3 Marks Question - Physics STD 12 Science Questions - Vidyadip

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3 Marks Question

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55 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer

Wavelength of incident light, $\lambda$ = 5000 $Å = 5000 \times 10^{-10} m$
Speed of light, $c = 3 \times 10^8 m$
Frequency of incident light is given by the relation,
$\text{v}=\frac{\text{c}}{\lambda}$
$=\frac{3\times10^8}{5000\times10^{-10}}=6\times10^{14}\text{Hz}$
The wavelength and frequency of incident light is the same as that of reflected ray.
Hence, the wavelength of reflected light is 5000 Å and its frequency is $6 \times 10^{14} Hz$. When reflected ray is normal to incident ray, the sum of the angle of Incidence, $\angle\text{i}$ and angle of reflection, $\angle\text{r}$ is 90°.
According to the law of reflection, the angle of incidence is always equa I to the angle of reflection. Hence, we can write the sum as:
$\angle\text{i}+\angle\text{r}=90$
$\angle\text{i}+\angle\text{i}=90$
$\angle\text{i}=\frac{90}{2}=45^\circ$
Therefore, the angle of incidence for the given condition is 45°.

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Question 23 Marks

The refractive index of glass is 1.5. What is the speed of light in glass? $($Speed of light in vacuum is $3.0 \times 10^8 ms^{–1}).$
Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

Answer

Refractive index of glass, µ = 1.5

Speed of light, $c = 3 \times 10^8 m/s$

Speed of light in glass is given by the relation,

$\text{v}=\frac{\text{c}}{\mu}$

$=\frac{3\times10^8}{1.5}=2\times10^8\ \text{m/s}$

Hence, the speed of light in glass is $2 \times 10^8 m/s.$

The speed of light in glass is not independent of the colour of light.

The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

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Question 33 Marks

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Answer

Wavelength of the light beam, $\lambda_1$ = 650 nm
Wavelength of another light beam, $\lambda_2$ = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d

Distance of the $n^{th}$ bright fringe on the screen from the central maximum is given by the relation,

$\text{x}=\text{n}\lambda_1\bigg(\frac{\text{D}}{\text{d}}\bigg)$

For third bright fringe, n = 3

$\therefore\ \text{x}=3\times650\frac{\text{D}}{\text{d}}=1950\bigg(\frac{\text{D}}{\text{d}}\bigg)\ \text{nm}$

Let the $n^{th}$ bright fringe due to wavelength, $\lambda_2$ and $(n - 1)^{th}$ bright fringe due to wavelength $\lambda_1$ coincide on the screen. We can equate the conditions for bright fringes as:

$\text{n}\lambda_2=\big(\text{n}-1\big)\lambda_1$

520n = 650n - 650

650 = 130n

$\therefore\ \text{n}=5$

Hence, the least distance from the central maximum can be obtained by the relation:

$\text{x}=\text{n}\lambda_2\frac{\text{D}}{\text{d}}$

$=5\times520\frac{\text{D}}{\text{d}}=2600\frac{\text{D}}{\text{d}}\text{nm}$

Note: The value of d and Dare not given in the question.

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Question 43 Marks

Answer the following question:
Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.

Answer

The size of the obstacle or aperture should be comparable to the wavelength for diffraction of waves by obstacles, through a large angle.
This follows from $\text{sin}\theta=\frac{\lambda}{\text{a}}.$
For light waves,
Wavelength, $\lambda=10^{-7} \text{m and},$
Size of the wall, a = 10 m
$\therefore\ \text{sin}\theta=\frac{10^{-7}}{10}=10^{-8}$
This implies, $\theta\ \rightarrow\ 0$
That is, light goes almost unbent and hence, the students are unable to see each other.
For sound waves,
Frequency, $\nu$ = 1000 Hz
$\text{i.e}\ \frac{\text{c}}{\nu}=\frac{330}{1000}=0.33\ \text{m}$
$\therefore\ \text{sin}\theta=\frac{\lambda}{\text{a}}=\frac{0.33}{10}=0.033\ \text{m}$
Here, $\theta$ has a definite value and the waves will bend around the partition. Hence, students can converse easily.

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Question 53 Marks

For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

Answer

Sound waves require a medium for propagation. Thus, even though both the situations may correspond to the same relative motion (between the source and the observer), they are not identical physically since, the motion of the observer, relative to the medium is different in the two situations. Therefore, we cannot expect Doppler formulas for sound to be identical for (i) and (ii).
For light waves in vacuum, there is clearly nothing to distinguish between two cases given. Here only the relative motion between the source and the observer counts and the relativistic Doppler formula is the same for both cases. For light propagation in a medium, once again like for sound waves, the two situations are not identical and we should expect the Doppler formulas for this case to be different for the two situations (i) and (ii).

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Question 63 Marks

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Answer

Distance between the slits, $d = 0.28\ mm = 0.28 \times 10^{-3}\ m$
Distance between the slits and the screen, $D = 1.4 m$
Distance between the central fringe and the fourth (n = 4) fringe,
$u = 1.2 cm = 1.2 \times 10^{-2} m$
In case of a constructive interference, we have the relation for the distance between the two fringes as:
$\text{u}=\text{n}\lambda\frac{\text{D}}{\text{d}}$
Where,
n = Order of fringes = 4
$\lambda$ = Wavelength of light used
$\therefore\ \lambda=\frac{\text{ud}}{\text{nD}}$
$=\frac{1.2\times10^{-2}\times0.28\times10^{-3}}{4\times1.4}$
$=6\times10^{-7}$
= 600 nm
Hence, the wavelength of the light is 600 nm.

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Question 73 Marks

Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?

Answer

According to Newton's Corpuscular theory of light, when corpuscles of light strike the interface XY, separating a denser medium from a rarer medium, the component of their velocity along XY remains the same.

If $\mathrm{v}_1$ is velocity of light in rarer medium (air), $\mathrm{v}_2$ is velocity of light in denser medium (water), i is angle of incidence, and $r$ is angle of refraction then, Component of $v_1$ along $X Y=v_1 \sin i$ Component of $v_2$ along $X Y=v_2 \sin r A s, v_1 \sin i$ $=\mathrm{v}_2 \sin \mathrm{r} \therefore \frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mu$ Since, $\mu > 1 \therefore \mathrm{v}_2 > \mathrm{v}_1$ i.e., light should travel faster in water than in air. This prediction of Newtons theory is opposite to the experimental result. Huygens wave theory predicts that $\mathrm{v}_2 < \mathrm{v}_1$ which is consistent with experimental conclusion.

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Question 83 Marks

You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Answer

We are given a plane mirror XV and let, 0 be a point object at a distane OP, in front of the plane mirror. A part RPQ of the wavefront touches the plane mirror at P and from this point spherical wavefronts start emanating. Whereas disturbance from Rand Q continue moving forward, along the normal rays OR and OQ, that reflects back v. When, disturbances from R, P and Q reach the mirror at A, B'and C respectively, reflected spehrical wavefront is formed.

The reflected wavefront AB'C, appears to start from I. Hence, I becomes virtual image for O as real point object. Draw AN normal to XV, hence parallel to OP. Now, OA is the incident ray (being normal to incident wavefront ABC) and AD is reflected ray (being normal to reflected wavefront AB'C). Thus, $\angle\text{OAN}=\angle\text{DAN}=\text{e}\ \ [\text{i}=\text{r}]$ $\text{But},\ \angle\text{OAN}=\text{alternate}\ \angle\text{AIP}$ $\therefore\ \angle\text{AOP}=\angle\text{AIP}$ Now, in $\triangle\text{AIP}$ and $\triangle\text{AOP}$ $\angle\text{AIP}=\angle\text{AOP}\ (\text{each e})$ $\angle\text{API}=\angle\text{APO}=90^\circ\ \text{each}\ 90^\circ$ AP is common to both $\triangle_\text{s}$ become congruent Hence, Pl = PQ i.e., normal distance of image from the mirror = normal distance of object from the mirror. Thus, virtual image is formed as much behind the mirror as the object in front of it.

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Question 93 Marks

What is the shape of the wavefront in each of the following cases:

Light diverging from a point source.
Light emerging out of a convex lens when a point source is placed at its focus.
The portion of the wavefront of light from a distant star intercepted by the Earth.

Answer

The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

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Question 103 Marks

States law of Malus.
Draw a graph showing the variation of intensity (I) of polarised light transmitted by an with angle $(\theta)$between polariser and analyser.
What is the value of refractive index of a medium of polarising angle $60^\circ?$

Answer

When the pass axis of a polaroid makes an angle $\theta$ with the plane of polarisation of polarised light of intensity $I_o$ incident on it, then the intensity of the transmitted emergent light is given by $\text{I} = \text{I}_{o}\cos^{2}\theta.$
$\mu = \tan\text{i}_{\beta}$

$ =\tan60^{o} =\sqrt{3} = 1.7.$

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Question 113 Marks

Show, giving a suitable diagram, how unpolarized light can be polarised by reflection.
Two polaroids $P_1$ and $P_2$ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity $I_0$ is incident on $P_1 A$ third polaroid $P_3$ is kept in between $P_1$ and $P_2$ such that its pass axis makes an angle of $60^{\circ}$ with that of $P_1$. Determine the intensity of light transmitted through $P_1, P_2$ and $P_3$.

Answer

When unpolarised light is incident on the boundary between two transparent media, the reflected light gets plane polarized with its electric vector perpendicular to the plane of incidence.

The polarization is complete when the reflected and refracted rays are at right angles to each other. This condition occurs for an angle of incidence, $i_p,$ where tan $i_p = \mu$

Intensity of light through $\text{P}_1=\frac{I_0}{2}$

Intensity of light through $\text{P}_2=\frac{I_0}{2}\cos^260$
$\frac{I_0}{2}.\bigg(\frac{1}{2}\bigg)^2=\frac{I_0}{8}$
Intensity of light through $ \text{P}_3=\frac{I_0}{8}\cos^230=\frac{I_0}{8}\times\bigg(\frac{\sqrt{3}}{2}\bigg)^2=\frac{3I_0}{32}$

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Question 123 Marks

Show, with the help of a diagram, how unpolarised sunlight gets polarised due to scattering.
Two polaroids $P_1$ and $P_2$ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity $I_0$ is incident on $P_1$. A third polaroid $P_3$ is kept in between $P_1$ and $P_2$ such that its pass axis makes an angle of $45^{\circ}$ with that of P 1 . Determine the intensity of light transmitted through $\mathrm{P}_1, \mathrm{P}_2$ and $\mathrm{P}_3$.

Answer

Alternate Answer
The acceleration, of the charges, in the scattering molecules, due to the electric field of the incident radiation, can be in two mutually perpendicular directions.
The observer, however, receives the scattered light, corresponding to only one of these two sets of the accelerated charges.
This causes scattered light to get polarised. Alternatively: The observer receives scattered light corresponding to only one of the two sets of accelerated charges i.e. electrons oscillating perpendicular to the direction of propagation

Intensity of light through $\text{P}_1=\frac{I_0}{2}$

Intensity of light through $\text{P}_2=\frac{I_0}{2}\cos^245$
$\frac{I_0}{2}.\frac{1}{2}=\frac{I_0}{4}$
Intensity of light through $\text{P}_3=\frac{I_0}{4}\cos^245=\frac{I_0}{4}\times\frac{1}{2}=\frac{I_0}{8}$

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Question 133 Marks

In Young’s double slit experiment, the two slits are separated by a distance of 1·5 mm and the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes. Find.

The distance of the third bright fringe for$\lambda = 520$ nm on the screen from the central maximum.
The least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

Answer

Distance of third bright fringe $ - \text{y}_{3} = \frac{\text{n}\lambda\text{D}}{\text{d}}$

$ = \frac{3 \times 520 \times10^{−9}\times 1}{1.5 \times 10^{−3}}$
$ = 1.04 \times 10^{−3}\text{?}\simeq1\text{ ??}$.

Let $n^{th}$ maxima of 650?? coincides with the $(n + 1)^{th}$ maxima of 520 nm.

$\therefore \text{?} \times 650 \times10^{−9} =(\text{ n} + 1 )520 \times 10^{−9}$
$\Rightarrow\text{n} = 4$
$\therefore$ The least distance of the point is given by
$\text{y} =\frac{\text{nD}\lambda_{1}}{\text{d}}$
$ = \frac{4\times 1 \times 650 \times 10^{−9}}{1.5 \times 10^{−3}}\text{m} = 1.733 \times 10^{−3}\text{?}\simeq 1.7\text{??}$.

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Question 143 Marks

Using the phenomenon of polarisation, show how transverse nature of light can be demonstrated.
Two polaroids $\mathrm{P}_1$ and $\mathrm{P}_2$ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity lo is incident on $P_1$. A third polaroid $P_3$ is kept in between $P_1$ and $P_2$ such that its pass axis makes an angle of $30^{\circ}$ with that of $P_1$. Determine the intensity of light transmitted through $P_1, P_2$ and $P_3$.

Answer

Light from the sodium lamp passing through the single Polaroid sheet ( $\mathrm{P}_1$ ) does not show any variation in intensity when this sheet is rotated. However, if the light, transmitted by $\mathrm{P}_1$, is made to pass through another Polaroid sheet $\left(\mathrm{P}_2\right)$ the light intensity, coming out of $\mathrm{P}_2$, varies from a maximum to zero, and again to maximum, when $\mathrm{P}_2$ is rotated. These observations are consistent only with the transverse nature of light waves.

Intensity of light transmitted through $P_1 = I_0 / 2.$

Intensity of light transmitted through $P_3= (I_0/ 2) \times \cos^2 \ 30^0$.

$= 3 I_0 / 8$

Intensity of light transmitted through $\text{P}_{2} = \frac{3}{8}\text{I}_{o}\cos^{2}60^{o}$

$=\frac{3}{32}\text{I}_{o}.$

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Question 153 Marks

What is linearly polarized light? Describe briefly using a diagram how sunlight is polarised.
Unpolarised light is incident on a polaroid. How would the intensity of transmitted light change when the polaroid is rotated?

Answer

Molecules in air behave like a dipole radiator. When the sunlight falls on a molecule, dipole molecule does not scatter energy along the dipole axis, however the electric field vector of light wave vibrates just in one direction perpendicular to the direction of the propagation. The light wave having direction of electric field vector in a plane is said to be linearly polarised. In figure, a dipole molecule is lying along x-axis. Molecules behave like dipole radiators and scatter no energy along the dipole axis.

The unpolarised light travelling along x-axis strikes on the dipole molecule get scattered along y and z directions. Light traversing along y and z directions is plane polarised light.

In figure unpolarised light falls on the polaroid, and transmitted light has electric vibrations in the plane consisting of polaroid axis and direction of wave propagation as shown in Fig.

If polaroid is rotated the plane of polarisation will change, however the intensity of transmitted light remain uncharged.

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Question 163 Marks

How does an unpolarised light get polarised when passed through a polaroid?
Two polaroids are set in crossed positions. A third polaroid is placed between the two making an angle θ with the pass axis of the first polaroid. Write the expression for the intensity of light transmitted from the second polaroid. In what orientations will the transmitted intensity be (i) minimum and (ii) maximum?

Answer

A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid, then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules.
Expression for the intensity transmitted through second Polaroid:
$\text{l} = \text{(I}_{0}\cos^{2}\theta)\cos^{2}(90^{0} - \theta) = \text{I}_{0}(\cos\theta\sin\theta)^{2} = \text{I}_{0}\sin^{2}2 \theta/ 4 $
Where $I_0$ is the intensity of the polarized light after passing through the first polaroid.
Intensity will be maximum when θ = 45° and minimum when θ = 0°

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Question 173 Marks

In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength 450 nm. The screen is 1.0 m away from the slits.

Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum.
How will the fringe pattern change if the screen is moved away from the slits?

Answer

Distance of $n^{th}$ maxima from central maxima$\text{X}_{n} = \frac{\text{n}\lambda\text{D}}{\text{d}}$
Given: n = 2, d = 0.15 mm, λ = 450 nm and D = 1.0 m
$\text{x}_{2} = \frac{2\times450\times10^{-9}\times1.0}{0.15\times10^{-3}} = 6 \times10^{-3}\text{m} = 6\text{mm}$
Distance of $n^{th}$ minima from central maxima
$\text{y}_{2} = \frac{(2\text{n} - 1 )\lambda\text{D}}{2\text{d}} = \frac{(2\times2-1)450\times10^{-9}\times1}{2\times0.15\times10^{-3}} = 4.5\times10^{-3}\text{m} = 4.5\text{mm}$
When the screen is moved away from the slits fringes become farther apart.
(Fringe width α distance of screen.)

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Question 183 Marks

In Young’s double slit experiment, monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8.1 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7.2 mm. Find the wavelength of light from the second source.
What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light?

Answer

$\beta =\frac{\lambda\text{D}}{\text{d}}$
$\therefore\beta_{1}\bigg/\beta_{2} = \lambda_{1}\bigg/\lambda_{2}$
$\therefore\lambda_{2} = \lambda_{1}\beta_{2}\bigg/\beta_{1}$
= 630 × 7.2 / 8.1 = 560 nm
Effect on fringe:
The central fringe is white and there are a few coloured fringes on either side of the central fringe.

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Question 193 Marks

Distinguish between unpolarised and plane polarised light. An unpolarised light is incident on the boundary between two transparent media. State the condition when the reflected wave is totally plane polarised. Find out the expression for the angle of incidence in this case.

Answer

Unpolarised light: Vibrations are symmetrically distributed in all direction
Polarised light: Vibrations are restricted to one plane only.
Condition: $\tan\text{i}_{p} = \mu$
Alternate Answer
Reflected and refracted rays should be perpendicular to each other.

Alternate Answer
Light should be incident at polarizing angle.
$\text{Derivation:}\mu = \frac{\sin\text{i}_{p}}{\sin\text{r}} =\frac{\sin\text{i}}{\sin\bigg(\frac{\pi}{2} - \text{i}_{p}\bigg)}$
$\tan\text{i}_{p}.$

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Question 203 Marks

Draw the intensity distributions for (i) the fringes produced in interference, and (ii) the diffraction bands produced due to single slit. Write two points of difference between the phenomena of interference and diffraction.

Answer

S.no	Interference	Diffraction
1.	All fringes are equal in width	Central bright maxima is twice as wide as the other maxima.
2.	Intensity of all bright fringes is same.	Intensity falls as we go to successive maxima away from centre.
3.	Conditions for maxima and minima are opposite to diffraction pattern.	Condition for maxima and minima are opposite to interference pattern.
4.	Pattern is formed by superposing two waves originating from two narrow slits.	Diffraction pattern is a superposition of wavelets originating from different parts of a single wave front.

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Question 213 Marks

A monochromatic light of wavelength $\lambda$ is incident normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed at a distance D from the slit. With the help of a relevant diagram, deduce the conditions for obtaining maxima and minima on the screen. Use these conditions to show that angular width of central maximum is twice the angular width of secondary maximum.

Answer

The path difference
?? − ?? = ??
= ? sin ? ≃ ??
By dividing the slit into an appropriate number of parts, we find that points P for which

$\theta=\frac{n\lambda}{a}$ are points of minima.
$\theta=\big(n+\frac{1}{2}\big)\frac{\lambda}{a}$ are points of maxima

Angular width of central maxima, $\theta=\theta_1-\theta_{-1}$
$=\frac{\lambda}{a}-\big(-\frac{\lambda}{a}\big)$
$\theta=\frac{2\lambda}{a}$
Angular width of secondary maxima $=\theta_2-\theta_1$
$=\frac{2\lambda}{a}-\frac{\lambda}{a}=\frac{\lambda}{a}$
$=\frac{1}{2}\text{X}$ Angular width of central maxima

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Question 223 Marks

Answer the following questions:

In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is $0.1^\circ$. Find the spacing between the two slits.
Light of wavelength 5000 Å. Propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected?

Answer

Angular width of fringes

$\theta =\lambda/\text{d},$ where d = separation between two slits

Here $\theta = 0.1^{\circ} = 0.1\times\frac{\pi}{180}\text{radian}$

$\therefore\text{d} = \frac{600\times10^{-9}\times180}{0.1\times\pi}$

= 3.43 X 10-4 m

= 0.34 m

For Reflected light: avelength remains same. Frequency remains same.

For Refracted light: Wavelength decreases½ Frequency remains same.

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Question 233 Marks

In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture 2 \times $10^{–4}\ m.$ The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

Answer

When a plane wavefront of monochromatic light illuminates, the slit LN, each point in the slit LN becomes the source of secondary wavelets. The secondary wavelets originating from different points superpose on each, while travelling towards the point C and point P; at angle$\theta$. However the superposition of the secondary wavelets produces a diffraction pattern of varying intensity, as shown in fig.

For maxima other than central maxima

a.$\theta = \bigg(\text{n} + \frac{1}{2}\bigg)\lambda$
and $\theta = \frac{\text{y}}{\text{D}}$
$\therefore\text{a}.\frac{\text{y}}{\text{D}} = \bigg(\text{n} + \frac{1}{2}\bigg)\lambda$

For light of wavelength $\lambda_{1}$= 590 nm
$2 \times10^{-14}\times\frac{\text{y}_{1}}{1.5} =\big(1+ \frac{1}{2}\big)\times590$
$\text{y}_{1} = \frac{3}{2}\times\frac{590\times10^{-9}\times1.5}{2\times10^{-4}}$
= 6.64 mm
For light of wavelength $\lambda_{2}$=596 mm
$2 \times10^{-4}\times\frac{\text{y}_{2}}{1.5} =\big(1+ \frac{1}{2}\big)\times596\text{nm}$
$\Rightarrow\text{y}_{2} =\frac{3}{2}\times\frac{596\times10^{-9}\times1.5}{2\times10^{-4}}$
= 6.705 mm
Separation between two positions of first maxima
$\Delta\text{y} = \text{y}_{2} - \text{y}_{1}$
= 6.705 – 6.64
= 0.065 mm.

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Question 243 Marks

Use Huygens's principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band?

Answer

Explanation:
As per Huygen's Principle
Net effect at any point = sum total of contribution of all wavelets with proper phase difference At the central Point (O) Contribution from each half in $SS_1$ is in phase with that from the corresponding
part in $SS_2.$ Hence, O is a maxima

At the point M where $\mathrm{SM}-\mathrm{SM}_1=\lambda / 2$ Phase difference between each wavelet from $\mathrm{SS}_1$ and corresponding wavelet from $\mathrm{SS}_2=\lambda / 2$ Hence, M would be a minima. All such points (path difference $=\mathrm{n} \lambda / 2$ ) are also minima. Similarly, all points, for which path difference $=(2 n+1) \lambda / 2$, are maxima but with decreasing intensity. From the figure

Half angular width of central maxima= $\lambda$/a $\therefore$, Size of central maxima will be reduced to half and intensity of central maxima will be four times.

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Question 253 Marks

Why are coherent sources necessary to produce a sustained interference pattern?
In Young's double slit experiment using monochromatic light of wavelength$\lambda,$ the intensity of light at a point on the screen where path difference is$\lambda,$ is K units. Find out the intensity of light at a point where path difference is $\lambda/3.$

Answer

Coherent sources are needed to ensure that the positions of maxima and minima do not change with time.

Alternate Answer

Coherent sources have constant phase difference and, therefore, produce a sustained interference pattern.

$\text{I} = \text{I}_{1} + \text{I}_{2} + 2 \sqrt{\text{I}_{1}\text{I}_{2}}\cos\theta$

For path difference $\lambda,$ phase difference

$\Phi = 2\pi$

$\text{Hence}, \text{k} = 4\text{I}_{0}\cos^{2}\pi =4 \text{ I}_{0}$

For path difference $\lambda\big/3$

Phase difference $\Phi = 2\pi\big/3$

Intensity

$\text{I}' = 4\text{I}_{0}\cos^{2}\pi\big/3$

$ = 4\text{I}_{0}\bigg(\frac{1}{2}\bigg)^{2} = \text{I}_{0}$

Therefore, $\text{I}' = \frac{\text{k}}{4}.$

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Question 263 Marks

Describe briefly, with the help of suitable diagram, how the transverse nature of light can be demonstrated by the phenomenon of polarization.
When unpolarized light passes from air to a transparent medium, under what condition does the reflected light get polarized?

Answer

When a polaroid $P_1$ is rotated in the path of an unpolarised light. There is no change in transmitted intensity.
The light transmitted through polaroid $P_1$ is made to pass through polaroid $P_2$.
On rotating polaroid $P_2$ in path of light transmitted from $P_1$ we notice a change in intensity of transmitted light.This shows that light transmitted from $P_1$ is polarized. Since light can be polarized, it has transverse nature.

Whenever the reflected and refracted rays are perpendicular to each other.

Alternate Answer
Whenever unpolarised light is incident from air to a transparent medium at an angle of incidence equal to polarizing angle, the reflected light gets polarized.

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Question 273 Marks

What is an unpolarized light? Explain with the help of suitable ray diagram how an unpolarized light can be polarized by reflection from a transparent medium. Write the expression for Brewster angle in terms of the refractive index of denser medium.

Answer

In an unpolarised light the vibrations of electric field vector are in every plane perpendicular to the direction of propagation of light.

When unpolarised light is incident on the boundary between two transparent media, the reflected light is polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. Brewster angle: $\mu =\tan\text{i}_{p}.$

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Question 283 Marks

In a single slit diffraction experiment, when a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why? State two points of difference between the interference pattern obtained in Young’s double slit experiment and the diffraction pattern due to a single slit.

Answer

Explanation: Waves diffracted at the edge of circular obstacle interfere constructively at the centre of the shadow producing a bright spot.
Points of Differences:

In interference, All bright fringes are of equal intensity, while in diffraction intensity of secondary maxima keeps on decreasing.
Width of all fringes are equal in interference pattern but, in diffraction the width of central maximum and the secondary maxima are different.

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Question 293 Marks

How is a wavefront defined? Using Huygen’s construction draw a figure showing the propagation of a plane wave refracting at a plane surface separating two media. Hence verify Snell’s law of refraction.

Answer

A surface of constant phase/A locus of points which oscillate in same phase.

$\sin\text{i} = \frac{\text{BC}}{\text{AC}}$

$\sin\text{r} = \frac{\text{AE}}{\text{AC}}$
$\therefore \frac{\sin\text{i}}{\sin\text{r}} = \frac{\text{BC}}{\text{AE}}$
$ = \frac{c_1\text{t}}{c_2\text{t}} = \frac{c_1}{\text{c}_{2}} = n =\text{ constant}$

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Question 303 Marks

If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
What kind of fringes do you expect to observe if white light is used instead of monochromatic light?

Answer

$\text{I}_1=\frac{\text{I}}{2}$

$\text{I}_2=\text{I}$

$\frac{\text{I}_{\text{max}}}{\text{I}_{\text{min}}}=\bigg(\frac{\sqrt{\text{I}_1}+\sqrt{\text{I}_2}}{\sqrt{\text{I}_1}-\sqrt{\text{I}_2}}\bigg)^2$

$=\Bigg(\frac{\sqrt{\frac{\text{I}}{2}}+\sqrt{\text{I}}}{\sqrt{\frac{\text{I}}{2}}-\sqrt{\text{I}}}\Bigg)^2$

$=\bigg(\frac{1+\sqrt{2}}{1-\sqrt{2}}\bigg)^2$

$=\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$

$=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

If we use white light, central fringe will be white, then there will be colourful fringes on both sides, starting from violet to red (fringe width $\propto$ wavelength) and then there will be general illumination of screen.

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Question 313 Marks

A resistance R and a capacitor C are connected in series to an ac source $\text{V}=\text{V}_0\sin\omega\text{t}.$

Obtain the expression for the instantaneous current (I) in the circuit.
Show graphically variations of V and I as a function of $\omega\text{t}.$

Answer

$\text{z}=\sqrt{\text{R}^2+\text{X}^2\text{C}}=\text{impedance}$

$\therefore\text{I}=\frac{\text{V}}{\text{Z}}=\frac{\text{v}_1\sin\omega\text{t}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$

$\text{V}_\text{R}=\text{IR}$

$\text{V}_\text{C}=\text{IX}_\text{C}$

In phase of 1 lags I by $\frac{\pi}{2}$

$\text{V}=\sqrt{\text{V}^2\text{R}+\text{V}^2_\text{C}}$

$=\sqrt{(\text{IR})^2+(\text{IX}_\text{C})^2}$

$\Rightarrow\text{I}=\frac{\text{V}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}=\frac{\text{r}_0\sin\omega\text{t}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$

$\phi=$ phase diff bet V & I

$\tan\phi=\frac{\text{V}_\text{C}}{\text{V}_\text{R}}=\frac{\text{X}_\text{C}}{\text{R}}$

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Question 323 Marks

Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15m and an eyepiece of focal length 1.0cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is $3.48 \times 10^6m$, and the radius of lunar orbit is $3.8 \times 10^8m.$

Answer

When the final image is formed at the least distance of distinct vision

Magnifying power, $\text{M}=\frac{\beta}{\alpha}$
$\alpha$ and $\beta$ are small.
$\therefore\text{M}=\frac{\tan\beta}{\tan\alpha}\ ...(\text{i})$
In $\triangle\text{A}'\text{B}'\text{C}_2,$
$\tan\beta=\frac{\text{A}'\text{B}'}{\text{C}_2\text{B}'}$
In $\triangle\text{A}'\text{B}'\text{C}_1,$
$\tan\alpha=\frac{\text{A}'\text{B}'}{\text{C}_1\text{B}'}$
From equation (i), we have:
$\text{M}=\frac{\text{A}'\text{B}'}{\text{C}_2\text{B}'}\times\frac{\text{C}_1\text{B}'}{\text{A}'\text{B}'}$
$\text{M}=\frac{\text{C}_1\text{B}'}{\text{C}_2\text{B}'}$
Here,
$\text{C}_1\text{B}'=+\text{f}_0$
$\text{C}_2\text{B}'=-\text{u}_\text{e}$
$\text{M}=\frac{\text{f}_0}{-\text{u}_\text{e}}\ ...(\text{ii})$
Using the lens equation $\Big(\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Big)$ for the eyepieces, we get
$\frac{1}{-\text{D}}-\frac{1}{-\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$-\frac{1}{\text{D}}+\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}+\frac{1}{\text{D}}$
$\frac{\text{f}_0}{\text{u}_\text{e}}=\frac{\text{f}_0}{\text{f}_\text{e}}\Big(1+\frac{\text{f}_\text{e}}{\text{D}}\Big)$
$\frac{-\text{f}_0}{\text{u}}=\frac{-\text{f}_0}{\text{f}}\Big(1+\frac{\text{f}_\text{e}}{\text{D}}\Big)$
$\text{M}=-\frac{\text{f}_0}{\text{f}_\text{e}}\Big(1+\frac{\text{f}_\text{e}}{\text{D}}\Big)$
Angular magnification is given by,
$\text{m}_0=\Big|\frac{\text{f}_0}{\text{f}_\text{e}}\Big|=\Big|\frac{1500}{1}\Big|=1500$
where, $f_₀$ is the focal length of the objective lens, and $f_e$ is the focal length of the eye-piece.
Given, the diameter of the moon $= 3.48 \times 10^6m$
The radius of the lunar orbit $= 3.8 \times 10^8m$
The diameter of the image of the moon formed by the objective lens is given by, $\text{d}= \alpha\text{f}_₀$
$\text{d}=\frac{\text{Diameter}\ \text{of}\ \text{the}\ \text{moon}}{\text{Radius}\ \text{of}\ \text{the}\ \text{lunar}\ \text{orbit}}\times\text{f}_0$
$\text{d}=\frac{3.48\times10^6}{3.8\times10^8}\times15=13.74\text{cm}$

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Question 333 Marks

In a single slit diffraction experiment, light of wavelength $\lambda$ illuminates the slit of width ‘a’ and the diffraction pattern is observed on a screen.

Show the intensity distribution in the pattern with the angular position $\theta.$
How are the intensity and angular width of central maxima affected when,

Width of slit is increased,
Separation between slit and screen is decreased?

Answer

Intensity increases, angular width decreases.
Intensity increases, no effect on angular width.

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Question 343 Marks

What is the effect on the interference fringes in Young’s double slit experiment due to each of the following operations? Justify your answers.

The screen is moved away from the plane of the slits.
The separation between slits is increased.
The source slit is moved closer to the plane of double slit.

Answer

Linear fringe width increases $\beta=\frac{\lambda\text{D}}{\text{d}}$

$\Rightarrow\beta\propto\text{D}$

No effect on angular fringe width $\Big(\text{Q}=\frac{\lambda}{\text{d}}\Big).$

Both linear fringe width & angular fringe width decrease $\Big(\beta\propto\frac{1}{\text{d}},\text{Q}\propto\frac{1}{\text{d}}\Big).$
If condition $\frac{\text{s}}{\text{S}}<\frac{\lambda}{\text{d}}$ is satisfied, interference will be obtained otherwise, no interference will be obtained.

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Question 353 Marks

The human eye has an approximate angular resolution of $\phi=5.8\times10^{-4}$ rad and a typical photoprinter prints a minimum of 300dpi (dots per inch, 1inch = 2.54cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.

Answer

We are given that, angular resolution of human eye $(\phi)=5.8\times10^{-4}$ red.
The printer prints 300 dots per inch.
The linear distance (l) between two dots can be calculated as $\frac{2.54}{300}\text{cm.}$
$=0.84\times10^{-2}\text{cm}$
Now, at a distance of z cm, this subtends an angle, $\phi=\frac{\text{l}}{\text{Z}}$
Substituting $(\phi)=5.8\times10^{-4}\text{ red and }=0.84\times10^{-2}\text{cm}$, we get
$\text{Z}=\frac{0.84\times10^{-2}\text{cm}}{5.8\times10^{-4}}=14.5\text{cm}.$

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Question 363 Marks

A partially plane polarised beam of light is passed through a Polaroid. Show graphically the variation of the transmitted light intensity with angle of rotation of the Polaroid.

Answer

The partially polarised beam consists of unpolarised plus polarised light. The intensity of unpolarised part varies according to. $\text{I}_\text{p}=\text{(I}_0)_\text{p}\cos^2\theta$ $\therefore\text{I}=\text{(I}_0)_\text{un}+\text{(I}_0)_\text{p}\cos^2\theta$ The graph from $\theta=0\text{ to }\theta=2\pi$ (full rotation) is shown in fig.

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Question 373 Marks

Sketch a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits (keeping other parameters same) in Young’s double slit experiment. What information can one obtain from the slope of this graph?

Answer

We know that the fringe width is given by:$\beta=\lambda\frac{\text{D}}{\text{d}}$
$\Rightarrow \beta=\frac{\lambda}{\text{d}}\text{D}$
The graph between β and D is shown alongside
The slope of graph $=\frac{\lambda}{\text{d}}$

Knowing d, the wavelength of light used can be calculated to be
$\lambda= \text{Slope of graph}\times \text{d.}$

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Question 383 Marks

Find the range of frequency of light that is visible to an average human being $(400\text{nm}<\lambda<700\text{nm}).$

Answer

Given that, $400\text{nm}<\lambda<700\text{nm}.$
$\frac{1}{700\text{nm}}<\frac{1}{\lambda}<\frac{1}{400\text{nm}}$
$\Rightarrow\frac{1}{7\times10^{-7}}<\frac{1}{\lambda}<\frac{1}{4\times10^{-7}}$ $\Rightarrow\frac{3\times10^8}{7\times10^{-7}}<\frac{\text{c}}{\lambda}<\frac{3\times10^8}{4\times10^{-7}} ($Where, c = speed of light $= 3 \times 10^8m/s)$
$\Rightarrow4.3\times10^{14}<\frac{\text{c}}{\lambda}<7.5\times10^{14}$
$\Rightarrow4.3\times10^{14}\text{Hz}<\text{f}<7 .5\times10^{14}\text{Hz}.$

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Question 393 Marks

How does the resolving power of a compound microscope get affected on:

Decreasing the diameter of its objective?
Increasing the focal length of its objective?

Answer

Resolving limit of microscope $=\frac{\lambda}{2\text{n}\sin\theta} $
Resolving power $\alpha\frac{1}{\text{Resolving limit}}$
i.e., Resolving power $=\frac{2\text{n}\sin\theta}{\lambda}$

When diameter of objective lens decreases, θ and hence sinθ decreases; so the resolving power decreases.
The focal length of objective lens has no effect on resolving power of microscope.

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Question 403 Marks

Two identical coherent waves, each of intensity I, are producing an interference pattern. Find the value of the resultant intensity at a point of:

Constructive interference.
Destructive interference.

Answer

Resultant intensity at any point having a phase difference $\varphi$ is
$\text{I}_\text{R}=\text{I}_1+\text{I}_2+2\sqrt{\text{I}_1\text{I}_2}\cos\phi$
Here, $\text{I}_1=\text{I}_2=\text{I}$
$\therefore\text{I}_\text{R}=\text{I}_1+\text{I}_2+2\sqrt{\text{I}.\text{I}.}\cos\varphi=2\text{I}+2\text{I}\cos\varphi$

At a point of constructive interference:

$\varphi=2\text{n}\pi(\text{n}=0,1,2,\dots)\Rightarrow\cos\varphi=1$

$\therefore\text{I}_\max=2\text{I}+2\text{I}=4\text{I}$

At a point of denstructive interference:

$\Rightarrow\cos\varphi=0 $

$\therefore \text { I}_\min=\text{2I}-\text{2I}=0$

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Question 413 Marks

Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing $\lambda=400\text{nm}$). Describe the nature of the fringe pattern observed.

Answer

The violet filter will allow only violet light to pass through it. Now, if the double slit experiment is performed with the white light and violet light, the fringe pattern will not be the same as obtained by just using white light as the source. To have interference pattern, the light waves entering from the slits should be monochromatic. So, in this case, the violet light will superimpose with only violet light (of wavelength 400nm) in such a way that the bright bands will be of violet colour and the minima will be completely dark.

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Question 423 Marks

In Young’s double slit experiment, explain with reason in each case, how the interference pattern changes, when :

Width of the slits is doubled.
Separation between the slits is increased.
Screen is moved away from the plane of slits.

Answer

The fringe width $\beta= \frac{\text{D}\lambda}{\text{d}}.$
When the width of the slit is doubled; the intensity of interfering waves becomes four times, intensity of maxima becomes 16 times i.e., fringes become brighter.
When separation between the slits is increased the fringe width decreases, i.e., fringes come closer.
$\beta \alpha\text {D }$ when screen is moved away from the plane of the slits, the fringe width increases, i.e., fringes become farther away.

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Question 433 Marks

If the separation between the slits in a Young's double slit experiment is increased, what happens to the fringe-width? If the separation is increased too much, will the fringe pattern remain detectable?

Answer

The fringe width in Young's double slit experiment depends on the separation of the slits.
$\text{x}=\frac{\lambda\text{D}}{\text{d}},$
where
$\lambda$ = wavelength
x = fringe width
D = distance between slits and screen
d = separation between slits
On increasing d, fringe width decreases. If the separation is increased too much, the fringes will merge with each other and the fringe pattern won't be detectable.

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Question 443 Marks

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?

Answer

Consider the ray diagram in which the point image $\mathrm{I}_1$ due to $L_1$ is at the focal point. Now, we placed one more convergent lens of short focal length on the other side of lens $L_1$. Let the final image formed is I which is also a point image. Hence, the wavefront for this image will be of spherical symmetry.

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Question 453 Marks

Consider a two slit interference arrangements 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 $\lambda$ such that the first minima on the screen fall at a distance D from the centre O.

Answer

$\text{T}_2\text{P}=\text{D}+\text{x},\text{T}_1\text{P}=\text{D}-\text{x}$ $\text{S}_1\text{P}=\sqrt{(\text{S}_1\text{T}_1)^2+(\text{P}\text{T}_1)^2}=[\text{D}^2+(\text{D}-\text{x}^2)]^\frac{1}{2}$ $\text{S}_2\text{P}=[\text{D}^2+\text{(D+x)}^2]^\frac{1}{2}$ Minima will occur when: $[\text{D}^2+(\text{D}+\text{x})^2]^\frac{1}{2}-[\text{D}^2+(\text{D}-\text{x})^2]^\frac{1}{2}=\frac{\lambda}{2}$ $\text{if}\text{ x}=\text{D},(\text{D}^2+4\text{D}^2)^\frac{1}{2}-\text{D}=\frac{\lambda}{2}$ $\Rightarrow \text{D}(\sqrt5-1)=\frac{\lambda}{2}\Rightarrow\text{D}=\frac{\lambda}{2(\sqrt5-1)}$

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Question 463 Marks

Find the ratio of intensities at two points on a screen in Young’s double slit experiment when waves from the two slits have a path difference of (i) 0 and (ii) $\frac{\lambda}{4}.$

Answer

Intesity $\text{I}=\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\varphi$
Let $\text{a}_1=\text{a}_2=\text{a}$ say, then
$\text{I}=\text{a}^2+\text{a}^2+2\text{a}^2\cos\varphi=2\text{a}^2(1+\cos\varphi)$
$\frac{\text{I}_1}{\text{I}_2}=\frac{1+\cos\phi_1}{1+\cos\phi_2}$
When path difference is 0, phase difference $\varphi_1=0$
When path difference $\frac{\lambda}{4},$ is phase difference $\phi_2=\frac{2\pi}{\lambda}\times\frac{\lambda}{4}=\frac{\pi}{2} $
$\therefore\frac{\text{I}_1}{\text{I}_2}\frac{1+\cos\phi_1}{1+\cos\frac{\pi}{2}}=\frac{1+1}{1}=\frac{2}{1}.$

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Question 473 Marks

Is Huygen’s principle valid for longitudunal sound waves?

Answer

The point source propagates light in all directions symmetrically. Similarly, in case of any point source of sound wave, the disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of wavefront is in accordance with Huygen's principle.
Hence, Huygen's principle is valid for longitudinal sound waves.

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Question 483 Marks

The index of refraction of fused quartz is 1.472 for light of wavelength 400nm and is 1.452 for light of wavelength 760nm. Find the speeds of light of these wavelengths in fused quartz.

Answer

We know that, $\frac{\mu_2}{\mu_1}=\frac{\text{v}_1}{\text{v}_2}$So, $\frac{1472}{1}=\frac{3\times10^8}{\text{v}_{400}}\Rightarrow\text{v}_{400}=2.04\times10^8\text{m/sec}.$
[because, for air, $\mu=1\ \text{and v}=3\times10^8\text{m/s}]$
Again, $\frac{1452}{1}=\frac{3\times10^8}{\text{v}_{760}}\Rightarrow\text{v}_{760}=2.07\times10^8\text{m/sec}.$

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Question 493 Marks

Why is the diffraction of sound waves more evident in daily experience than that of light wave?

Answer

The frequencies of sound waves lie between 20Hz to 20kHz, their wavelength ranges between 15m to 15mm. The diffraction occurs if the wavelength of waves is nearly equal to slit width.
The wavelength of light waves is $7000 \times 10^{-10}m$ to $4000 \times 10^{-10}m$. For observing diffraction of light we need very narrow slit width. In daily life experience we observe the slit width very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

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Question 503 Marks

Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?

Answer

Young's double slit experiment can be performed with sound waves, as the sound waves also show interference pattern. To get a reasonable 'fringe pattern", the separation of the slits should be of the order of the wavelength of the sound waves used.
In this experiment, the bright and dark fringes can be detected by measuring the intensity of sound using a microphone or any other detector.

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Question 513 Marks

A narrow slit S transmitting light of wavelength $\lambda$ is placed a distance d above a large plane mirror as shown in figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen $\sum$ placed at a distance D from the slit.

What will be the intensity at a point just above the mirror, i.e., just above O?
At what distance from 0 does the first maximum occur?

Answer

Since, there is a phase difference of $\pi$ between direct light and reflecting light, the intensity just above the mirror will be zero.
Here, 2d = equivalent slit separation

D = Distance between slit and screen.

We know for bright fringe, $\Delta\text{x}=\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda$

But as there is a phase reversal of $\frac{\lambda}{2}.$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}+\frac{\lambda}{2}=\text{n}\lambda$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda-\frac{\lambda}{2}\Rightarrow\text{y}=\frac{\lambda\text{D}}{4\text{d}}$

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Question 523 Marks

What is the shape of the wavefront on earth for sunlight?

Answer

The sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. We can treat it like a point object as seen from the surface of earth.
Because of large distance, the radius of wavefront can be considered as Large (infinity) and hence, wavefront is almost plane.

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Question 533 Marks

A parallel beam of light of wavelength 600nm is incident normally on a slit of width ‘a’. If the distance between the slit and the screen is 0.8 and the distance of 2nd order maximum from the centre of the screen is 1.5mm, calculate the width of the slit.

Answer

Given: $\lambda=600\text{nm}=600\times10^{-9}\text{m}=6.0\times10^{-7}\text{m},\text{D}=0.8\text{m,}$
$\text{y}_2=1.5\text{mm}=1.5\times10^{-3}\text{m},\text{n}=2,\text{a=?}$
Position of $n^{th}$ maximum in diffraction of a single slit
$\text{y}_n=\text{(n}+\frac{1}{2})\frac{\lambda\text{D}}{\text{a}}\Rightarrow\text{a}=(\text{n}+\frac{1}{2})\frac{\lambda\text{D}}{\text{y}_\text{n}}$
Substituting given value $\text{a}=(2+\frac{1}{2})\frac{6.0\times10^{-7}\times0.8}{1.5\times10^{-3}}$
$\frac{5}2\times4.0\times0.8\times10^{-4}\text{m}=0.8\times10^{-3}=0.8\text{mm}.$

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Question 543 Marks

Two coherent waves of equal amplitude produce interference pattern in Young’s double slit experiment. What is the ratio of intensity at a point where phase difference is $\frac{\pi}{2}$ to intensity at centre?

Answer

$\text{I}_\frac{\pi}{2}=\text{a}^2_1+\text{a}^2_2+\text{2a}_1\text{a}_2\cos\varphi$
$=\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}=2\text{a}^2$
$\text{I}_\max=(\text{a}_2+\text{a}_2)^2=(\text{a}+\text{a})^2=4\text{a}^2$
$\therefore\frac{\text{I}_\frac{\pi}{2}}{\text{I}_\max}=\frac{2\text{a}^2}{\text{4a}^2}=\frac{1}{2}.$

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Question 553 Marks

Two narrow slits emitting iight in phase are separated by a distance of 1.0cm. The wavelength of the light is $5.0 \times 10^{-7}m.$ The interference pattern is observed on a screen placed at a distance of 1.0m.

Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima?
Find the separation between the sources which will give a separation of 1.0mm between the consecutive maxima.

Answer

Given that, $d = 1cm = 10^{-2}m, \lambda=5\times10^{-7}\text{m}$ and D = 1m

Separation between two consecutive maxima is equal to fringe width.

So, $\beta=\frac{\lambda\text{D}}{\text{d}}=\frac{5\times10^{-7}\times1}{10^{-2}}\text{m}=5\times10^{-5}\text{m}=0.05\text{mm}.$

When, $\lambda=1\text{mm}=10^{-3}\text{m}$

$10^{-3}\text{m}=\frac{5\times10^{-7}\times1}{\text{D}}\Rightarrow\text{D}=5\times10^{-4}\text{m}=0.50\text{mm}.$

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