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.