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** MULTIPLE CHOICE QUESTIONS **

**Q.1. ****Consider a light beam incident from air to a glass slab at Brewsterâ€™s angle as shown in Fig.****A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.(a) For a particular orientation there shall be darkness as observed through the polaoid.(b) The intensity of light as seen through the polaroid shall be independent of the rotation.(c) The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the polaroid.(d) The intensity of light as seen through the polaroid shall go through a minimum for four orientations of the polaroid.**

From figure, it is clear that Î¸

Also n = tan Î¸

(i) For i < Î¸

Both reflected and refracted rays becomes partially polarised.

(ii) For glass Î¸

If a light beam is incident on a glass slab at Brewsterâ€™s angle, the transmitted beaiji is unpolarised and reflected beam is polarised.

In the given figure, the light beam is incident from air to the glass slab at Brewsterâ€™s angle (i

As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

light: The phenomenon of limiting the vibrating of electric field vector in one direction in a plane perpendicular to the direction of propagation of light wave is called polarization of light.

(i) The plane in which oscillation occurs in the polarised light is called plane of oscillation.

(ii) The plane perpendicular to the plane of oscillation is called plane of polarisation.

(iii) Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.

(A) Transmission axes of the polariser and analyser arc parallel to each other, so whole of the polarised light passes through analyser

(B) Transmission axis of the analyser is perpendicular to the polariser. hence no light passes through the analyser

(a) Be a fine sharp slit white in colour at the center.

(b) A bright slit white at the center diffusing to zero intensities at the edges.

(d) Only be a diffused slit white in colour.

Width of slit 10

Wavelength of visible light varies from 4000 to 8000 Ã…. As the width of slit 10000 Ã… is comparable to that of wavelength of visible light i.e. 8000 Ã….

Hence the diffraction occurs with maxima at the centre. So at the centre all colours appear i.e. white colour at the centre appear.

Consider a ray of light ABCD through prism, and reflected rays BE and CF from incidence points B and C as shown in figure.

The time difference between two reflected ray BE and CF is equal to the time taken by ray to travel from B to C.

âˆ´ Time difference dt between two reflected rays BE and CF are

v

Substitute (II) and (III) in (I),

The phase diff. between ray AB and BC after refraction is Ï€

âˆ´ Net phase difference = dÏ•' + Ï€

[it is very near to option (a).]

(a) There shall be alternate interference patterns of red and blue.

(b) There shall be an interference pattern for red distinct from that for blue.

(c) There shall be no interference fringes.

(d) There shall be an interference pattern for red mixing with one for blue.

For sustained interference, the source must be coherent and should emit the light of same frequency.

In this problem one hole is covered with red and other with blue, which has different frequency, so no interference takes place.

(a) There would be no interference pattern on the second screen but it would be lighted.

(b) The second screen would be totally dark.

(d) There would be a regular two slit pattern on the second screen.

At P

S

(a) S

(b) S

(c) S

(d) S

(i) As the intensity at dark fringe is zero so intensities of S

(ii) As the graph of maxima and minima is symmetric. So the waves from S

(a) A sharp white ring.

(b) Different from a geometrical image.

(c) A diffused central spot, white in colour.

(d) Diffused coloured region around a sharp central white spot

The width of pinhole 10

So light from pinhole will diffract from the hole. Due to the diffraction pattern of fringes, the shape are quite different from hole.

(a) The size decreases.

(b) The intensity increases.

(c) The size increases.

(d) The intensity decreases.

We know that width (B

So on increasing width of hole of pinhole, â€˜dâ€™ increase. Hence the size of central maxima decreases verifies the option (a).

As the energy passing through hole increased on increasing the size of hole. So the intensity of pattern will increase. Hence verifies the option (b).

(a) The wavefront is spherical.

(b) The intensity decreases in proportion to the distance squared.

(c) The wavefront is parabolic.

(d) The intensity at the wavefront does not depend on the distance.

Light from point source emits in all around the source with same speed so forms a spherical surface of wavefront or spherical wavefront.

As the intensity (I) always decreases as the reciprocal of square of distance.

r = radius of spherical wavefront at any time (r = vt)

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.10. Is Huygenâ€™s principle valid for longitudunal sound waves?****Ans. **Consider a source of sound formed with the compressions and rarefactions forward in all directions with same velocity. So longitudinal waves propagate with spherical symmetry in all directions as the wavefront in light waves. So Huygenâ€™s principle is valid for longitudinal sound waves also.

On a surface of sphere there will be either compression or rarefaction and that part can also behave like a source of sound but with low intensity.**Q.11. ****Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?****Ans. **Consider a point â€˜Fâ€™ on focus of converging lens L_{1}. The light rays from F, becomes parallel after refraction through L_{1}. When these parallel rays falls on converging lens L_{2 }placed co-axial on the other side of F of L_{1}. L_{2} converges the rays at itâ€™s focus at I. It now behave like a point source of rays and form a spherical wave front.**Q.12. What is the shape of the wavefront on earth for sunlight?****Ans. **As the sun is very-very far from the earth so can be considered at infinity and sun can be considered as a point source which gives spherical wavefront. The size of the earth is very small as compared to distance of sun from earth and size of the sun so the plane wavefront reaches on earth as shown in figure here.**Q.13. ****Why is the diffraction of sound waves more evident in daily experience than that of light wave?****Ans. **We know that frequencies of sound waves varies from 20 Hz to 20,000 Hz, so its corresponding wavelength varies from 15m to 15mm respectively. The size of slit (almost) becomes comparable to wavelength of sound so diffraction of sound wave takes place easily.

But the wavelength of visible light varies from 0.4 to 0.7 micron which is very small. So the size of most of the slits in not comparable with wavelength of visible light, due to this diffraction of light cannot take place.**Q.14. ****The human eye has an approximate angular resolution of Ï† = Ã— 5.8 10 ^{-4} rad and a typical photo printer prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.**

The average distance between any two dots

At the distance z cm, angle subtended

Resolution angle for human

Maximum distance up to which human eye cannot see

Which is less than distance of distinct vision.

So a normal person cannot see the dots.

Now polariod (II) is rotated till no light passes. In this situation the pass axis of polariod (II) is perpendicular to polariod (I), then (I) and (II) are set in crossed positions. No light passes through polaroid-II.

Now third polaroid (III) is now placed in between (I) and (II). Only in the special case s w hen the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging. In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Now polariod (II) is rotated till no light passes. In this situation the pass axis of polariod (II) is perpendicular to polariod (I), then (I) and (II) are set in crossed positions. No light passes through polaroid-II.

Now third polaroid (III) is now placed in between (I) and (II). Only in the special cases when the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging. In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

**SHORT ANSWER TYPE QUESTIONS**

**Q.16. ****Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?****Ans. **If angle of incidence is equal to Brewsterâ€™s angle, the transmitted light is slightly polarised and reflected light is plane polarised.

Polarisation by reflection occurs when the angle of incidence is the Brewsterâ€™s angle.

i.e.,

When the light rays travels in such a medium, the critical angle is

As |tan i_{B}| > |sin i_{C}| for large angles i_{B} < i_{C}.

Thus, the polarisation by reflection occurs definitely.**Important point:** Brewsterâ€™s angle (also known as the polarization angle) is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incidents this angle, the light that is reflected from the surface is therefore perfectly polarized. This special angle of incidence is named after the Scottish physicist Sir David Brewster.**Q.17. ****For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Ã… and electrons accelerated through 100V used as the illuminating substance.****Ans. **Î» = 5000 Ã… = 5000Ã—10^{-10}m

In microscope,

Limit of resolution by light of 5000 Ã…

d_{min} = or d_{min} =

The de Broglie wavelength Î»_{d} of illuminated light =

The limit of resolution by 100 V light d'_{min} =

d'_{min} = **Q.18. ****Consider a two slit interference arrangements (Fig.) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of Î» such that the first minima on the screen falls at a distance D from the centre O.****Ans. **According to Î¸,

x = D (Given) ....(I)

d = 2D

Path difference at P = S_{2} P-S_{1} P

Substitute the value of d and x from I and II

The path difference for nth dark fringe from central maxima â€˜Oâ€™ is

âˆ´ For Ist minima p = Î»/2

Put the value of p in (III)

Rationalising the denominator, we get,

**SHORT ANSWER TYPE QUESTIONS**

**Q.19. ****Figure shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I _{0} is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.Ans. **The resultant amplitude of wave reaching on screen will be the sum of amplitude of either wave in perpendicular and parallel polarisation.

Amplitude of the wave in perpendicular polarisation

Amplitude of the wave in parallel polarisation

â‡’

âˆ´ Average Intensity on the screen

â‡’

With polariser P.

Given, = Intensity without polariser at principal maxima.

Intensity at first minima with polariser

âˆ´ Î”x = 2d sin Î¸ + (Î¼ - 1)L

For principal maxima, path difference Î”x = 0

âˆ´ 2d sinÎ¸

For central maxima,

For small

For the first minima the path difference = +Î»/2

(For both upper and lower side from O]

For diffraction, Î»=d (half the slit dist.)

For positive direction side, sin

For negative direction side,

The distance of first minima from principal maxima on either side are

[above point O on screen]

The first minima starts after the end of central maxima. So the Ist minima starts at distance of D tan above O in positive direction.

The distance of first minima on the negative side is

(ii) Which of the two receivers picks up the larger signal when B is turned off?

(iii) Which of the two receivers picks up the larger signal when D is turned off?

Let the equation of wave at R

y

The path difference of the signal from A with that from B is Î»/2 and hence, the phase difference

Thus, the wave equation at R

y

The path difference of the signal from C with that from A is A and hence the phase difference

Thus, the wave equation a R

y

The path difference between the signal from D with that of A is

Therefore, phase difference is Ï€.

âˆ´ y

The resultant signal picked up at R

Thus, the signal picked up at R

Now let us consider the resultant signal received at R

y

The path difference of the signal from D with that from B is Î»/2 and hence, the phase difference

Thus, the wave equation at R

y

The path difference between signal at A and that at B is

As d â‰« Î», therefore this path differences

Hence, y

Similarly, y

The resultant signal picked up at R

âˆ´ Signal picked up by

âˆ´

Thus. R

(ii) If B is switched off.

R

R

Thus, R

(iii) If D is switched off,

R

R

Thus. R

(iv) Thus, a signal at R

(i) According to the description above show that if rays of light enter such a medium from air (refractive index =1) at an angle Î¸ in 2

(ii) Prove that Snellâ€™s law holds for such a medium.

Again consider figure (i), let A B represent the incident wavefront and D E represent the refracted wavefront. All points on a wavefront must be in same phase and in turn, must have the same optical path length.

Thus

or

As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig. 2)

Then

or

If BC > 0, then CD > AE

which is obvious from Fig. (i). Hence, the postulate is reasonable.

However, if the light proceed in the sense it does for ordinary material, (going from 2nd quadrant to 4th quadrant) as shown in Fig. (i). then proceeding as above,

or

As AE > CD, therefore BC < 0 which is not possible. Hence, the given postulate is correct.

(ii) From Fig. (i),

and CD - AE = AC sin Î¸_{r}

As

âˆ´

or

which proves Snellâ€™s law.**Q.23. ****To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens).A typically used dielectric film is MgF _{2} (n = 1.38). What should the thickness of the film be so that at the center of the visible spectrum (5500 Ã…) there is maximum transmission.**

The amplitude (intensity) of wave during refraction and reflection decreases.

If the interference due to two reflected rays AR

Both reflections are from lower to higher refractive index surfaces So, optical path difference between AR

Î¼(AD + CD) - AB ...(I)

If d is the thickness of film then,

AB = AC sin i

âˆ´ d tan r = AC/2

AC = 2d tan r

Or

AB = 2d tan r sin i ...(III)

So the optical path difference from (I)

Optical path difference between AR

For two rays AR

âˆ´ 2d Î¼ cos r = Î»/2

Î¼d cos r = Î»/4

For photographic lenses the sources are vertical planes i.e., rays incident at very small angle.

d = 1000Ã…

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