Physics M.C.Q (1 Marks)

$\rho=\frac{1}{\rho}=\frac{\text{m}}{\text{ne}^2\tau}$
As we increase temperature, average speed of the electrons, which act as the carriers of current, increases resulting in more frequent collisions. The average time of collisions $\tau,$ thus decreases with temperature.

Key Concept: According to Couchy relationship,
$\lambda\propto\frac{1}{\mu}$
Smaller the wavelengh higher the refractive index and consequently smaller the critical angle.

We know $\text{v}=\text{f}\lambda$ the frequency of wave remains unchanged with medium hence $\text{v}\propto\lambda$.
The critical angle, sin $\text{C}=\frac{1}{\mu}$
Also, velocity of light, $\text{v}\propto\frac{1}{\mu}$
According to $\text{VIBGYOR}$, among all given sources of light, the blue light have smallest wavelength. As $\lambda_{\text{blue}}<\lambda_\text{yellow}$ hence $\text{v}_{\text{blue}}<\text{v}_\text{yellow}$, it means $\mu_{\text{blue}}<\mu_\text{yellow}$.
It means critical angle for blue is less than yellow colour, the critical angle is least which facilitates total internal reflection for the beam of blue light.

Flint$-$glass lens has a larger chromatic aberration because the dispersive power of flint$-$glass is higher.

$\frac{1}{\text{f}}​=(\text{n}−1)\Big( \frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
where, n is refractive index of lens material with respect to its surrounding medium.
If lens is immersed in the liquid of refractive index equal to its own then $n = 1$ and hence above equation becomes: $\frac{1}{\text{f}}=0$
$\Rightarrow f = \infty$

By Rayleigh's Criterion intensity of scattered light is given by, $\text{I}\propto\frac{1}{\lambda^4}​$
Intensity of scattered light is more for smaller wavelengths of light.
In the visible spectrum, the blue side of the spectrum has a smaller wavelength compared to the red side. Hence blue light is scattered more by air molecules present in earth's atmosphere.

As we know,
$\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{\text{u+v}}{\text{uv}}$
$\text{f}=\frac{\text{uv}}{\text{u+v}}$

Radio waves are reflected by a layer of atmosphere called the Ionosphere, so they can reach distant parts of the Earth. The reflection of radio waves by ionosphere is due to total internal reflection. It is the same as total internal reflection of light in air during a mirage, i.e., angle of incidence is greater than critical angle.
Important point: The ionized part of the Earth’s atmosphere is known as the ionosphere. Ultraviolet light from the sun collides with atoms in this region knocking electrons loose. The creates ions, or atoms with missing electrons. This is what gives the Ionosphere its name$-$ and it is the free electrons that cause the reflection and absorption of ratio waves.

As the magnifying lens $($convex lens$)$ is moved far away from the eye, the image formed is real and inverted and is formed inside the human eye and hence it blurs. Since the object is far away from the mirror, the image formed is diminished and hence magnification reduces.

In astronomical telescope $2$ convex lens called eyepiece and objective lens are used and object is placed before eyepiece lens, such that final image inverted, a camera and eye also form inverted image on the screen.
Whereas simple microscope gives an erect, virtual and enlarged image of the object placed between first principal focus and the optic nerve of the convex lens.
In a projector, the image formed is real, inverted magnified on the other side of the lens. This inverted image is again inverted by the film.

A concave lens always produces virtual, erect and diminished images and the decrease in the size of the image depends on the position of the object.
Concave lens will shrink the size of the already small letters.
A convex lens produces real and virtual, erect and inverted, diminished, same sized and magnified image of the object, depending upon the position of the object on the principal axis.
When the object is placed between $F$ and $2F$ of convex lens, an enlarged but inverted image of the object is formed. The magnified image makes it easier to read small letters but the inverted image is undesirable.
When the object is placed at a distance less than the focal length of convex lens, an enlarged and erect image of the object is formed, which makes it easier to read small letters.

Key concept:
In thin prisms, the distance between the refracting surfaces is ineligible and the angle of prism $(A)$ is very small. Since $A=r_1+r_2$, therefore if $A$ is small then both $r_1$, and $r_2$ are also small, and the same is true for $i_1$ and $i_2$.

According to Snell's law, $1 \sin i _1=\mu$. $\sin r _1 \Rightarrow i _1=\mu . r _1$
Also, $1 . \sin i _2=\mu \cdot \sin r _2 \Rightarrow i _2=\mu \cdot r _2$
Therefore, deviation, $\delta=\left( i _1- r _1\right)+\left( i _2- r _2\right)$
$\Rightarrow \delta=\left( i _1+ i _2\right)-\left( r _1+ r _2\right)=\left( r _1+ r _2\right)(\mu-1)$
$\Rightarrow \delta= A (\mu-1)$
Since, deviation $\delta=(\mu-1) A$
$=(1.5-1) \times 5^{\circ}=2.5^{\circ}$
The angle of the prism is $5^{\circ}$. The ray emerges from refracting face of a prism normally.
Then, $i _2= r _2=0$
As $A=r_1+r_2 \Rightarrow A$ or $r_1=5^{\circ}$
But $i _1=\mu . r _1=\frac{3}{2} \times 5=7.5^{\circ}$

Here $P, P_1 \ P_2$ are the Power of Lenses.
$P=P_1+P_2$
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}=\frac{1}{\text{f}_2}$
$(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$
$(1.2-1)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$
$\frac{1}{\text{f}}=\frac{2}{5\text{R}}-\frac{1}{3\text{R}}$
$\frac{1}{\text{f}}=\frac{6-5}{15\text{R}}$
$\text{f}=15\text{R}$
Focal lenght of combined is positive, but it's magnitude in capair to $f_1 \ f_2$ is High. So it will be hare like a divergent lens.

If an object approaches a convergent lens from the left of the lens with a uniform speed of $5m/s$, then the image moves away from lens with a non$-$uniform acceleration.

Person $B$ has stronger ciliary muscles than person $A$. So, the muscles in his case can be strained more and the focal length of his eye can be reduced more compared to those of person $A$. While seeing far objects, the muscles are relaxed, so their strength will not affect the far point of the eye.

Magnification of image created by mirror is defined as
$\text{m}=\frac{\text{size of object}}{\text{size of image}}$​
and in case of inverted image. Size of image is negative whereas size of object is positive. Hence , magnification produced is negative and it can be unity when object is placed at center of curvature and infinity when object is at focus.

$\text{Magnification}=\frac{\text{Image Height}}{\text{Object Height}}​ $
because unit of image height and object height is same hence unit of magnification is none because it is a constant number. So, it has got to unit.

As aeroplane is at higher altitude, the passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.
Key Concept:
If an object move with constant speed ($V_0$) towards a convex lens from infinity to focus, the image will move slower in the beginning and then faster. Also $\text{V}_\text{i}=\Big(\frac{\text{f}}{\text{f}+\text{u}}\Big)^2\text{V}_0$.
If an object approaches the lens, the image moves away from lens with a non$-$uniform acceleratiion.

Refraction is a phenomenon in which when a ray passes from one medium to another it bends away from its straight$-$line path due to the difference in optical densities or refractive indices of the two mediums. This bending is known as deviation.

As a glass has refractive index of $\frac{3}{2}$ with respect to air, the convergent light will bend towards the normal in the glass and again bend away from the normal, but in the whole process the beam is displaced parallel to initial path as shown in figure and converge at a greater distance from the glass slab.

Here initially $A\ \&\ B$ is parallel to each other after reflection by teh plane mirror $A'\ \&\ B'$ goes Parallel to each other.

Lateral displacement due to a slab is given by $\delta=\text{t}(\mu−1)$; where t is thickness of the medium and $\mu $ is refractive index of medium.

In combination $($refractive angles of prisms reversed with respect to each other$)$, the deviations through two prisms cancel out each other and the net deviation is due to the third prism only.

If $\mu$ is the refractive index and A is the angle of prism, then the angular dispersion produced by the prism will be given by $\delta=(\mu-1)\text{A}.$
Because the relative refractive index of glass with respect to water is small compared to the refractive of glass with respect to air, the dispersive power of the glass prism is more in air than that in water.

Before cut
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{2}{\text{R}}\Big)=4\text{D}$
After cut
$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_1$
$\&\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_2$
Power of a divided lens will be = $P_1 + P_2$
$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)$
$=4\text{D}$

Magnifaction of a mirror is defined as
$\text{m}=\frac{\text{size of image​}}{\text{size of object}}$
and since, in case of virtual and errect image, size of image and object both are positive. Hence magnification created by mirror is positive.

$\text{v}=(40-4)$
$\frac{1}{\text{f}}=\frac{1}{40-4}-\frac{1}{(-\text{u})}$
$\frac{\text{df}}{\text{du}}=0$ for f minimum.
$\frac{\text{df}}{\text{du}}=1-\frac{\text{u}}{20}=0$
$\text{u}=20$
$\text{f}_{\text{min}}=10\text{cm}$

If $\mu$ is the average refractive index and A is the angle of prism, then the angular dispersion produced by the prism is given by $\delta=(\mu-1)\text{A}.$

As velocity of wave is given by the relation $\text{v}=\text{f}\lambda$. When light ray goes from one medium to other medium, the frequency of light remains unchanged. Hence $\text{v}\propto\lambda$ or greater the wavelength, greater the speed.
The light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerges first.

Principal section is also defined as the normal 'side view' of the mirror for a ray diagram. In the diagram $AB$ is the principal section.

The ray $PQ$ of light passes through focus $F$ and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray $2$ in the figure.
Important points: We can locate the image of any extended object graphically by drawing any two of the following four special rays:
A ray initially parallel to the principal axis is reflected through the focus of the mirror $(1).$
A ray passing through the center of curvature is reflected back along itself $(3).$
A ray initially passing through the focus is reflected parallel to the principal axis $(2).$
A ray incident at the pole is reflected symmetrically.

Shift by which the object appears to be raised depends on these $4$ factors. The shift increases with increase in refraction index of medium and also with the thickness of denser medium.

The apparent vertical shift is given by the equation $\text{AD}=\frac{\text{RD}}{\mu}$, that is, ​$\mu=\frac{\text{RD}}{\text{AD}}$, where $AD$ is the apparent vertical shift, $RD$ is the real depth of the tank and $n$ is the refractive index of the denser medium, that is, water.
As real depth and refractive index is going to be constant always, the apparent shift or depth will be independent of viewing angle.
From the above equation, we can see that the apparent vertical shift is inversely proportional to the refractive index. That is, $\frac{1}{\mu}$​.

Here $P, P_1\ \&\ P_2$ are the power of Lenses.
$P = P_1+P_2$
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$
$=\Big(\frac{3}{2}-1\Big)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$
$\frac{1}{\text{f}}=\frac{1}{\text{R}}-\frac{1}{3\text{R}}$
$\frac{1}{\text{f}}=\frac{3-1}{3\text{R}}$
$\text{f}=\frac{3\text{R}}{2}$
focal length of combined is positive means it will behave like a canvergent lens.

When water is poured then it is due to refraction of light the coin becomes visible. When the light rays travel from the water medium $($denser$)$ to the air medium $($rarer$)$ it bends away from the normal due to the refraction. Therefore, an image of the coin is formed at a smaller depth causing it to be visible $($as shown in the figure$)$.

Since, $\mu_\text{V}​>\mu_\text{R}$​, the shift is more for he violet light than the red light in a given medium.

We know that, $\text{f}=\frac{\text{R}}{\mu-1}$
Substituting $\text{R}=20\text{cm},\mu=1.5,$ we get
$\text{f}=\frac{20}{1.5-1}=40\text{cm}$
Since, the focal length is greater than zero.
Therefore, lens act as a convex lens irrespective of the side on which the object lies.

$\text{refractive index} = \frac{\text{actual depth​}}{\text{apparent depth}}$

critical angle: It is the angle of incidence for which angle of refraction is $90^\circ . ​$

$\Rightarrow\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\Rightarrow\frac{1}{\text{f}}=\Big(\frac{3}{2}-1\Big)\Big(\frac{1}{\text{R}}-\Big(-\frac{1}{\text{R}}\Big)\Big)$
$\frac{1}{\text{f}}=\frac{1}{\text{R}}$
$\text{f}=\text{R}$

As shown in figure $A$, a virtual image is formed because refracted rays only appear to meet at the image.
As shown in figure $B$, a real image is formed because refracted rays actually meet at the image.

$m_1$, Image is $O^1$
$m_3$, Image is $O^{11}$
Spherical Surface formula
$\frac{\mu^{11}}{\text{v}}-\frac{\mu_1}{\mu}=\frac{\mu^{11}-\mu^1}{\text{R}}$
If ray goes to $m^2$ to $m^1$ than Image is formed at $O^1$ and if ray goes to $m_2$ to $m_3$ than Image is formed at $O^{11}$.

Key concept: The Snell's law describes the relation between angle of incidence $\theta_1$ and angle of refraction $\theta_2$:
$\mu\sin\theta_1=\mu_2\sin\theta_2=\text{constant}\ .....(\text{i})$
where $\mu_1$ and $\mu_1$ are refractive indices of the two media.

When a light ray goes from $($optically$)$ denser medium to $($optically$)$ rarer medium, then it bends away the normal, i.e., $\theta_1<\theta_2$ and vice-versa.
Here, light ray goes from $($optically$)$ rarer medium air to optically denser medium turpentine, then it bends towards the normal, i.e., $\theta_1>\theta_2$ whereas when it goes from to optically denser medium turpentine to rarer medium water, then it bends away the normal.

When the image is formed at infinity
$m = Df = DPm = Df = DP$
when the image is formed at the near point
$m = (1+Df) = 1 + DP$

Lateral displacement produced by a glass slab of thickness $t$, incident on it at angle of incidence i is
Lateral displacement $ \delta=\frac{\text{t}}{\text{cosr}}​\text{sin}(\text{i - r})$

Focal length is inversely proportional to refractive index and refractive index is inversely proportional to $\lambda^2.$
So, keeping other parameters the same, we can say:
$\text{f}\propto\frac{1}{\lambda^2}\ \ (\because\lambda_\text{r}<\lambda_\upsilon)$
$\therefore\text{f}_\text{v}<\text{f}_\text{r}$

A lens is a piece of glass $($plastic$)$ with two refracting surfaces, which are either curved $($e.g., a segment of a sphere$)$ or plain.
Lenses are used to form images by refraction in optical instruments $($microscopes, telescopes, cameras, etc.$)$
There are two types of lenses: converging $($thickest in the middle$)$ and diverging $($thickest at the edges$)$.

The distance $CP$ is the radius of curvature.

Refraction is a phenomenon in which a light ray incident on a surface separating two transparent media bends at the change of medium.
Snell's law gives the relation between angle of incidence and refraction and the Refractive index of the respective medium.
$\mu_1 \sin (i) = \mu_2 ​\sin(r)$
where, i, r are the angles of incidence and refraction respectively.

Real images can be obtained on a screen. On$-$screen rays meet in real.
Virtual images can not be obtained on a screen. Because there is no meaning of screen for the virtual image.

The phenomena when light passes through the object $($a medium$)$ is known as refraction. Refraction is defined as the bending of light ray when it passes from one medium to another.

If operators using a microscope usually wear spectacles $($glasses$)$ for activities such as working at their $PC$, they often need to remove them when looking through a microscope so they can align their eyes correctly with the eyepieces.

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

Step $1:$ Virtual Image
It is formed when ray of light appear to meet at a point.
Step $2:$ Ray Diagram

​The angular magnification is the ratio of the angle subtended by the image to the angle subtended by the object on an unaided eye.
In a simple microscope,
$\text{m}=\frac{\frac{\text{h}}{\text{x}}}{\frac{\text{h}}{\text{D}}}$
Here,
$u =$ Object distance from the lems
$D =$ Image distance form the lens
$h =$ Height of the object
In normal adjustment, the object is placed at a distance equal to focal length $(f)$ from the lens and then magnification is given by m
$=\frac{\text{D}}{\text{f}}$
for $\text{u}<\text{f},\text{ m}=\frac{\text{D}}{\text{f}}+1$

Given:
Near point of the human eye, $u = -25\ cm$
Distance between the retina and the eye lens, $v = 2\ cm ($approximately$)$
thus, we have the focal length, $f.$
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{f}}\cong\frac{1}{2}-\frac{1}{-25}$
$\Rightarrow\frac{1}{\text{f}}\cong\frac{27}{50}$
$\Rightarrow\text{x}\cong2\text{cm}$

Before cut
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{2}{\text{R}}\Big)=4\text{D} \ ...(1)$
After cut
$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)+\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{2}\Big) \ ...(2)$
From eq. $(1)$ we get Power of $f_1$ = power of $f_2$
$\text{P}=\frac{1}{\text{f}_1}=\frac{1}{\text{f}_2}=2\text{D}$

The Intensity on the other side of the lans is maximum at a distance $2f.$

As shown in the diagram,
The due to the refraction of the light, the light will move away from the normal $(NN')$ and consequently, the fish $($denoted as $F)$ will see a longer image of the fisherman $(AB)$ as $AB\ '.$