Physics 3 Marks Question

1. $\mu=\frac{\sin\big(\frac{A+\delta_m}{2}\big)}{\sin\big(\frac{A}{2}\big)}$

$=\frac{\sin\big(\frac{60+30}{2}\big)}{\sin\big(\frac{60^\circ}{2}\big)}=\sqrt{2}$

Also $\mu=\frac{c}{v}\Rightarrow v=\frac{3\times10^8}{\sqrt{2}}\text{m/s}$

$=2.122\times10^8\text{m/s}$

At face AC, let the angle of incidence be r₂. For grazing ray, e = 90⁰

^{$\Longrightarrow\mu=\frac{1}{sin r_2}\Longrightarrow r_2=\sin^{-1}\big(\frac{1}{\sqrt{2}}\big)=45^\circ$}

Let angle of refraction at face AB be $r_1$. Now $r_1+r_2=A$

$\therefore r_1=A-r_2=60^\circ-45^\circ=15^\circ$

Let angle of incidence at this face be $i$

$\mu=\frac{\sin i}{\sin r_1}$

$\Longrightarrow\sqrt{2}=\frac{\sin i}{\sin 15^\circ}$

$\therefore i=\sin^{-1}(\sqrt{2}.\sin 15^\circ)$

1. When the telescope is in normal adjustment, its magnifying power is given as:

When the final image is formed at d, the

$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}$

$=\frac{140}{5}=28$

2. magnifying power of the telescope is given as:

$\frac{\text{f}_0}{\text{f}_\text{e}}\Big[1+\frac{\text{f}_\text{e}}{\text{d}}\Big]$

$=\frac{140}{5}\Big[1+\frac{5}{25}\Big]$

= 28[1 + 0.2]

= 28 × 1.2 = 33.6

1. Focal length of the convex lens, f = 20 cm

Image distance = v

According to the lens formula, we have the relation:

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

$\frac{1}{\text{v}}-\frac{1}{12}=\frac{1}{20}$

$\frac{1}{\text{v}}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}$

$\therefore \ \text{v}=\frac{60}{8}=7.5 \ \text{cm}$

Hence, the image is formed 7.5 cm away from the lens, toward its right.

2. Focal length of the concave lens, f = –16 cm

Image distance = v

According to the lens formula, we have the relation:

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

$\frac{1}{\text{v}}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48}$

$\therefore \ \text{v}=48\ \text{cm}$

Hence, the image is formed 48 cm away from the lens, toward its right.

1. The angular magnification of a telescope is given as:

$\alpha=\frac{\text{f}_0}{\text{f}_\text{e}}$

$=\frac{15\times10^2}{1.0}=1500$

Hence, the angular magnification of the given refracting telescope is 1500.

2. Diameter of the moon, d = 3.48 × 10⁶ m

Radius of the lunar orbit, r₀ = 3.8 × 10⁸ m

Let d' be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image.

$\frac{\text{d}}{\text{r}_0}=\frac{\text{d}'}{\text{d}_0}$

$\frac{3.48\times10^6}{3.8\times10^8}=\frac{\text{d}'}{15}$

$\therefore \ \text{d}'=\frac{3.48}{3.8}\times10^{-2}\times15$

= 13.74 × 10^-2 m = 13.74 cm

1. The angular magnification of a telescope is given as:

$\alpha=\frac{\text{f}_0}{\text{f}_\text{e}}$

$=\frac{15\times10^2}{1.0}=1500$

Hence, the angular magnification of the given refracting telescope is 1500.

2. Diameter of the moon, d = 3.48 × 10⁶ m

Radius of the lunar orbit, r₀ = 3.8 × 10⁸ m

Let d' be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image.

$\frac{\text{d}}{\text{r}_0}=\frac{\text{d}'}{\text{d}_0}$

$\frac{3.48\times10^6}{3.8\times10^8}=\frac{\text{d}'}{15}$

$\therefore \ \text{d}'=\frac{3.48}{3.8}\times10^{-2}\times15$

= 13.74 × 10^-2 m = 13.74 cm

1. Angular dispersion produced by two prisms should be zero so that, no dispersion take place.

The sum of angular dispersion by crown glass prism and angular dispersion by flint glass prism is equal to zero.

i.e., $(\mu_\text{b}-\mu_\text{r})\text{A}+(\mu_\text{b}'-\mu\text{r}')\text{A}'=0.$

Since $(\mu_\text{b}'-\mu_\text{r}')$ for flint glass is more than that for crown glass, therefore,

A' < A

Hence, the combination of prism is such that the, flint glass prism of smaller angle has to be suitably combined with crown glass prism of larger angle.

2. For almost no deviation,

$(\mu_\text{y}-1)\text{A}'+(\mu_\text{y}'-1)\text{A}'=0.$

To disperse without deviation, we need flint glass prism of greater so that the deviation due to two prisms are equal and opposite. In the final combination however, angle of flint glass prism will be smaller than the angle of crown glass prism because the refractive index is more for flint glass as compared to the refractive index of crown glass.

1. Ray Diagram for compound microscope:

• Objective: Lens L₃
• Eye Piece: Lens L₂

3. $\text{R}_\text{p}=\frac{2{\mu}\sin \beta}{1.22\lambda}$

$\therefore$ v = -15cm

The final image is, therefore, at a distance of 15cm to the left of the concave mirror.

1. $\lambda=\frac{589\text{ }\text{nm}}{1.33}=442.8\text{ }\text{nm}$

Frequency  $\nu=\frac{3\times10^8\text{ }\text{ms}^{-1}}{589\text{ }\text{nm}}=5.09\times10^{12}\text{Hz}$

Speed $\nu=\frac{3\times10^8}{1.33}\text{m/s}=2.25\times10^{8}\text{m/s}$

2. $\frac{1}{f}=\bigg[\frac{\mu_2}{\mu_1}-1\bigg]\bigg[\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\bigg]$

$\therefore\text{ }\frac{1}{20}=\bigg[\frac{1.55}{1}-1\bigg]\frac{2}{\text{R}}$

$\therefore\text{ }\text{R}=(20\times1.10)\text{cm}=22\text{ }\text{cm}$

1. Image is not free from chromatic aberration and spherical aberration.
2. Aperture of the objective lens should be large for high resolving power.

1. When object is between f and 2f, we have 1 < n < 2

$\therefore\text{v} \text{ is -ve real image }$

(For n =1 and n =2), magnitude of v becomes $ \infty$ and 2c, respectively

$\therefore$ Real image is formed beyond 2F.

2. Obiect between pole and F we have 0 < 0 < 1

v is + ve $\Rightarrow$(virtual image) and | v | > c

$\therefore$ We get a virtual, enlarged image

3. Convex Mirror

f = + d Let u = - pd (p can have any value)

1/v = 1/d + 1/pd = (1+p)/pd

$\text{v} = \frac{\text{pd}}{(\text{p} + 1) }$

$\therefore$ v is always +ve and always less than d

$\therefore$ Convex mirror always produces a virtua1 image between pole and focus.

Alternate Answer

1.  
   1. Diverging lens or concave lens.
   2. Converging lens or convex lens.
2.  
   1. Focal length will become negative and its magnitude would increase.
   2. Focal length increases.

Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image at infinity.

Calculation of magnifying power:

Given: Power of eyepiece = 10 D

Power of objective = 1 D

Magnifying power in normal adjustment: $\text{m} = \frac{\text{f}_{o}}{\text{f}_{e}} = \frac{\text{p}_{e}}{\text{p}_{o}} = \frac{10}{1} = 10 $

Alternate Answer

Alternate Answer

1. Given R = -20 cm, and magnification m= -2 

Focal length of the mirror $\text{f} = \frac{\text{R}}{2} = - 10\text{cm}$

Magnification $(\text{m}) = -\frac{\text{v}}{\text{u}}$

$ - 2 = - \frac{\text{v}}{\text{u}}$

$ = > {v} = 2\text{u}$

Using mirror formula

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

$\Rightarrow - \frac{1}{10} = \frac{1}{2\text{u}} + \frac{1}{\text{u}}$

$\Rightarrow\text{u} = - 15 \text{cm}$

$\therefore\text{v} = 2 \times -15 \text{cm} = - 30 \text{cm}$

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

Using sign convention, for convex mirror, we have 

$\text{f} > 0 , \text{u} < 0$

From the formula

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

$\because \text{f}$ is positive and $\text{u}$ is negative,

$\Rightarrow\text{v}$ is always positive, hence image is always virtual.

1. Increases:

Resolving power $ = \frac{\text{d}}{1.22\lambda}$

$\therefore$ R.P is directly proportional to ‘d’ (For a given $\lambda$)

2. Decreases:

$\therefore$ R.P is inversely proportional to $'\lambda'$ (For a given d).