Physics M.C.Q (1 Marks)

The resolving power of a telescope can be given as:
Resolving power $=\frac{1}{\text{d}(\theta)}=\frac{1}{1.22\lambda/ \text{D}}=\frac{\text{D}}{1.22} ($wavelength$)$
So, resolving power can be increased by decreasing the wavelength and increasing the diameter of objective.

To observe an interference pattern
$\frac{\text{dy}}{\text{D}}​=\text{n}\lambda$
$\text{y}=\frac{\text{nD}\lambda}{\text{d}}​$
i.e, for the bright fringes nearest the central achromatic fringe, wavelength must be minimum and in white light wavelength is minimum for violet.

Resolving power, $\text{R}=\frac{\text{a}}{1.22\lambda}$
where, a is diameter of objective $\lambda$ is wavelength of light magnifying power
$\text{m}=\frac{\text{-f}_0}{\text{f}_\text{e}}\Big(1+\frac{\text{f}_\text{e}}{\text{D}}\Big)$
so, decreasing the wavelength of light increases the resolving power and magnifying power of telescope.

Key concept:
Wave front,
Every point on the given wave front acts as a source of new disturbance called secondary wavelets which travel in all directions with the .velocity of light in the medium.
A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wave front at that instant. This is called secondary wave front.In the given question, there is a hole at point which is a maxima point. From Huygen’s principle, wave will propagate from the sources $S_1$ and $S_2$. Each point on the screen will act as secondary sources of wavelets.

The wave front emitted bu a narrow source is divided in two parts by reflection, refraction or diffraction. The coherent soutces so obtained are imaginary.

As the two slits $S_1$​ and $S_2$​ are not equidistant from the slit s the distance traversed by light through $S_1$​ and $S_2$​ may not differ by an integral multiple of wavelength. Thus it need not be bright . similarly it need not be dark.

In $1678,$ Dutch physcist, christian Huygens beived that light was made up of waves vibrating up and down perpendicular to the direction of the light travels, and therefore formulated a way of visualising wave propagation. This became known as Huygens Principe.

Ratio of maximum intensity and minimum intensity is given by
$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(\sqrt{\text{I}_1}+\sqrt{\text{I}_2})^2}{(\sqrt{\text{I}_1}-\sqrt{\text{I}_2})^2}=\frac{25}{1}$
$\Rightarrow\sqrt{\text{I}_1}=3 \ \text{and}\ \sqrt{\text{I}_2}=2$
$\Rightarrow\text{I}_1=9\ \text{and}\ \text{I}_2=4$
Then,
$\frac{\text{I}_1}{\text{I}_2}=\frac{9}{4}$

Intensity of a point source obeys the inverse square law.
Intensity of light at distance r from the point source is given by
$\text{I}=\frac{\text{S}}{(4\pi\text{r}^2)}$
Where S is the source strength.

None of the option is correct

Consider the diagram, the ray $(P)$ is incident at an angle $\theta$ and gets reflected in the direction $P'$ and refracted in the direction P. Due to reflection from the glass medium, there is a phase change of $\pi$.
According to snell's law, we have $\text{n}=\frac{\sin\theta}{\sin\text{r}}$
$\Rightarrow\sin\text{r}=\frac{\sin\theta}{\text{n}}$
$\Rightarrow\ \cos\text{r}=\sqrt{1-\sin2\text{r}}$
$\Rightarrow\ \cos\text{r}=\sqrt{1-\frac{\sin^2\theta}{\text{n}^2}}$
The time taken to travel along $OP"$ is given by
$\Delta\text{t}=\frac{\text{OP}''}{\text{v}}$
$=\frac{\frac{\text{d}}{\cos\text{r}}}{\frac{\text{c}}{\text{n}}}\ \Big[\because\ \text{PO}''=\frac{\text{d}}{\cos\text{r}}\text{ and }\text{v}=\frac{\text{c}}{\text{n}}\Big]$
$=\frac{\text{nd}}{\text{c}\cos\text{r}}$
$=\frac{\text{nd}}{\text{c}\Big(1-\frac{\sin^2\theta}{\text{n}^2}\Big)^{\frac{1}{2}}}\ \ \bigg[\because\ \cos\text{r}=\sqrt{1-\frac{\sin^2\theta}{\text{n}^2}}\bigg]$
$=\frac{\text{nTd}}{\lambda}\Big(1-\frac{\sin^2\theta}{\text{n}^2}\Big)^\frac{-1}{2}$
Now, the phase difference $(\Delta\phi)$ is given by
$\frac{2\pi}{\text{T}}\times\Delta\text{t}\times\frac{2\pi}{\text{T}}\times=\frac{\text{nTd}}{\lambda}\Big(1-\frac{\sin^2\theta}{\text{n}^2}\Big)^\frac{-1}{2}$
$=\frac{2\pi\text{nd}}{\lambda}\Big(1-\frac{\sin^2\theta}{\text{n}^2}\Big)^\frac{-1}{2}$
Therefore, the net phase difference $=\Delta\phi+\pi$
$=\frac{2\pi\text{nd}}{\lambda}\Big(1-\frac{\sin^2\theta}{\text{n}^2}\Big)^\frac{-1}{2}+\pi$

The fringe width of a double slit interference pattern is given by
$\beta=\frac{\text{D}\lambda}{\text{d}}​$
where $D$ is the distance between screen and the plane containing slits $d$ is distance between the slits $S_2$ and $S_3$. To make the fringe pattern visible, the fringe width must be increased, for which $d$ should be decreased.

Path difference at the central fringe will be the same as the path difference at the slits i.e. $\lambda$.
Hence, the waves reaching the central fringe will be $180^\circ $ out of phase and will result in destructive interference. Hence, the intensity of the central fringe will be $0.$

Resolving power is given by the distance between two objects to be distinguished per unit distance of objects from the object distinguishing them.
Hence, $\theta=\frac{\text{d}}{\text{D}}​$
Hence, $\theta\text{D}=\frac{1}{60}\times\frac{\pi}{180}​​\times110000$
$= 3.2m$

The photoelectric effect.
In $1905$, Albert Einstein published a paper in advancing hypothesis that the light energy $I'd$ being carried in discrete quantized packets to explain experimental data from photoelectric effect. This model contributed the development of quantum mechanics.
Photoelectric effect refers to the emission, or rejection of electrons grim the surface of generally a metal in response to the incident light.

As we know the visibility of the human eye is limited up to a distance.
It is known to us that the normal pupil size of any human being is $4\ mm$. This measurement sets a minimum resolution approximately $1'$ to $2′.$
When we want to pull small objects closer to our eyes, we aim to see them properly. But it is often seen that after crossing a certain distance the particles become unclear no matter how much closer it is to our eyes.
This signifies that there is a minimum distance of comfortable viewing. This distance is roughly calculated as $25\ cm.$

The resolution of the human eye is the smallest object our eye can see. This is limited by the diffraction limit, which is approximated by the angular size ratio of the object's size versus the distance to the object.
The normal pupil size of a human eye is $4\ mm,$ which sets a minimum angular resolution of the eye and to able to see the small objects we bring them as close to our eyes as possible, but there is a minimum distance for comfortable viewing which is roughly at $25\ cm.$
But quoted figure for the smallest resolvable size is $0.1\ mm,$ showing that the diffraction limit is a crucial factor in visual resolving power.

For very thin films the distance travelled inside the film is insignificant and so the two reflected waves are almost exactly out of phase with each other $($due to the phase change at one surface$)$; they interfere destructively and the film appears 'black'.

As fringe width is proportional to the wavelength and wavelength of light is inversely proportional to the refractive index of the medium,
Here,
$\lambda_\text{M}=\frac{\lambda}{\eta}$
$\lambda_\text{M}$ = wavelength in medium
$\lambda$ = wavelength in vacuum
$\eta$ = refractive index of medium
Hence, fringe width decreases when Young's double slit experiment is performed under water.

The corpuscular theory of light is proposed by newton in $1704$. It is referred as particle theory or newton’s theory of light.
According to this theory,
Light is made up of tiny particles called corpuscles having negligible mass. These particles $($corpuscles$)$ are perfectly elastic.
These tiny particles always travel in straight line in all directions.
These corpuscles ravel at very high velocity.
These corpuscles are of different sizes. The different color of light is due to different sizes.

Fringe width, $\beta=\frac{\lambda\text{D}}{\text{d}}$
Wavelength of red light is greater than wavelength of violet light; so, the fringe width will reduce.

Key concept:
Diffraction of Light is the phenomenon of bending of light around the comers of an obstacle/aperture of the size of the wavelength of light.

Size of the slit is very large compared to wavelength

Size of the slit is comparable to wavelenght
In figure $(A)$, no diffraction phenomenon is observed as the size of slit is weary large compared to wavelength. But in figure$(B),$ there will be diffraction of light as size of slit is compared to the wavelength of light incident.
Here in the question it is given, width of the slit
$\text{b}=10^4\mathring{\text{A}}=10^4\times10^{-10}\text{m}$
$=10^{-6}\text{m}=1\text{pm}$
Wavelength of $($visible$)$ sunlight varies from $4000\mathring{\text{A}}\text{ to }8000\mathring{\text{A}}$.
Hence the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear, i.e., mixing of colours form white patch at the centre.

When ordinary $($unpolarised$)$ light is reflected from a surface, it gets partial polarisation, it means some of the electric vectors in some planes are cut$-$off. When this partially polarized light is incident on polarized glasses, glass acts as an analyser and therefore block out some reflected light.

In $1678$, Dutch physicist, Christian Huygens, believed that the light was made up of waves vibrating up and down perpendicular to the direction of the light travels and therefore formulated away of visualising wave propagation.
This becomes know as Huygens principle. Huygens theory was the successful theory of light in three dimensions. Huygens suggested that the light wave peaks form from surfaces like the layers of onion. In a vacuum or other uniform mediums the light waves are spherical and these wave surfaces advance or spread out as they travel at the speed if light.
This theory explains why light shining through a pinhole or slitwall spread out rather than going in straight lines.

On writing the given equation in the plane equation form $lx + my + nz = p,$
Where $l^2+m^2+n^2$ and $p>0$, we get:
$\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=\frac{\text{c}}{\sqrt{3}}$
If $\theta$ is the angle between the normal and $+x$ axis, then
$\cos\theta=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big) $

Interference effect is produced by a thin film $($coating of a thin layer of a translucent material on a medium of different refractive index which allows light to pass through it$)$. ln the present case, oil floating on water forms a thin film on the surface of water, leading to the display of beautiful colours in daylight because of the interference of sunlight.

waves front are plane perpendicular to the direction of rays.
so, as light is traveling along $y -$ axis, plane perpendicular to $y -$ axis is the $x-z$ plane with any constant value of $y.$
so, $y =$ constant is the wave front plane

Resolving power of a telescope $=\frac{\text{a}}{1.22\lambda}$
where a is the aperture of the telescope.
Thus resolving power∝ aperture.
Hence, if the aperture of telescope decreases, the resolving power decreases.

We know that, for the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength.
In a Young's double$-$slit experiment, if one of the holes is covered by a red filter and another by a blue filter, then only red and blue lights are present due to alteration. In Young's double$-$slit experiment, a monochromatic light is used for the formation of fringes on the screen.
Therefore, there shall be no interference fringes.

Due to the large distance, the radius of the wavefront can be considered as large $($infinity$)$ and hence, a wavefront is almost plane.

Unequal width of slits will cause unequal intensity of lights entering from both slits.
As a result, during interference complete cancelling of light intensity will not take place at regions of otherwise dark fringe.
As the value of $\beta$ does not depend on intensity of light, there will be no shifting of fringes as well as no change in fringe width.

Frequency of a light wave, as it travels from one medium to another, always remains unchanged, while wavelength decreases.
Decrease in the wavelength of light entering a medium of refractive index $\mu$, is given by,
$\lambda_\text{M}=\frac{\lambda}{\mu},$
Where $\lambda_\text{M}$ = wavelength in medium
$\lambda =$ wavelength in vacuum
$\mu =$ refractive index

The setup shown is that of Young's Double Slit Experiment.
The fringe width in the experiment is given as $\beta=\frac{\text{D}\lambda}{\text{d}}​$
where d is the distance between the slits, $D$ is the distance between screen and slits.
Hence the central fringe also widens when distance between screen and slits increases.

The Rayleigh criterion is the generally accepted criterion for the minimum resolvable detail $-$ the imaging process is said to be diffraction$-$limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.

Intensity is proportional to the area of the slit.
As slit widths are in the ratio of $1:9$
The areas are also in the ratio $1:9$
Thus Intensities are in the ratio $1:9$
amplitudes are square root of Intensities
Thus amplitudes are in ratio $1:3$
Let amplitudes be x and $3x$
At maxima the amplitudes get added up $x + 3x = 4x$
At minima they become $x − 3x = −2x$
Intensity of maxima to minima is $\frac{16\text{x}^2}{4\text{x}^2}=\frac{4}{1}$

The speed of the current in the conductor is the same as the speed of light in the vacuum which is $3\times 10^8$
$\frac{\text{m}}{\text{s}}$

When the film is thin,$ t \rightarrow 0$, path diff. $= \frac{\lambda}{2}$. Therefore, in reflected light, the film appears black.

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point $($closely spaced$)$ as distinct as possible.

Polarizers are used in industry to reveal stress patterns in machinery and tools. Sunglasses are used to protect the eyes by polarizing the light to reduce glare using the tailor$-$made material of the glasses.

An electromagnetic wave bends round the corners of an obstacle if the size of the obstacle is comparable to the wavelength of the wave. An $AM$ wave has less frequency than an $FM$ wave, So, an $AM$ wave has a higher wavelength than an $FM$ wave and it bends round the comers of a $1m \times 1m$ board.

​The wave is travelling along the $X-$axis. So, it'll have planar wavefront perpendicular to the $X-$axis.

The locus of all particles in a medium, vibrating in the same phase is called wave front.
The direction of propagation of light $($ray of light$)$ is perpendicular to the wave front.
A wave front is an imaginary surface where all particles lying on this vibrate in the same phase.

Radius of nth order Newton's ring is proportional to $\sqrt{\lambda}$ which decreases in oil since $\lambda=\frac{\lambda_\text{vacuum}}{\mu}$​​. Thus the radius of newton's ring decreases in oil film.

when you considered it a large distance and measuring justice Mall section of it then it can be considered to be plane wavefront source at Infinity example the one coming from sun to earth surface is considered to be plain $VU$ friend from light diverging from a point source will be spherical.
So, the wave front due to a source situated at infinity is planar.

In $1669,$ another Danish scientist, Erasmus Bartholinus discovered the polarization of light by double refraction in Iceland spar $($calcite$).$

Resolving power of telescope $\text{R}=\frac{1}{\Delta\theta}=\frac{\text{a}}{1.22\lambda}$
where, $\Delta\theta$ is angular separation between two objects.
a is the diameter of the objective.
$\lambda$ is wavelength of light.
So, clearly resolving power of a telescope depends on diameter of the objective.

A wavefront is the locus of points characterized by propagation of position of the same phase:
a propagation of a line in $1D$, a curve in $2D$ or a surface for a wave in $3D.$

Hint: If a light beam incidents at Brewster's angle, then the transmitted beam is always unpolarised and reflected beam is always polarised.
In the given diagram, the light beam incident from air to the glass slab at Brewster's angle $(ip).$ Therefore, the incident ray represented by dot $(.),$ is unpolarised and the reflected light represented by arrow, is plane polarized.
Since, the emergent ray is unpolarised.
Hence, the intensity cannot be zero when passes through polaroid.

For a fringe to appear, the two wavelengths must interfere to give a maxima, which appears at a distance where the phase of both the rays are same, that is, at the least count multiple of the waves.
Therefore the first fringe appears at $5200 \mathring{\text{A}}$.
To accommodate $250$ fringes, the thickness of the film is $250$ times the distance where the first fringe occurs.
Thus the thickness of the air film is $1.3\ mm$

The range for visible light is between $4 \times 10^{-7} \mathrm{~m}$ to $7 \times 10^{-7} \mathrm{~m}$. The only wavelength that is present in this range given as an option is $6 \times 10^{-7} \mathrm{~m}$.

For constructive interference,
$A_R=A_1+A_2$
Due to the constructive interference of the superimposing waves, resultant wave of larger amplitude is obtained and hence constructive interference results in the larger wave.

The length of telescope $=$ focal length of object $(-f_0) +$focal length of eyepiece $(f_e​) = 100 + 5 = 105\ cm$

Fringe width $\beta= \frac{\lambda\text{D}}{\text{d}}$ where $\lambda $ is the wavelength of the light used
$\Rightarrow\beta\propto\lambda$
As wavelength of the red light is the largest among the visible light, thus fringe width is the greatest formed due to red light.

The corpuscular theory was largely developed by Sir Isaac newton. Newton's theory remained in force for more than $100$ years and took precedence over Huygen's wave front theory, partly because of Newton’s great prestige.
When the corpuscular theory failed to adequately explain the diffraction, interference and polarization of light it was abandoned in favour of Huygens' wave theory.

Resolving power of a telescope:
$\text{R}=\frac{\text{a}}{1.22\lambda}$
where, a is diameter of the objective
so, $R$ increases when a is increased and a increases when aperture of objective is increased.

$\text{Coherent time} =\frac{\text{Coherence length}}{\text{Velocity of light}}​=\frac{\text{L}}{\text{c}}​$

Thomas Young demonstrated the phenomenon of interference in water waves.
In $1801$, he presented a famous paper to the Royal Society entitled "On the Theory of Light and Colours" which described various interference phenomena, and in $1803$ he performed his famous double$-$slit experiment.

In photometry, luminous intensity is a measure of the wavelength$-$weighted power emitted by a light source in a particular direction per unit solid angle, based on the luminosity function, a standardized model of the sensitivity of the human eye. The $SI$ unit of luminous intensity is the candela $(cd),$ an $SI$ base unit.

$\beta$ are stream of particles comprising of electrons moving with very high velocity, hence it cannot be polarized, while others can as they are electromagnetic waves.

Resolving limit
$\text{d}\theta=\frac{1.22\lambda}{\text{a}}$
$=\frac{1.22\times4538\times10^{-10}}{1}$
$= 5.54 \times 10^{-7} \mathrm{rad}.$

Newton's particle theory remained accepted for a long time when it was challenged by the double$-$slit experiment of Thomas Young $(1773-1829)$ in $1801$. This experiment clearly established that light coming from two coherent sources interferes and produces maxima and minima depending on the path difference.

Only the transverse waves can be polarized.
As the sound waves are longitudinal waves in nature whereas visible light rays and $X-$rays are transverse waves.
Thus sound waves cannot be polarized.

When exposed to sunlight, thin films of oil on water often exhibit brilliant colors due to the phenomenon of interference.
In the case of a thin oil film, a layer of oil sits atop a layer of water. The oil may have an index of refraction near $1.5 $ and the water has an index of $1.33.$
The materials on either side of the oil film $($air and water$)$ both have refractive indices that are less than the index of the film.