5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

Questions

5 Marks Questions

Take a timed test

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

Question 15 Marks

In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?

Answer

Let $I_1$ and $I_2$ be the Intensity of the two light waves. Their resultant intensities can be obtained as:
$\text{I}'=\text{I}_1+\text{I}_2+2\sqrt{\text{I}_1{\text{I}_2}}\ \text{cos}\phi$
Where,
$\phi$ = Phase difference between the two waves
For monochromatic light waves,
$\text{I}_1=\text{I}_2$
$\therefore\ \text{I}'=\text{I}_1+\text{I}_1+2\sqrt{\text{I}_1\text{I}_1}\ \text{cos}\phi$
$=2\text{I}_1+2\text{I}_1\text{cos}\phi$
Phase difference $=\frac{2\pi}{\lambda}\times$ Path difference
Since path difference = $\lambda,$
Phase difference, $\phi=2\pi$
$\therefore\ \text{I}'=2\text{I}_1+2\text{I}_1=4\text{I}_1$
Given,
I' = K
$\therefore\ \text{I}_1=\frac{\text{K}}{4}.....(1)$
When path difference $=\frac{\lambda}{3},$
Phase difference, $\phi=\frac{2\pi}{3}$
Hence, resultant intensity, $\text{I}'_R=\text{I}_1+\text{I}_1+2\sqrt{\text{I}_1\text{I}_1}\ \text{cos}\frac{2\pi}{3}$
$=2\text{I}_1+2\text{I}_1\bigg(-\frac{1}{2}\bigg)=\text{I}_1$
Using equation (1), we can write:
$\text{I}_\text{R}=\text{I}_1=\frac{\text{K}}{4}$
Hence, the intensity of light at a point where the path difference is $\frac{\lambda}{3}\ \text{is}\ \frac{\text{K}}{4}$ units.

View full question & answer→

Question 25 Marks

Let us list some of the factors, which could possibly influence the speed of wave propagation:

Nature of the source.
Direction of propagation.
Motion of the source and/or observer.
Wavelength.
Intensity of the wave.

On which of these factors, if any, does

The speed of light in vacuum,
The speed of light in a medium (say, glass or water), depend?

Answer

Speed of light in vacuum is an absolute constant, according to Einstein's theory of relativity. It does not depend upon any of the factors listed above or any other factor.
The speed of light in a medium like water or glass:

Does not depend upon the nature of the source.
Does not depend upon the direction of propagation, when the medium is isotropic.
Does not depend upon motion of the source w.r.t. the medium, but depends on motion of the observer relative to the medium.
Depends on wavelength of light, being lesser for shorter wavelength and vice-versa.
Does not depend upon intensity of light.

View full question & answer→

Question 35 Marks

Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

Answer

Wavelength of incident monochromatic light,
$\lambda$ = 589 nm $= 589 \times 10^{-9} m$
Speed of light in air, $c = 3 \times 10^8 m/s$
Refractive index of water, $µ = 1.33$

The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.

Frequency of light Is given by the relation,
$ \text{v}=\frac{\text{c}}{\lambda}$
$=\frac{3\times{10}^8}{589\times{10}^{-9}}$
$= 5.09 \times 10^{14}$Hz
Hence, the speed, frequency, and wavelength of the reflected light are $3 \times 10^8 m/s, 5.09 \times 10^{14}$ Hz, and 589 nm respectively.

Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water wdl be equal to the frequency of the incident or reflected light in air.

$\therefore$ Refracted frequency, $v = 5.09 \times 10^{14}$ Hz
Speed of light in water is related to the refractive index of water as:
$\text{v}=\frac{\text{c}}{\mu}$
$\text{v}=\frac{3\times{10}^8}{1.33}=2.26\times10^8\ \text{m/s}$
Wavelength of light in water is given by the relation,
$\lambda=\frac{\text{v}}{\text{v}}$
$=\frac{2.26\times{10}^8}{5.09\times10^{14}}$
$= 444.007 \times 10^{-9} m$
$= 444.01 nm$
Hence, the speed, frequency, and wavelength of refracted light are $2.26 \times 10^8$ m/s, $444.01$ nm, and $5.09 \times 10^{14}$ Hz respectively.

View full question & answer→

Question 45 Marks

A parallel beam of white light is incident normally on a water film $1.0 \times 10^{-4}$cm thick. Find the wavelength in the visible range (400nm - 700nm) which are strongly transmitted by the film. Refractive index of water = 1.33.

Answer

For strong transmission, $2\mu\text{d}=\text{n}\lambda\Rightarrow\lambda=\frac{2\mu\text{d}}{\text{n}}$
Given that, $\mu=1.33,\text{d}=1\times10^{-4}\text{cm}=1\times10^{-6}\text{m}$
$\Rightarrow\lambda=\frac{2\times1.33\times1\times10^{-6}}{\text{n}}=\frac{2660\times10^{-9}}{\text{n}}\text{m}$
When, $\text{n}=4,\lambda_1=665\text{nm}$
$\text{n}=5,\lambda_2=532\text{nm}$
$\text{n}=6,\lambda_3=443\text{nm}$

View full question & answer→

Question 55 Marks

A thin paper of thickness 0.02mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits $\frac{4}{9}$ of the light energy falling on it.

Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern.
How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600nm.

Answer

Given that, $t = 0.02mm = 0.02 × 10^{-3}m$, $\mu_1=1.45,\ \lambda=600\text{nm}=600\times10^{-9}\text{m}$

Let, $I_1 =$ Intensity of source without paper = I
Then $I_2 =$ Intensity of source with paper $=\Big(\frac{4}{9}\Big)\text{I}$

$\Rightarrow\frac{\text{I}_1}{\text{I}_2}=\frac{9}{4}\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{3}{2}\ [\because\text{I}\propto\text{r}^2]$
where, $r_1$ and $r_2$ are corresponding amplitudes.
So, $\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(\text{r}_1+\text{r}_2)^2}{(\text{r}_1-\text{r}_2)^2}=25:1$
No. of fringes that will cross the origin is given by,
$\text{n}=\frac{(\mu-1)\text{t}}{\lambda}=\frac{(1.45-1)\times0.02\times10^{-3}}{600\times10^{-9}}=15.$

View full question & answer→

Question 65 Marks

A glass surface is coated by an oil film of uniform thickness $1.00 \times 10^{-4}cm$. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400nm - 750nm) which are completely transmitted by the oil film under normal incidence.

Answer

For the thin oil film,
$\text{d}=1\times10^{-4}\text{cm}=10^{-6}\text{m},$ $\mu_{\text{oil}}=1.25\ \text{and}\ \mu_\text{x}=1.50$
$\lambda=\frac{2\mu\text{d}}{\Big(\text{n}+\frac{1}{2}\Big)}\frac{2\times10^{-6}\times1.25\times2}{2\text{n}+1}=\frac{5\times10^{-6}\text{m}}{2\text{n}+1}$
$\Rightarrow\lambda=\frac{5000\text{nm}}{2\text{n}+1}$
For the wavelengths in the region (400nm - 750nm)
When, $\text{n}=3,\lambda=\frac{5000}{2\times3+1}=\frac{5000}{7}=714.3\text{nm}$
When, $\text{n}=4,\lambda=\frac{5000}{2\times4+1}=\frac{5000}{9}=555.6\text{nm}$
When, $\text{n}=5,\lambda=\frac{5000}{2\times5+1}=\frac{5000}{11}=454.5\text{nm}$

View full question & answer→

Question 75 Marks

In a Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron $(1$ micron $= 10^{-6}m)$ is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.

Answer

In the given Young’s double slit experiment,
$\mu=1.6,t = 1.964$ micron $= 1.964 \times 10^{-6}m$
We know, number of fringes shifted $=\frac{(\mu-1)\text{t}}{\lambda}$
So, the corresponding shift = No.of fringes shifted × fringe width
$=\frac{(\mu-1)\text{t}}{\lambda}\times\frac{\lambda\text{D}}{\text{d}}=\frac{(\mu-1)\text{tD}}{\text{d}}\ ...(1)$
Again, when the distance between the screen and the slits is doubled,
Fringe width $=\frac{\lambda(2\text{D})}{\text{d}}\ ...(2)$
From (1) and (2), $\frac{(\mu-1)\text{tD}}{\text{d}}=\frac{\lambda(2\text{D})}{\text{d}}$
$\Rightarrow\lambda=\frac{(\mu-1)\text{t}}{\lambda}$
$=\frac{(1.6-1)\times(1.964)\times10^{-6}}{2}=589.2\times10^{-9}=589.2\text{nm}.$

View full question & answer→

Question 85 Marks

The wavelength of sodium light in air is 589nm.

Find its frequency in air.
Find its wavelength in water (refractive index = 1.33).
Find its frequency in water.
Find its speed in water.

Answer

Given that, for sodium light, $\lambda=589\text{nm}=589\times10^{-9}\text{m}$

$\text{f}_\text{a}=\frac{3\times10^8}{589\times10^{-9}}=5.09\times10^{14}\sec\Big[\because\text{f}=\frac{\text{c}}{\lambda}\Big]$
$\frac{\mu_\text{a}}{\mu_\text{w}}=\frac{\lambda_\text{w}}{\lambda_\text{a}}\Rightarrow\frac{1}{1.33}=\frac{\lambda_\text{w}}{589\times10^{-9}}\Rightarrow\lambda _\text{w}=443\text{nm}$
$\text{f}_\text{w}=\text{f}_\text{a}=5.09\times10^{14}\sec^{-1}$ [Frequency does not change]
$\frac{\mu_\text{a}}{\mu_\text{w}}=\frac{\text{v}_\text{w}}{\text{v}_\text{a}}\Rightarrow\text{v}_\text{w}=\frac{\mu_\text{a}\text{v}_\text{a}}{\mu_\text{w}}=\frac{3\times10^{10^8}}{1.33}=2.25\times10^8\text{m/sec}.$

View full question & answer→

Question 95 Marks

Define the term ‘resolving power’ of an astronomical telescope. How does it get affected on:

Increasing the aperture of the objective lens?
Increasing the wavelength of light used?
Increasing the focal length of the objective lens?

Justify your answer in each case.

Answer

Resolving Power of an Astronomical Telescope:The resolving power of an astronomical telescope is its ability to form separate images of two neighbouring astronomical objects (e.g. stars).
The least distance between two neighbouring objects for which astronomical telescope can form separate images is called the resolving limit. The angular limit of resolution is given by:
$\theta_\min=\frac{1.22\lambda}{\text{d}}$
Where $\lambda $ is wavelength and d is diameter of aperture objective lens. Smaller the resolving limit, greater is the resolving power.
$\therefore$ Resolving power $=\frac{\text{d}}{1.22\lambda}$

RP $\alpha$ d, so by increasing aperture of objective lens, the resolving power of telescope increases.
RP $\alpha\frac{1}{\text{y}},$ so by increasing the wavelength of light, the resolving power of telescope decreases.
Resolving power of telescope is independent of its focal length, so there is no effect on resolving power if focal length of objective lens is increased.

View full question & answer→

Question 105 Marks

Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?

Answer

If angle of incidence is equal to Brewster’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 Breswster's angle.
i.e., $\tan\text{i}_\text{B}=^{1}\mu_2=\frac{\mu_1}{\mu_2}\text{ where }\mu_2<\mu_1$
When the light rays travels in such a medium, the critical angle is
$\sin\text{i}_\text{c}=\frac{\mu_2}{\mu_1},\text{ where }\mu_2<\mu_1$
As $|\tan\text{i}_\text{B}|>|\sin\text{i}_\text{C}|$ for large angles $i_B < i_C$
Thus, the polarisation by reflection occure definitely.
Important Point: Brewster'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 isw incident at this angle, the light that is reflected from the surface it therefore perfectly polarized. This special angle of incidence is named after the Scottish physicist Sir David Brewster.

View full question & answer→

Question 115 Marks

For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000A and electrons accelerated through 100V used as the illuminating substance.

Answer

Resolving power of a microscope $=\frac{2\sin\beta}{1.22\lambda}$ where $\mu$ is refractive index of the medium, $\lambda$ is the wavelength of light and $\beta$ is the angle subtended by the objective at the object.
Now, $\frac{1}{\text{d}}=\frac{2\sin\beta}{1.22\lambda}$
$\Rightarrow\ \text{d}_\text{min}=\frac{1.22\lambda}{2\sin\beta}$
For the light of wavelength 5000A,
$\text{d}_\text{min}=\frac{1.22\times5000\times10^{-10}}{2\sin\beta}\Big[\because\ 1\text{A}=10^{-10}\text{m}\Big]\ .....(\text{i})$
Now, the de-Broglie wavelenght is given by $\lambda=\frac{12.27}{\sqrt{\text{V}}}$
Substituting V = 100 in $\lambda=\frac{12.27}{\sqrt{\text{V}}}$, we get
$\frac{12.27}{\sqrt{100}}=1.227\times10^{-10}\text{m}$
$\therefore\ \text{d}'_\text{min}=\frac{1.22\times1.227\times10^{-10}}{2\sin\beta}$
Ratio of the least separation
$\frac{\text{d}_\text{min}}{\text{d}'_\text{min}}=\frac{\frac{1.22\times5000\times10^{-10}}{2\sin\beta}}{\frac{1.22\times1.227\times10^{-10}}{2\sin\beta}}$
$=\frac{5000}{1.227}$
$=4075.$

View full question & answer→

Question 125 Marks

What are coherent sources of light? Why are coherent sources required to produce interference of light? Give an example of interference in everyday life.

Answer

Coherent Sources: Two sources giving light waves of same frequency and zero or constant initial phase are called coherent sources.
Necessity of coherent sources to produce interference of light: Intensity at any point in the region of superposition is $\text{I}=\text{a}^2_1+\text{a}^2_2+\text{2a}_1\text{ a}_2\cos\text{f}$ If the interfering sources are not coherent, then the phase difference at any point in the region of superposition will go on changing with time. Then we shall get time-averaged intensity which is equal to $\text{I}=\text{a}^2_1+\text{a}^2_2+\text{i.e.,}$ sum of intensities of individual waves. This means that there will be no modifications in intensity in the region of superposition and hence no interference will be observed.
Example of Interference in Daily Life: The bubbles of colourless soap solution appear coloured in white light. This is due to interference of white light rays reflected from upper and lower surface of soap film. The colours of soap solution observed are those which satisfy the condition of maxima in reflected light.

View full question & answer→

Question 135 Marks

Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let $\text{BP}_0-\text{AP}=\frac{\lambda}{3}$ and $\text{D}>\lambda.$

Show that in this case $\text{d}=\sqrt{\frac{2\lambda\text{D}}{3}}.$
Show that the intensity at P, is three times the intensity due to any of the three slits individually.

Answer

Since $S_1, S_2$_ are in same phase, at O there will be maximum intensity.
Given that, there will be a maximum intensity at P.

⇒ path difference $=\Delta\text{x}=\text{n}\lambda$
From the figure,
$(\text{S}_1\text{P})^2-(\text{S}_2\text{P})^2=\Big(\sqrt{\text{D}^2+\text{X}^2}\Big)^2-\Big(\sqrt{(\text{D}-2\lambda)^2+\text{X}^2}\Big)^2$
$=4\lambda\text{D}-4\lambda^2=4\lambda\text{D}$ ($\lambda^2$ is so small and can be neglected)
$\Rightarrow\text{S}_1\text{P}-\text{S}_2\text{P}=\frac{4\lambda\text{D}}{2\sqrt{\text{x}^2+\text{D}^2}}=\text{n}\lambda$
$\Rightarrow\frac{2\text{D}}{\sqrt{\text{x}^2+\text{D}^2}}=\text{v}$
$\Rightarrow\text{n}^2(\text{X}^2+\text{D}^2)=4\text{D}^2=\Delta\text{X}=\frac{\text{D}}{\text{n}}\sqrt{4-\text{n}^2}$
when $\text{n}=1,\text{x}=\sqrt{3}\text{D}$ $(1^{st}$ order$)$
$\text{n}=2,\text{x}=0$ $(2^{nd}$ order$)$
$\therefore$ When $\text{X}=\sqrt{3}\text{D},$ at P there will be maximum intensity.

View full question & answer→

Question 145 Marks

Consider the arrangement shown in figure (17-E4). The distance D is large compared to the separation d between the slits.

Find the minimum value of d so that there is a dark fringe at 0.
Suppose d has this value. Find the distance x at which the next bright fringe is formed.
Find the fringe-width.

Answer

From the diagram, it can be seen that at point O.

Path difference = (AB + BO) - (AC + CO) = 2(AB - AC) [Since, AB = BO and AC = CO] $=2\Big(\sqrt{\text{d}^2+\text{D}^2}-\text{D}\Big)$ For dark fringe, path difference should be odd multiple of $\frac{\lambda}{2}.$ So, $2\Big(\sqrt{\text{d}^2+\text{D}^2}-\text{D}\Big)=(2\text{n}+1)\Big(\frac{\lambda}{2}\Big)$ $\Rightarrow\sqrt{\text{d}^2+\text{D}^2}=\text{D}+(2\text{n}+1)\Big(\frac{\lambda}{4}\Big)$ $\Rightarrow\text{D}^2+\text{d}^2=\text{D}^2+(2\text{n}+1)^2\frac{\lambda^2}{16}+(2\text{n}+1)\frac{\lambda\text{D}}{2}$ Neglecting, $(2\text{n}+1)^2\frac{\lambda^2}{16},$ as it is very small We get, $\text{d}=\sqrt{(2\text{n}+1)\frac{\lambda\text{D}}{2}}$ For minimum ‘d’, putting $\text{n}=0\Rightarrow\text{d}_\text{min}=\sqrt{\frac{\lambda\text{D}}{2}}.$

View full question & answer→

Question 155 Marks

A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50mm and the separation between the slits is 0.12cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590nm which is used in the experiment. The interference is observed on a screen a distance one meter away.

What would be the fringe-width?
At what distance from the centre will the first maximum be located?

Answer

Given that, $t_1 = t_2 = 0.5mm = 0.5 \times 10^{-3}m, \mu_\text{m}=1.58$ and $\mu_\text{p}=1.55,$
$\lambda=590\text{nm}=590\times 10^{-9}\text{m}, \text{d}=0.12\text{cm},\\ \mu_{\text{m}}=1.58\text{ and }\mu_{\text{p}}=1.55,$
Fringe width $=\frac{\text{D}\lambda}{\text{d}}=\frac{1\times590\times10^{-9}}{12\times10^{-4}}=4.91\times10^{-4}\text{m}.$
When both the strips are fitted, the optical path changes by
$\Delta\text{x}=(\mu_\text{m}-1)\text{t}_1-(\mu_\text{p}-1)\text{t}_2=(\mu_\text{m}-\mu_\text{p})\text{t}$
$=(1.58-1.55)\times(0.5)(10^{-3})=0.015\times10^{-13}\text{m}$
So, No. of fringes shifted $=\frac{0.015\times10^{-3}}{590\times10^{-3}}=25.43$
⇒ There are 25 fringes and 0.43 th of a fringe.
⇒ There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe.
So, position of first maximum on both sides will be given by,
$\therefore x = 0.43 \times 4.91 \times 10^{-4} = 0.021cm$
$ x' = (1 - 0.43) \times 4.91 \times 10^{-4} = 0.028cm($since, fringe width $= 4.91 \times 10^{-4}m)$

View full question & answer→

Question 165 Marks

Two Polaroids are placed 90° to each other and the transmitted intensity is zero. What happens when one more Polaroid is placed between these two bisecting the angle between them. Take intensity of unpolarised light 10. How will the intensity of transmitted light vary on further rotating the third Polaroid?

Answer

If ordinary unpolarised light of intensity I0' is incident on first Polaroid (A, say)
Intensity of light transmitted from first Polaroid is $\text{I}_0=\frac{\text{I}_{0'}}{2}$
Given angle between transmission axes of two polaroids A and B is initially 90°.
According to Malus law, intensity of light transmitted from second polaroid (B, say) is
$\text{I}=\text{I}_0\cos^2\theta\Rightarrow\text{I}=\text{I}_0\cos^290^\circ=0$
When one more polaroid (C say) is placed between A and B making an angle of 45° with the transmission axis of either of polaroids, then intensity of light transmitted from A is
$\text{I}=\text{I}_0\cos^2\theta\Rightarrow\text{I}=\text{I}_0\cos^290^\circ=0$
Intensity of light transmitted from C is:
$\text{I}_\text{c}=\text{I}_0\cos^245^\circ=\frac{\text{I}_0}{2} $
Intensity of light transmitted from B is:
$\text{I}_\text{B}=\text{I}_\text{C}\cos^245^\circ=\frac{\text{I}_0}{2}\times\frac{1}{2}=\frac{I_0}{4}$
This means that the intensity becomes one-fourth of intensity of light that is transmitted from first Polaroid.
On further rotating the Polaroid C such that if angle between their transmissions axes increases, the intensity decreases and if angle decreases, the intensity increases.

View full question & answer→

Question 175 Marks

Define resolving power of a compound microscope. How does the resolving power of a compound microscope change when:

Refractive index of medium between the object and objective lens increases.
Wavelength of radiation used is increased?

Answer

Resolving Power of a Microscope:The resolving power of a microscope is defined as its ability to form separate images of two close objects placed near the microscope.
The minimum distance between close objects for which microscope can just form separate images of the objects is called the limit of resolution of microscope. Smaller the limit, larger the resolving power.
The angular resolving limit of microscope is $\text{d}\theta=\frac{\lambda}{2\text{n}\sin\theta},$
Where n is the refractive index of medium between object and objective.
Resolving power $\frac{1}{\text{d}\theta}=\frac{2\text{n}\sin\theta}{\lambda}$

Resolving power $\alpha$ n; therefore resolving power of a compound microscope increases when refractive index (n) between the object and objective lens increases.
Resolving power $\alpha \frac{1}{\lambda}$ ; therefore, resolving power of a compound microscope decreases with the increase of wavelength of light used.

View full question & answer→

Question 185 Marks

Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between the sources is $2 \lambda$. Consider a line passing through $\mathrm{S}_2$ and perpendicular to the line $S_1 S_2$. What is the smallest distance from $S_2$ where a minimum of intensity occurs?

Answer

For minimum intensity

$\therefore\text{S}_1\text{P}-\text{S}_2\text{P}=\text{x}=(2\text{n}+1)\frac{\lambda}{2}$
From the figure, we get,
$\Rightarrow\sqrt{\text{X}^2+(2\lambda)^2}-\text{Z}=(2\text{n}+1)\frac{\lambda}{2}$
$\Rightarrow\text{Z}^2+4\lambda^2=\text{Z}^2+(2\text{n}+1)^2\frac{\lambda^2}{4}+\text{Z}(2\text{n}+1)\lambda$
$\Rightarrow\text{Z}=\frac{4\lambda^2-(2\text{n}+1)^2\Big(\lambda^2{4}\Big)}{(2\text{n}+1)\lambda}=\frac{16\lambda^2-(2\text{n}+1)^2\lambda^2}{4(2\text{n}+1)\lambda}\ ...(1)$
Putting, $\text{n}=0\Rightarrow\text{Z}=\frac{15\lambda}{4}$
$\text{n}=-1\Rightarrow\text{Z}=\frac{-15\lambda}{4}$
$\text{n}=1\Rightarrow\text{Z}=\frac{7\lambda}{12}$
$\text{n}=2\Rightarrow\text{Z}=\frac{-9\lambda}{20}$
$\therefore\text{Z}=\frac{7\lambda}{12}$ is the smallest distance for which there will be minimum intensity.

View full question & answer→

Question 195 Marks

In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to,

Half the maximum.
One fourth of the maximum.

Answer

When intensity is half the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{2}$

$\Rightarrow\frac{4\text{a}^2\cos^2\big(\frac{\phi}{2}\big)}{4\text{a}^2}=\frac{1}{2}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{2}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{\sqrt{2}}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{4}\Rightarrow\phi=\frac{\pi}{2}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{4}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{4\text{d}}$

When intensity is $\frac{1}{4}\text{th}$ of the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{4}$

$\Rightarrow\frac{4\text{a}^2\cos^2\Big(\frac{\phi}{2}\Big)}{4\text{a}^2}=\frac{1}{4}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{4}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{2}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{3}\Rightarrow\phi=\frac{2\pi}{3}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{3}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{3\text{d}}$

View full question & answer→

Question 205 Marks

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 λ such that the first minima on the screen falls at a distance D from the centre O.

Answer

Key concept: Young's experiment to show interference of light passing through two slits. A pattern of bright and dark areas appears on the screen (as shown in figure (i)).

The condition for destructive interference is $\Delta\text{x}=\text{S}_2\text{P}-\text{S}_1\text{P}=\pm\Big(\frac{2\text{n}-1}{2}\Big)\lambda\text{ where n}= 1, 2,\ .....$ For nth minima to be formed on the screen path difference $(\Delta\text{x})$ between the rays coming from $S_1$ and $S_2$ must be $\Big(\frac{2\text{n}-1}{2}\Big)\lambda$. The minima will occur when $\Delta\text{x}=\text{S}_2\text{P}-\text{S}_1\text{P}=\Big(\frac{2\text{n}-1}{2}\Big)\lambda \ .....(\text{i})$

From the given figure, $\text{S}_1\text{P}=\sqrt{(\text{S}_1\text{T}_1)^2+(\text{PT}_1)^2}=\sqrt{\text{D}^2+(\text{D}-\text{x})^2}$ and $\text{S}_1\text{P}=\sqrt{(\text{S}_2\text{T}_2)^2+(\text{T}_2\text{P})^2}=\sqrt{\text{D}^2+(\text{D}+\text{x})^2}$$\text{T}_2\text{P}=\text{T}_2\text{O}+\text{OP}=\text{D}+\text{x}$
And $\text{T}_1\text{P}=\text{T}_1\text{O}-\text{OP}=\text{D}-\text{x}$ Hence, $[\text{D}^2+(\text{D}+\text{x})^2]^{\frac{-1}{2}}-[\text{D}^2-(\text{D}-\text{x})^2]^{\frac{-1}{2}}=\frac{\lambda}{2}$ [for first minima n = 1] If $\text{x}=\text{D}$ we can write, $[\text{D}^2+4\text{D}^2]^{\frac{1}{2}}-[\text{D}^2+0]^{\frac{1}{2}}=\frac{\lambda}{2}$ $\Rightarrow\ [5\text{D}^2]^{\frac{1}{2}}-[\text{D}^2+0]^{\frac{1}{2}}=\frac{\lambda}{2}$ $\Rightarrow\ \sqrt{5\text{D}}-\text{D}=\frac{\lambda}{2}\text{ or }\text{D}=\frac{\lambda}{2(\sqrt{5}-1)}=0.404\lambda$

View full question & answer→

Question 215 Marks

$AC = CO = D, S_1C = S_2C = d < < D$
A small transparent slab containing material of $\mu = 1.5$ is placed along $AS_2 $(Fig). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.

Answer

Key concept:Shifting of fringe pattern in YDSE:
When we introduce a thin transparent plate in front of one of the slits in YDSE, the fringe pattern shifts toward the isde where the plate is present.

$(\text{S}_2\text{O}')_\text{new}=(\text{S}_2\text{O}'-\text{t})_\text{air}+\text{t}_\text{plate}=(\text{S}_2\text{O}'-\text{t})_\text{air}+\mu\text{t}_\text{air}$
$(\text{S}_2\text{O}')_\text{new}=\text{S}_2\text{O}'+[\mu(\text{t}-1)]$
Path difference, $\Delta\text{x}=(\text{S}_2\text{O}')_\text{new}-\text{S}_\text{O}'=(\text{S}_2\text{O}'-\text{S}_1\text{O}')+[\mu(\text{t}-1)]$
$\Delta\text{x}=\text{d}\sin\theta+[\mu(\text{t}-1)]$
$\Rightarrow\ \text{Additional path difference}=(\mu-1)\text{t}$
Here the separation between the slits $= S_1S_2= 2d$.
Hence for calculating path diffrence, equation (i) becomes $\Delta\text{x}=2\text{d}\sin\theta+[\mu(\text{t}-1)]$
For the principal maxima, (path difference is zero)
$\Delta\text{x}=2\text{d}\sin\theta_0+[\mu(\text{t}-1)]=0$
$\sin\theta_0=-\frac{\text{L}(\mu-1)}{2\text{d}}=\frac{-\text{L}(0.4)}{2\text{d}}\ \ \Big[\because\ \text{L}=\frac{\text{d}}{4}\Big]$
or $\Rightarrow\ \sin\theta_0=\frac{-1}{16}$
$\theta_0$ is the angular position corresponding to the principal maxima.
$\Rightarrow\ \text{OP}=\text{D}\tan\theta_0\approx\text{D}\sin\theta_0=\frac{-\text{D}}{16}$
For the first minima, the path difference is $\pm\frac{\lambda}{2}$
$\Delta\text{x}=2\text{d}\sin\theta_1+0.5\text{L}=\pm\frac{\lambda}{2}$
$\sin\theta_1=\frac{\pm\frac{\lambda}{2}-0.5\text{L}}{2\text{d}}=\frac{\pm\frac{\lambda}{2}-\frac{\text{d}}{8}}{2\text{d}}$
$\Rightarrow\ \frac{\pm\frac{\lambda}{2}-\frac{\lambda}{8}}{2\lambda}=\pm\frac{1}{4}-\frac{1}{16}$
[$\because$ The diffraction occurs if the wavelength of waves is nearly equal to the side width (d).]
On the positive side $\sin\theta_1^+=+\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$
On the negative side $\sin\theta_1^-=\frac{1}{4}-\frac{1}{16}=-\frac{5}{16}$
The first principal maxima on the positive side is at distance
$\text{D}\tan\theta_1'^+=\text{D}\frac{\sin\theta_1'^+}{\sqrt{1-\sin^2\theta_1'}}=\text{D}\frac{3}{\sqrt{16^2-3^2}}=\frac{3\text{D}}{\sqrt{247}}$ above point O
The first principal minima on the negative side is at distance
$\text{D}\tan\theta_1^+=\text{D}\frac{3\text{D}}{\sqrt{16^2-5^2}}=\frac{5\text{D}}{\sqrt{231}}$ below point O.

View full question & answer→

Question 225 Marks

The optical properties of a medium are governed by the relative permitivity $(\in_\text{r})$ and relative permeability $(\mu_\text{r})$. The refractive index is defined as $\sqrt{\mu_\text{r}\in_\text{r}}=\text{n}$.For ordinary material $\in_\text{r}=0\text{ and }\mu_\text{r}<0$ and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with $\in_\text{r}=0\text{ and }\mu_\text{r}<0$. Since then such 'metamaterials' have been produced in the laboratories and their optical properties studied. For such materials $\text{n}=-\sqrt{\mu_1\in_1}$. As light enters a medium of such refractive index the phases travel away from the direction of propagation.

According to the description above show that if rays of light enter such a medium from air (refractive index = 1) at an angle ? In $2^{nd}$ quadrant, them the refracted beam is in the $3^{rd}$ quadrant.
Prove that Snell's law holds for such a medium.

Answer

If given postulate is true, then two parallel rays would proced as shown in the figure (i).

Again consider figure (i), let AB represent the incident wavefront and DE represent wavefront. All point on a wavefront must be in same phase and in turn, must have the same optical path length.
Thus $-\sqrt{\in_\text{r}\mu_\text{r}}\text{AE}=\text{BC}-\sqrt{\in_\text{r}\mu_\text{r}}\text{CD}$
or $\text{BC}=\sqrt{\in_\text{r}\mu_\text{r}}(\text{CD}-\text{AE})$
$\text{BC}>0,\text{CD}>\text{AE}$
As showing that the postulate is reasonable. If however, the light proceeeded in the sence it does for ordineray material (viz. in the fourth quadrant, Fig. 2)
Then, $-\sqrt{\in_\text{r}\mu_\text{r}}\text{AE}=\text{BC}-\sqrt{\in_\text{r}\mu_\text{r}}\text{CD}$
or $\text{BC}=\sqrt{\in_\text{r}\mu_\text{r}}(\text{CD}-\text{AE})$
if $\text{BC}>0,\text{ then }\text{CD}>\text{AE}$
which is obvious from Fig. (i). Hence, the postulate is reasonable.
However, if the light proced in the sense it does for ordinary material, (going from $2^{nd}$ quadrant to $4^{th$} quadrant) as shown on Fig. (i). then proceeding as above,
$-\sqrt{\in_\text{r}\mu_\text{r}}\text{AE}=\text{BC}-\sqrt{\in_\text{r}\mu_\text{r}}\text{CD}$
or $\text{BC}=\sqrt{\in_\text{r}\mu_\text{r}}(\text{CD}-\text{AE})$
As AE > CD, therefore BC < 0 which is not possible. Hence, the given postulate is correct.

From Fig. (i),

$\text{BC}=\text{AC}\sin\theta_\text{i}$
and $\text{CD}-\text{AE}=\text{AC}\sin\theta_\text{r}$
As $\text{BC}=-\sqrt{\mu_\text{r}\in_\text{r}}(\text{CD}-\text{AE})$
$\therefore\ \text{AC}\sin\theta_\text{i}=\sqrt{\in_\text{r}\mu_\text{r}}\text{AC}\sin\theta_\text{r}$
or $\frac{\sin\theta_\text{i}}{\sin\theta_\text{r}}=\sqrt{\in_\text{r}\mu_\text{r}}=\text{n}$
which proves Snell's law.

View full question & answer→

Question 235 Marks

A polariod (I) is placed in front of a monochromatic source. Another polatiod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.

Answer

A polaroid (I) is placed in front of a monochromatic source. Another polariod (II) is placed in front of this polaroid (I) when the pass axis of (II) is parallel to (I), light passes through polaroid-II is unaffected.

Now polariod (II) is rotated till no light passes. In this situation the pass axis of polariod (II) is perpedicular 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) of (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).

View full question & answer→

Question 245 Marks

What is interference of light? Write two essential conditions for sustained interference pattern to be produced on the screen. Draw a graph showing the variation of intensity versus the position on the screen in Young’s experiment when (a) both the slits are opened and (b) one of the slits is closed. What is the effect on the interference pattern in Young’s double slit experiment when:

Screen is moved closer to the plane of slits?
Separation between two slits is increased?

Explain your answer in each case.

Answer

Interference of light: When two waves of same frequency and constant initial phase difference travel in the same direction along a straight line simultaneously, they superpose in such a way that the intensity of the resultant wave is maximum at certain points and minimum at certain other points. This phenomenon of redistribution of energy due to superposition of two waves of same frequency and constant initial phase difference is called interference
Conditions for Sustained Interference of Light Waves:
To obtain sustained (well-defined and observable) interference pattern, the intensity must be maximum and zero at points corresponding to constructive and destructive interference. For the purpose following conditions must be fulfilled:

The two interfering sources must be coherent and of same frequency, i.e., the sources should emit light of the same wavelength or frequency and their initial phase should remain constant. If this condition is not satisfied the phase difference between the interfering waves will vary continuously. As a result the resultant intensity at any point will vary with time being alternately maximum and minimum, just like the phenomenon of beats in sound.
The interfering waves must have equal amplitudes. Otherwise the minimum intensity will not be zero and there will be general illumination.

The variation of intensity I versus the position x on the screen in Young’s experiment. Fringe width, $\beta=\frac{\text{D}\lambda}{\text{d}}.$

$\text{B}\alpha\text{D},$ therefore with the decrease of separation between the plane of slits and screen, the fringe width decreases.
On increasing the separation between two slits (d), the fringe separation decreases as β is inversely proportional to d $(\text{i.e.,}\beta\alpha\frac{1}{\text{d}}.)$

View full question & answer→

Question 255 Marks

White coherent light (400nm - 700nm) is sent through the slits of a Young's double slit experiment (figure.) The separation between the slits is 0.5mm and the screen is 50cm away from the slits. There is a hole in the screen at a point 1.0mm away (along the width of the fringes) from the central line.

Which wavelength (s) will be absent in the light coming from the hole?
which wavelength (s) will have a strong intensity?

Answer

Given that, $\lambda= (400\text{nm to } 700\text{nm}),d = 0.5mm = 0.5 \times 10^{-3}m,$
$D = 50cm = 0.5m$ and on the screen $y_n = 1mm = 1 \times 10^{-3}m$

We know that for zero intensity (dark fringe)

$\text{y}_\text{n}=\Big(\frac{2\text{n}+1}{2}\Big)\frac{\lambda_\text{n}\text{D}}{\text{d}}$ where n = 0, 1, 2, ...
$\Rightarrow\lambda_\text{n}=\frac{2}{(2\text{n}+1)}\frac{\lambda_\text{n}\text{d}}{\text{D}}=\frac{2}{2\text{n}+1}\times\frac{10^{-3}\times0.5\times10^{-3}}{0.5}$
$\Rightarrow\frac{2}{(2\text{n}+1)}\times10^{-6}\text{m}=\frac{2}{(2\text{n}+1)}\times10^3\text{nm}$
If $\text{n}=1,\lambda_1=\Big(\frac{2}{3}\Big)\times1000=667\text{nm}$
If $\text{n}=1,\lambda_2=\Big(\frac{2}{5}\Big)\times1000=400\text{nm}$
So, the light waves of wavelengths 400nm and 667nm will be absent from the out coming light.

For strong intensity (bright fringes) at the hole

$\text{y}_\text{n}=\frac{\text{n}\lambda_\text{n}\text{D}}{\text{d}}\Rightarrow\lambda_\text{n}=\frac{\text{y}_\text{n}\text{d}}{\text{nD}}$
When, $\text{n}=1,\lambda_1=\frac{\text{y}_\text{n}\text{d}}{\text{D}}$
$=\frac{10^{-3}\times0.5\times10^{-3}}{0.5}=10^{-6}\text{m}=1000\text{nm}.$
1000nm is not present in the range 400nm - 700nm
Again, where $\text{n}=2,\lambda_2=\frac{\text{y}_\text{n}\text{d}}{2\text{D}}=500\text{nm}$
So, the only wavelength which will have strong intensity is 500nm.

View full question & answer→

Question 265 Marks

In a Young's double slit experiment, the separation between the slits = 2.0mm, the wavelength of the light = 600nm and the distance of the screen from the slits = 2.0m. If the intensity at the centre of the central maximum is $0.20W/ m^2$, what will be the intensity at a point 0.5cm away from this centre along the width of the fringes?

Answer

Given that, $d = 2mm = 2 \times 10^{-3}m, \lambda=600\text{nm}=6\times10^{-7}\text{m},I_{max} = 0.20W/ m^2, D = 2m$
For the point, $y = 0.5cm$
We know, path difference $=\text{x}=\frac{\text{yd}}{\text{D}}=\frac{0.5\times10^{-2}\times2\times10^{-3}}{2}=5\times10^{-6}\text{m}$
So, the corresponding phase difference is,
$\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi\times5\times10^{-6}}{6\times10^{-7}}\Rightarrow\frac{50\pi}{3}=16\pi+\frac{2\pi}{3}\Rightarrow\phi=\frac{2\pi}{3}$
So, the amplitude of the resulting wave at the point y = 0.5cm is,
$\text{A}=\sqrt{\text{r}^2+\text{r}^2+2\text{r}^2\cos(\frac{2\pi}{3})}=\sqrt{\text{r}^2+\text{r}^2-\text{r}^2}=\text{r}$
$\frac{\text{I}}{\text{I}_\text{max}}=\frac{\text{A}^2}{(2\text{r})^2}$ [since, maximum amplitude = 2r]
$\Rightarrow\frac{\text{I}}{0.2}=\frac{\text{A}^2}{4\text{r}^2}=\frac{\text{r}^2}{4\text{r}^2}$
$\Rightarrow\text{I}=\frac{0.2}{4}=0.05\text{W/ m}^2.$

View full question & answer→

Question 275 Marks

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 speetrum $(5000\mathring{\text{A}})$ there is maximum transmission.

Answer

Consider the figure,

In the given figure, a dielectric film of thickness d deposited on a lgasslens. we are given that, $\lambda=5500\mathring{\text{A}}$ refractive index of film = 1.38 and refractive index of glass = 1.5 Let us consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface and a part refracted inside. This is partly reflected at the film-glass interface and a part transmitted. a part of the reflcrted ray is reflected at the film-air interface and a part transmitted as $r_2$ parallel to $r_2$ of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hecne, rays $r_1$ and $r_2$ shall dominate the behaviour, if incident light is to be transmitted If incident light is to be transmitted through the lens, $r_1$ and $r_2$ should interfere destructively.Buth the reflcetions at A and D path difference between $r_2$ and $r_1$, is given by $n (AD + CD) - AB$ If d is the thickness of the film, then n $(AD + CD) - AB$ if d is the thickness of the film, then $\text{AD}=\text{CD}=\frac{\text{d}}{\cos\text{r}}$ $\text{AB}=\text{AC}\sin\text{i}$ $\frac{\text{AC}}{2}=\text{d}\tan\text{r}$ $\therefore\ \text{AC}=2\text{d}\tan\text{r}$ Substituting $\text{AC}=2\text{d}\tan\text{r}\text{ in }\text{AB}=\text{AC}\sin\text{i},$ we get $\text{AB}=2\text{d}\tan\text{r}\sin\text{i}$ Therefore, the optical path difference is given by $\frac{2\text{nd}}{\cos\text{r}}-2\text{d}\tan\text{r}\sin\text{i}$. $=2\frac{\sin\text{id}}{\sin\text{r}\cos\text{r}}-2\text{d}\frac{\sin\text{r}}{\cos\text{r}}\sin\text{r}$ $=2\text{d}\sin\text{i}\bigg[\frac{1-\sin^2\text{r}}{\sin\text{r}\cos\text{r}}\bigg]$ $=2\text{nd}\cos\text{r} \ \bigg[\because\ \text{n}=\frac{\sin\text{i}}{\sin\text{r}}\bigg]$ For these waves to interface destructively, the path difference should be $\frac{\lambda}{2}$. $\therefore\ 2\text{nd}\cos\text{r}=\frac{\lambda}{2}$ $\Rightarrow\ \text{nd}\cos\text{r}=\frac{\lambda}{4} \ .....(\text{i})$ For photographic lenses, the sources are normally in vertical palne Therefore, $\text{i}=\text{r}=0^\circ$ From Eq. (i), $\text{nd}\cos0^\circ=\frac{\lambda}{4}$ $\Rightarrow\ \text{d}=\frac{\lambda}{4\text{n}}$ $=\frac{5500\mathring{\text{A}}}{4\times1.38}\approx1000\mathring{\text{A}}$

View full question & answer→

Question 285 Marks

Consider the situation shown in figure (17-E6). The two slits $S_1$ and $S_2$ p laced symmetrically around the central line are illuminated by a monochromatic light of wavelength $\lambda$. The separation between the slits is d . The light transmitted by the slits falls on a screen $E_1$ placed at a distance $D$ from the slits. The slit $S_3$ is at the central line and the slit $S_4$ is at a distance $z$ from $S_3$. Another
screen $\sum_2$ is placed a further distance D away from $\sum_1$. Find the ratio of the maximum to minimum intensity observed on $\sum_2$, if z is equal to,

$\text{z}=\frac{\lambda\text{D}}{2\text{d}}$
$\frac{\lambda\text{D}}{\text{d}}$
$\frac{\lambda\text{D}}{4\text{d}}$

Answer

When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$ at $S_4$, minimum intensity occurs

⇒ Amplitude = 0,
At $S_3$, path difference = 0
⇒ Maximum intensity occurs.
⇒ Amplitude = 2r.
So, on $\sum_2$ screen,
$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+0)^2}{(2\text{r}-0)^2}=1$

When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$ At $S_4$, minimum intensity occurs.

⇒ Amplitude = 0.
At $S_3$, path difference = 0
⇒ Maximum intensity occurs.
⇒ Amplitude = 2r.
So, on $\sum_2$ screen,
$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+2\text{r})^2}{(2\text{r}-0)^2}=\infty$

When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$, At $S_4$, intensity $=\frac{\text{I}_\text{max}}{2}$

⇒ Amplitude $=\sqrt{2\text{r}}$
$\therefore$ At $S_3$, intensity is maximum
⇒ Amplitude = 2r
$\therefore\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+\sqrt{2\text{r}})^2}{(2\text{r}-\sqrt{2\text{r}})^2}=34.$

View full question & answer→

Question 295 Marks

Four identical monochromatic sources A,B,C,D as shown in the (Fig.) produce waves of the same wavelength λ and are coherent. Two receiver $R_1$ and $R_2$ are at great but equal distaces from B.

Which of the two receivers picks up the larger signal?
Which of the two receivers picks up the larger signal when B is turned off?
Which of the two receivers picks up the larger signal when D is turned off?
Which of the two receivers can distinguish which of the sources B or D has been turned off?

Answer

Consiter the disturbances at rhe receiver $R_1$ which is at a distanced from B. Let the wave at $R_1$ because of A be $\text{Y}_\text{A}=\text{a}\cos\omega\text{t}.$ Since, the path difference of the signal from A with that from B is $\frac{\lambda}{2}$. Hence, the phase difference is $\pi$. Therefore, the wave at $R_1$ because of B is given by $\text{y}_\text{B}=\text{a}\cos(\omega\text{t}-\pi)$ $=-\text{a}\cos\omega\text{t}$ Since, the path difference of the signal from C with that from A is $\lambda$. Hence, the phase difference is $2\pi$. Therefore, the wave at $R_1$ because of C is given by $\text{Y}_\text{c}=\text{a}\cos(\omega\text{t}-2\pi)$ $=\text{a}\cos\omega\text{t}$

The path difference between the signal from D with that of A is given by $\sqrt{\text{q}^2+\Big(\frac{\lambda}{2}\Big)^2}-\Big(\text{d}-\frac{\lambda}{2}\Big)$ $=\text{d}\Big(1+\frac{\lambda}{4\text{d}^2}\Big)^\frac{1}{2}-\text{d}+\frac{\lambda}{2}$ $=\text{d}\Big(1+\frac{\lambda^2}{8\text{d}^2}\Big)-\text{d}+\frac{\lambda}{2}\approx\frac{\lambda}{2}\ \ (\because\ \text{d} < < \lambda)$ Therefore, the phase difference is $\pi$ $\therefore\ \text{Y}_\text{D}=\text{a}\cos(\omega\text{t}-\pi)=-\text{a}\cos\omega\text{t}$ Thus, the signal picked up at $R_1$, from all the four source is given by $\text{Y}_{\text{R}_1}=\text{y}_\text{A}+\text{y}_\text{B}+\text{y}_\text{C}+\text{y}_\text{D}$ $=\text{a}\cos\omega\text{t}-\text{a}\cos\omega\text{t}+\text{a}\cos\omega\text{t}-\text{a}\cos\omega\text{t}$ $=0$

Let the signal picked up at $R_2$ from S be $\text{y}_\text{B}=\text{a}_1\cos\omega\text{t}.$

The path difference between signal at D and that at B is $\frac{\lambda}{2}$.
$\therefore\ \text{y}_\text{D}=\text{a}_1\cos\omega\text{t}$
The path difference between signal at A and that at B is
$\sqrt{(\text{d})^2+\Big(\frac{\lambda}{2}^2\Big)}-\text{d}=\text{d}\bigg(1+\frac{\lambda^2}{4\text{d}^2}\bigg)^{\frac{1}{2}}-\text{d}\simeq\frac{1\lambda^2}{8\text{d}^2}$
As $\text{d}>>\lambda,$ therefore this path difference is 0
And phase difference $=\frac{2\pi}{\lambda}\Big(\frac{1}{8}\frac{\lambda^2}{\text{d}^2}\Big)$ is 0
Hence, $\text{y}_\text{A}=\text{a}_1\cos(\omega\text{t}-\phi)$
Similarly, $\text{y}_\text{C}=\text{a}_1\cos(\omega\text{t}-\phi)$
$\therefore$ Signal picked up by $R_2$ is given by $y_A + y_B + y_C + y_D$
$=\text{y}$
$=2\text{a}_1\cos(\omega\text{t}-\phi)$
Now, $|\text{y}|^2=4\text{a}_1^2\cos^2(\omega\text{t}-\phi)$
$\therefore\ <\text{I}>=2\text{a}_1^2$
Thus, $R_1$ picks up the larger signal.

If B is switched off.

$R_1$ picks up $\text{y}=\text{a}\cos\omega\text{t}$
$\therefore\langle\text{I}_{\text{R}_1}\rangle=\frac{1}{2}\text{a}^2$
$R_2$ picks up $\text{y}=\text{a}^2<\cos^2\omega\text{t}>=\frac{\text{a}^2}{2}$
$\therefore\ \langle\text{I}_{\text{R}_1}\rangle=\frac{1}{2}=\text{a}^2<\cos^2\omega\text{t}>=\frac{\text{a}^2}{2}$

Thus, $R_1$ and $R_2$ pick up the same signal.

If D si switched off.
$R_1$ picks up $\text{y}=\text{a}\cos\omega\text{t}$
$\therefore\ \langle\text{I}_{\text{R}_1}\rangle=\frac{1}{2}\text{a}^2$
$R_2$ picks up $\text{y}=3\text{a}\cos\omega\text{t}$
$\therefore\ \langle\text{I}_{\text{R}_1}\rangle=\frac{1}{2}=9\text{a}^2<\cos^2\omega\text{t}>=\frac{9\text{a}^2}{2}$

Thus, a signal $R_1$ indicated B has been switched off and an enhanced signal at $R_2$ indicated D has been swithced off.

View full question & answer→

Question 305 Marks

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.

Answer

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
$\text{A}_\bot=\text{A}_\bot^1+\text{A}_\bot^2=\text{A}_\bot^0\sin(\text{kx}-\omega\text{t})+\text{A}_\bot^2\sin(\text{kx}-\omega\text{t}+\phi)$
$\Rightarrow\ \text{A}_\bot=\text{A}_\bot^0(\sin(\text{kx}-\omega\text{t})+\sin(\text{kx}-\omega\text{t}+\phi))$
Amplitude of the wave in parallel polarisation
$\text{A}_{||}=\text{A}_{||}^{(1)}+\text{A}_{||}^{(2)}$
$\Rightarrow\ \text{A}_\bot=\text{A}_\bot^0[\sin(\text{kx}-\omega\text{t})+\sin(\text{kx}-\omega\text{t}+\phi)]$
$\therefore$ Average Intensity on the screen
$\text{I}=\begin{Bmatrix} \Big|\text{A}_\bot^0\Big|^2+\Big|\text{A}_\bot^0\Big|^2\end{Bmatrix}\big[\sin(\text{kx}-\omega\text{t})(1+\cos^2\phi+2\sin\phi)+\sin^2(\text{kx}-\omega\text{t})\sin^2\phi\big]_\text{average}$
$=\begin{Bmatrix} \Big|\text{A}_\bot^0\Big|^2+\Big|\text{A}_\bot^0\Big|^2\end{Bmatrix}\Big(\frac{1}{2}\Big)2(1+\cos\phi)$
$\Rightarrow\ \text{I}=2\Big|\text{A}_{||}^0\Big|^2(1+\cos\phi)\text{ since},\Big|\text{A}_\bot^0\Big|_\text{av}+\Big|\text{A}_\bot^0\Big|_\text{av}$
With polariser P,
Assume $\text{A}_\bot^2$ is blocked
Intensity $=(\text{A}_{||}^1+\text{A}_{||}^2)^2+({\text{A}_{||}^2})^2$
$=\Big|\text{A}_{\bot}^0\Big|(1+\cos\phi)+\Big|\text{A}_{\bot}^0\Big|.\frac{1}{2}$
Given, $\text{I}_0=4\Big|\text{A}_{\bot}^0\Big|=$ Intensity without polariser at principal maxima.
Intensity at first minima with polariser
$=\Big|\text{A}_{\bot}^0\Big|(1-1)+\frac{\Big|\text{A}_{\bot}^0\Big|}{2}=\frac{\text{I}_0}{8}.$

View full question & answer→

Question 315 Marks

Using Huygens’ Principle, draw a diagram to show propagation of a wavefront originating from a monochromatic point source. Explain briefly.

Answer

Propagation of Wavefront from a Point Source:This principle is useful for determining the position of a given wavefront at any time in the future if we know its present position. The principle may be stated in three parts as follows:

Every point on a given wavefront may be regarded as a source of new disturbance.
The new disturbances from each point spread out in all directions with the velocity of light and are called the secondary wavelets.
The surface of tangency to the secondary wavelets in forward direction at any instant gives the new position of the wavefront at that time.

Let us illustrate this principle by the following example:
Let AB shown in the fig. be the section of a wave front in a homogeneous isotropic medium at t = 0. We have to find the position of the wave front at time t using Huygens’ Principle. Let v be the velocity of light in the given medium.

Take the number of points 1, 2, 3, … on the wave front AB. These points are the sources of secondary wavelets.
At time t the radius of these secondary wavelets is vt. Taking each point as centre, draw circles of radius vt.
Draw a tangent A1B1 common to all these circles in the forward direction.

This gives the position of new wave front at the required time t.
The Huygens’ construction gives a backward wave front also shown by dotted line $A_2B_2$ which is contrary to observation. The difficulty is removed by assuming that the intensity of the spherical wavelets is not uniform in all directions; but varies continuously from a maximum in the forward direction to a minimum of zero in the backward direction.
The directions which are normal to the wave front are called rays, i.e., a ray is the direction in which the disturbance is propagated.

View full question & answer→

Question 325 Marks

Consider the arrangement shown in figure. By some mechanism, the separation between the slits $S_3$ and $S_4$ can be changed. The intensity is measured at the point P which is at the common perpendicular bisector

of $S_1S_2$ and $S_3S_4$. When $\text{z}=\frac{\text{D}\lambda}{2\text{d}},$ intensity measured at P is I. Find this intensity when z is equal to:

$\frac{\text{D}\lambda}{\text{d}}$
$\frac{3\text{D}\lambda}{2\text{d}}$
$\frac{2\text{D}\lambda}{\text{d}}$

Answer

When, $\text{z}=\frac{\text{D}\lambda}{\text{d}}$

So, $\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{2\text{d}}$
$\Rightarrow$ Dark fringe at $S_3 $ and $S_4.$
$\Rightarrow$ At $S_3$, intensity at $S_3 = 0 \Rightarrow I_1 = 0$
At $S_4$, intensity at $S_4 = 0 \Rightarrow I_2 = 0$
At P, path difference = 0 ⇒ Phase difference = 0.
$\Rightarrow\text{I}=\text{I}_1+\text{I}_2+\sqrt{\text{I}_1\text{I}_2}\cos0^\circ=0+0+0=0$
⇒ Intensity at P = 0.

Given that, when $\text{z}=\frac{\text{D}\lambda}{2\text{d}}$, intensity at P = I

Here, $\text{OS}_3=\text{OS}_4=\text{y}=\frac{\text{D}\lambda}{4\text{d}}$
$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{\pi}{2}.$
$\Big[$Since, x = path difference $=\frac{\text{yd}}{\text{D}}\Big]$
Let, intensity at $S_3$ and $S_4 = I'$
$\therefore$ At P, phase difference = 0
So, $\text{I}'+\text{I}'+2\text{I}'\cos0^\circ=\text{I}$
$\Rightarrow4\text{I}'=\text{I}\Rightarrow\text{I}'=\frac{1}{4}$
When, $\text{z}=\frac{3\text{D}\lambda}{2\text{d}},\Rightarrow\text{y}=\frac{3\text{D}\lambda}{4\text{d}}$
$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{3\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{3\pi}{2}$
Let, I'' be the intensity at $S_3$ and $S_4$ when, $\phi=\frac{3\pi}{2}$
Now comparing,
$\frac{\text{I}"}{\text{I}}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos\Big(\frac{3\pi}{2}\Big)}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{2\text{a}^2}{2\text{a}^2}=1$ $\Rightarrow\text{I}"=\text{I}'=\frac{\text{I}}{4}$
$\therefore$ Intensity at $\text{P}=\frac{\text{I}}{4}+\frac{\text{I}}{4}+2\times\Big(\frac{\text{I}}{4}\Big)\cos0^\circ=\frac{\text{I}}{2}+\frac{\text{I}}{2}=1$

When $\text{z}=\frac{2\text{D}\lambda}{\text{d}}$

$\Rightarrow\text{y}=\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{\text{d} }$
$\therefore\phi=\frac{2\pi\text{X}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{\text{d}}\times\frac{\text{d}}{\text{D}}=2\pi.$
Let, I"' = intensity at $S_3$ and $S_4$ when, $\phi=2\pi.$
$\frac{\text{I}'''}{\text{I}'}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos2\pi}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{4\text{a}^2}{2\text{a}^2}=2$
$\Rightarrow\text{I}'''=2\text{I}'=2\Big(\frac{\text{I}}{4}\Big)=\frac{\text{I}}{2}$
At P,$ I_{resultant} =\frac{\text{I}}{2}+\frac{\text{I}}{2}+2\Big(\frac{\text{I}}{2}\Big)\cos0^\circ=\text{I}+\text{I}=2\text{I}$
So, the resultant intensity at P will be 2I.

View full question & answer→

Generate Paper Free All Wave Optics topics