 1·5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.


a) Diffraction of light at a Single slit

A single narrow slit is illuminated by a monochromatic source of light. The diffraction pattern is obtained on the screen placed in front of the slits. There is a central bright region called as central maximum. All the waves reaching this region are in phase hence the intensity is maximum. On both side of central maximum, there are alternate dark and bright regions, the intensity becoming weaker away from the center. The intensity at any point P on the screen depends on the path difference between the waves arising from different parts of the wave-front at the slit.



According to the figure, the path difference (BP – AP) between the two edges of the slit can be calculated.

Path difference, BP – AP = NQ = a sin straight theta space almost equal to space aθ

Angle straight theta  is zero at the central point C on the screen, therefore all path difference are zero and hence all the parts of slit are in phase. Due to this, the intensity at C is maximum. If this path difference is, (the wavelength of light used), then P will be point of minimum intensity. This is because the whole wave-front can be considered to be divided into two equal halves CA and CB and if the path difference between the secondary waves from A and B is , then the path difference between the secondary waves from A and C reaching P will be straight lambda /2, and path difference between the secondary waves from B and C reaching P will again be straight lambda /2. Also, for every point in the upper half AC, there is a corresponding point in the lower half CB for which the path difference between the secondary waves, reaching P is straight lambda/2. Thus, destructive interference takes place at P and therefore, P is a point of first secondary minimum. 

Width of secondary maximum is given by, 

space space space space space straight beta straight space equals straight space straight y subscript straight n straight space minus straight space straight y subscript straight n minus 1 end subscript straight space equals straight space nDλ over straight a minus straight space fraction numerator left parenthesis straight n minus 1 right parenthesis Dλ over denominator straight a end fraction

therefore space straight space straight beta straight space equals straight space Dλ over straight a 

Width of secondary minima is given by,  


straight beta '  straight space equals straight space nDλ over straight a minus straight space fraction numerator left parenthesis straight n minus 1 right parenthesis Dλ over denominator straight a end fraction
therefore space beta apostrophe space equals space fraction numerator D lambda over denominator a end fraction

Since angular width is independent of n. all secondary minima and maxima are of the same width.

b) Given,

lambda subscript 1 space equals space 590 space n m space
lambda subscript italic 2 space equals space 596 space n m
Aperture of the slit, a = 2  10 -6 m

Distance between slit and screen = 1.5 m

Therefore, linear separation, between the first maxima (n=1) of the two wavelengths, on the screen, is given by, 

                            fraction numerator 3 left parenthesis straight lambda subscript 2 minus straight lambda subscript 1 right parenthesis over denominator straight a end fraction cross times straight D
Therefore, separation is given by, 


Error converting from MathML to accessible text.

2446 Views

32.

(a) In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.

(b) The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9: 25. Find the ratio of the widths of the two slits.


a)

Let S1 and S2 be two coherent sources separated by a distance d.

Let the distance of the screen from the coherent sources be D. Let point M be the foot of the perpendicular drawn from O, which is the midpoint of S1 and S2 on the screen. Obviously point M is equidistant from S1 and S2. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity.

Now, Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2 P. 

The path difference between two waves reaching at P from S1 and S2 is given by, 

increment space equals space S subscript 2 P minus S subscript 1 P space almost equal to space S subscript 2 N 

Since, D >> d, so angle S subscript 2 S subscript 1 N space equals space theta is very small.

angle S subscript 2 S subscript 1 N space equals space angle M O P space equals space theta 

In space increment space S subscript 1 S subscript 2 N italic comma

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space sin space theta space equals space fraction numerator S subscript italic 2 N over denominator S subscript italic 1 S subscript italic 2 end fraction
I n italic space italic increment M O P italic comma italic space

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space sin space theta space equals space theta space equals space tan space theta italic space

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space fraction numerator S subscript italic 2 N over denominator S subscript italic 1 S subscript italic 2 end fraction space equals space fraction numerator M P over denominator O M end fraction
italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space S subscript italic 2 N space equals space S subscript italic 1 S subscript italic 2 space cross times space fraction numerator M P over denominator O M end fraction space equals space d. y over D

Therefore comma space

Path space difference comma space increment equals S subscript 2 P minus S subscript 1 P space equals S subscript 2 N space equals space fraction numerator y d over denominator D end fraction italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic. italic. italic space italic left parenthesis italic 1 italic right parenthesis

F o r italic space b r i g h t italic space f r i n g e italic comma italic space w e italic space h a v e italic space

fraction numerator y d over denominator D end fraction equals space n lambda italic comma italic space n italic space italic equals italic space italic 0 italic comma italic 1 italic comma italic 2 italic comma italic 3 italic comma italic. italic. italic.

T h e r e f o r e italic comma italic space

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space y subscript n equals fraction numerator n D lambda over denominator d end fraction  

Error converting from MathML to accessible text.

For dark fringe, path difference is an odd multiple of half wavelength.

So, we have  

Error converting from MathML to accessible text. 


Let, straight y subscript straight n plus 1 end subscript straight space and straight space straight y subscript straight n italic spacebe the distance of two consecutive fringes. Then, we have

y subscript n plus 1 end subscript space equals space left parenthesis n plus 1 right parenthesis fraction numerator D lambda over denominator d end fraction a n d space space y subscript n equals fraction numerator n D lambda over denominator d end fraction

So comma space fringe space width space is comma space
space y subscript n plus 1 end subscript italic space minus space space y subscript n space end subscript equals space left parenthesis n plus 1 right parenthesis fraction numerator D lambda over denominator d end fraction minus fraction numerator n D lambda over denominator d end fraction equals fraction numerator D lambda over denominator d end fraction 

Fringe width is same for both bright and dark fringe.

b)

Give, ratio of minima to maxima intensity is given by, 9:25

That is, 

I subscript m i n end subscript over I subscript m a x end subscript equals space fraction numerator left parenthesis a subscript 1 minus a subscript 2 right parenthesis squared over denominator left parenthesis a subscript 1 plus a subscript 2 right parenthesis squared end fraction equals space 9 over 25 

rightwards double arrow straight space fraction numerator straight a subscript 1 minus straight a subscript 2 over denominator straight a subscript 1 plus straight a subscript 2 end fraction equals straight space 3 over 5 semicolon

where space straight a space is space the space amplitude space of space the space slit space and space straight I space the space intensity.

So comma space
space space space space space straight a subscript 1 over straight a subscript 2 equals straight space 4 over 1
fraction numerator straight space straight w subscript 1 over denominator straight w subscript 2 end fraction equals straight space fraction numerator left parenthesis straight a subscript 1 right parenthesis squared over denominator left parenthesis straight a subscript 2 right parenthesis end subscript squared end fraction equals straight space 16 over 1 semicolon space is space the space ratio space of space the

space width space of space the space slit.

1800 Views

33.

(a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.

(b) Answer the following:

(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer?

(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.


b) Moving coil galvanometer:

  

Principle: The underlying principle of moving coil galvanometer is that a current carrying coil, placed in a uniform magnetic field, experiences torque.

Consider a rectangular coil for which no. of turns = N

Are of cross-section is A = lb

Intensity of the uniform magnetic field = B

Current through the coil = I

Therefore,

Deflecting torque is given by,

BIL x b = BIA

For N number of turns,

straight tau= NBIA

Restoring torque in the spring = kstraight theta

Therefore, 

        NBIA space equals space kθ

This space implies comma space

straight I thin space equals space open parentheses straight k over NBA close parentheses straight theta

That space is comma space

straight I space proportional to space straight theta  

b) i) The soft iron coil in a galvanometer will make the field radial. Also, it increases the strength of the magnetic field.

ii) Current sensitivity in the galvanometer is given by, straight theta over straight I space equals space fraction numerator N B A over denominator k end fraction
Voltage sensitivity in the galvanometer is given by, 

straight theta over straight V space equals space fraction numerator theta over denominator I R end fraction space equals space open parentheses fraction numerator n B A over denominator k end fraction close parentheses. begin inline style 1 over R end style

The above two equations imply that increasing the current sensitivity may not necessarily increase the voltage sensitivity.

3288 Views

NCERT Solutions
Textbook Solutions | Additional Questions