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'.