### Sample Papers

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

### 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

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

Angle   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  /2, and path difference between the secondary waves from B and C reaching P will again be  /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 /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,

Width of secondary minima is given by,

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

b) Given,

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,

Therefore, separation is given by,

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,

Since, D >> d, so  is very small.

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

So, we have

Let, be the distance of two consecutive fringes. Then, we have

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,

1800 Views

33.

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

(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,

= NBIA

Restoring torque in the spring = k

Therefore,

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,
Voltage sensitivity in the galvanometer is given by,

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

3288 Views