Physics M.C.Q [1M]

2. Decrease in relaxation time.

Explanation:

$\rho=\frac{1}{\rho}=\frac{\text{m}}{\text{ne}^2\tau}$

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.

1. Infinity.

Explanation:

By mirror formula:

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

Here u = -30cm

f = +30cm

So, $\frac{1}{\text{V}}-\frac{1}{30}=\frac{1}{30}$

Þ v= 15cm behind the mirror.

1. 2D

Explanation:

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 f₁ = power of f₂

$\text{P}=\frac{1}{\text{f}_1}=\frac{1}{\text{f}_2}=2\text{D}$

1. 7.5º.

Solution:

Key concept:

In thin prisms, the distance between the refracting surfaces is ineligible and the angle of prism (A) is very small. Since A = r₁ + r₂, therefore if A is small then both r₁, and r₂ are also small, and the same is true for i₁and i₂.

According to Snell's law, $1\sin\text{i}_1=\mu.\sin\text{r}_1\Rightarrow\text{i}_1=\mu.\text{r}_1$

Also, $1.\sin\text{i}_2=\mu.\sin\text{r}_2\Rightarrow\text{i}_2=\mu.\text{r}_2$

Therefore, deviation, $\delta=(\text{i}_1-\text{r}_1)+(\text{i}_2-\text{r}_2)$

$\Rightarrow\ \delta=(\text{i}_1+\text{i}_2)-(\text{r}_1+\text{r}_2)=(\text{r}_1+\text{r}_2)(\mu-1)$

$\Rightarrow\ \delta=\text{A}(\mu-1)$

Since, deviation $\delta=(\mu-1)\text{A}$

$=(1.5-1)\times5^\circ=2.5^\circ$

The angle of the prism is 5º. The ray emerges from refracting face of a prism normally.

Then, i₂ = r₂ = 0

As A = r₁ + r₂ ⇒ A or r₁ = 5º

But $\text{i}_1=\mu.\text{r}_1=\frac{3}{2}\times5=7.5^\circ$

4. Decreases.

Explanation:

$\frac{1}{\text{f}}=\frac{1}{\text{f}_{\text{L}_1}}+\frac{1}{\text{f}_{\text{L}_2}}$

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

Local length of the combination.

$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{-2}{\text{R}}\Big)+(\mu-1)\Big(\frac{-2}{\text{R}}\Big)$

$\frac{1}{\text{f}}=-4(\mu-1)\Big(\frac{1}{\text{R}}\Big)$

$\text{f}=\frac{\text{R}}{4(\mu-1)}$

Where $\text{f}_{\text{L}_1}=\text{f}_{\text{L}_2}=\frac{\text{R}}{2(\mu-1)}$

$(\text{f}_{\text{L}_1}=\text{f}_{\text{L}_2})>\text{f}$

2. May see a primary and a secondary rainbow as concentric circles.

Solution:

As aeroplane is at higher altitude, the passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

Key Concept:

1. If an object move with constant speed (V₀) towards a convex lens from infinity to focus, the image will move slower in the beginning and then faster. Also $\text{V}_\text{i}=\Big(\frac{\text{f}}{\text{f}+\text{u}}\Big)^2\text{V}_0$.
2. If an object approaches the lens, the image moves away from lens with a non-uniform acceleratiion.

2. Moves away from the glass slab

Explanation:

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.

2. Decreases.

Explanation:

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.

3. 1

We have,

h = Object height

u = Object distance = 25cm

D = Near point = 25cm

Now,

1. a convergent lens.

Explanation:

Here P, P₁ & P₂ are the power of Lenses.

P = P₁ + P₂

$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$

$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$

$=\Big(\frac{3}{2}-1\Big)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$

$\frac{1}{\text{f}}=\frac{1}{\text{R}}-\frac{1}{3\text{R}}$

$\frac{1}{\text{f}}=\frac{3-1}{3\text{R}}$

$\text{f}=\frac{3\text{R}}{2}$

focal length of combined is positive means it will behave like a canvergent lens.

1. Shift is more for violet colour

Explanaiton:

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

3. Act as a convex lens irrespective of the side on which the object lies.

Solution:

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.

2. 2.5cm

Explanation:

$\tan\theta=\frac{5}{40}=\frac{\text{h}}{20}$

$\text{h}=\frac{5}{2}=\frac{\text{h}}{20}$

$\text{h}=\frac{5}{2}=2.5\text{cm}$

The maximum value of "h =2.5cm" to see the Image of the object.

2. $\text{f}=\text{R}$

Explanation:

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

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

$\frac{1}{\text{f}}=\frac{1}{\text{R}}$

$\text{f}=\text{R}$

3. Infinite

Explanation:

$\frac{1}{\text{f}}​=(\text{n}−1)\Big( \frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$

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: $\frac{1}{\text{f}}=0$

⇒ f = $\infty$ 

3. $\frac{1}{\text{v}-\text{t}}-\frac{1}{\text{u}+\text{t}}=\frac{1}{\text{f}}$

Explanation: