Science Answer the questions.[Phy-3M]

1. The incident ray, the refracted ray and the normal to the interface of two media at the point of incidence, all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media.

$\frac{\sin\text{i}}{\sin\text{r}}=\text{constant}={^1}\text{n}_2=\frac{\text{n}_2}{\text{n}_1}$

This law is also known as Snell’s law.

The constant, written as ¹n₂, is called the refractive index of the second medium with respect to the first medium.

Absolute refractive index of a medium:

When a beam of light is going from vacuum to another medium, the value of the refractive index is called absolute refractive index.

The refractive index of vacuum (air) is always 1. The absolute refractive index is denoted by the symbol n₂.

$\text{n}_2=\frac{\text{Refractive index of medium}}{\text{Refractive index of vacuum (air)}}=\frac{\text{Speed of light in vacuum}}{\text{Speed of light in the medium}}=\frac{\text{c}}{\text{v}}$

$\therefore\ \text{n}_2=\frac{\text{c}}{\text{v}}$

The power of a lens is the measure of degree of convergence or divergence of light rays falling on it. The SI unit of power of a lens is dioptre denoted by ‘D’.

The power of a lens is given by,

$\text{P}=\frac{1}{\text{f (in metres)}}$

$\therefore$ For lens of focal length 40cm, f = 0.4m

$\therefore\ \text{P}=\frac{1}{0.4}=+2.5\text{D}$

The lens is convex lens and is used to correct hypermetropia.

For lens of focal length -20cm, f = 0.2m

$\therefore\ \text{P}=\frac{1}{-0.2}=-5.0\text{D}$

The lens is concave lens and is used to correct myopia.

$\frac{\sin\text{i}}{\sin\text{r}}=\text{n}$

$\text{u}= -50\text{cm},$

$\text{m}=-\frac{1}{2}$ 

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

$-\frac{1}{2}=\frac{\text{v}}{-50}$

$\text{v}=-25\text{cm}$

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

$\Rightarrow\frac{1}{-25}+\frac{1}{-50}=\frac{1}{\text{f}}$

$\Rightarrow\frac{-3}{-50}=-\frac{1}{\text{f}}$

$\Rightarrow\text{f}=\frac{-50}{3}\text{cm}$

Case 2:

$\text{m}=\frac{1}{2}$ 

$\text{f}=\frac{-50}{3}$

$\text{m}=-\frac{1}{5}=-\frac{\text{v}}{\text{u}}$

$\text{v}=-\frac{\text{u}}{5}$

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

$\frac{5}{\text{u}}+\frac{1}{\text{u}}=\frac{-3}{50}$

$\frac{6}{\text{u}}=-\frac{-3}{50}$

$\text{u}=\frac{600}{-3}=-100\text{cm}$

u = -20cm, m = -3cm, for the real image

1. We know that

$\text{m}=-\frac{\text{m}}{\text{u}}$ 

$\therefore \text{m}=-3=-\frac{\text{v}}{(-20)}$

$\Rightarrow\text{v}= -60\text{cm}$

We have

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

$\Rightarrow \frac{1}{(-60)}+\frac{1}{(-20)}=\frac{1}{\text{f}}$

$\frac{1}{\text{f}}=-\frac{1}{60}-\frac{1}{20}=\frac{-1-3}{60}=\frac{1}{15}$

$\text{f}=-15\text{cm}$

2. For virtual image m = 3 , and f = 15cm

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

$\therefore \text{m}=-3=-\frac{\text{v}}{(-20)}\Rightarrow \text{v}=-3\text{u}$

We have

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

$\Rightarrow \frac{1}{\text{(3u)}}+\frac{1}{\text{u}}=\frac{1}{(-15)}$

$\Rightarrow \frac{-1+3}{\text{3u}}=-\frac{1}{15}$

$\Rightarrow \text{u}=-\frac{2\times15}{3}=-10\text{cm}$

So object should be placed 10cm from the concave mirror.

1. Given that,

$\text{p}=+2\text{D}$

Using the the formula,

We get,

$\frac{1}{\text{f}}=2$

${\text{f}}=\frac{1}{2}$

${\text{f}}=0.5\text{m}$

$\text{f}=50\text{cm}$

The lens is convex lens of focal length = 0.5m = 50cm

2. Given that,

$\text{p}=+4\text{D}$

Using the the formula,

We get,

$\frac{1}{\text{f}}=-4$

${\text{f}}=\frac{1}{-4}$

${\text{f}}=-0.25\text{m}$

$\text{f}=-25\text{cm}$

The lens is concave lens of focal length = -0.25m = -25cm

$\text{f (in m)}=\frac{1}{\text{p}}$

• Convex Lens: Positive (+)
• Concave Lens: Negative (-)

1. Lens A is a convex lens (positive focal length) and lens B is a concave lens (negative focal length).

2. $\text{P}_\text{A}=\frac{1}{\text{f}_\text{A}}=\frac{1}{+0.2}=+5\text{D}$

$\text{P}_\text{B}=\frac{1}{\text{f}_\text{B}}=\frac{1}{-0.1}=-10\text{D}$

3. Power of combination,

P = PA + PB = +5D + (-10D) = -5D

1. R = 50cm

f = ?

We know that

$\text{f}=\frac{\text{R}}{2}=\frac{50}{2}=25\text{cm}$

2. Convex mirror
3. It will diverge light rays.

$\text{f}=\frac{\text{R}}{2}=1\text{m}$ 

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

$\Rightarrow\frac{1}{\text{v}}+\frac{1}{(-3.5)}=\frac{1}{1}$

$\Rightarrow\frac{1}{\text{v}}=1+\frac{1}{3.5}=1+\frac{10}{30}=1+\frac{2}{7}=\frac{9}{7}$ 

$\therefore \text{v}=\frac{7}{9}=0.77\text{m}$

$\text{u}=-2\text{v}=-2\times10=-20\text{cm}$

So, the image is fomed 0.77m in behind the mirror.

2. Now, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{0.77}{(-3.5)}=\frac{\frac{7}{9}}{3.5}=\frac{1}{4.5}$ 

AS m is positive, so image formed is virtual and efect.

1. $\text{P}=\text{P}_\text{x}+\text{P}_\text{y}$

$\text{P}=\frac{1}{\text{f}_\text{x}}+\frac{\text{1}}{\text{f}_\text{y}}$

$5=\frac{100}{15}+\frac{\text{1}}{\text{f}_\text{y}}$

$\frac{\text{1}}{\text{f}_\text{y}}=5-\frac{100}{15}$

$=\frac{-25}{15}$

${\text{f}_\text{y}}=-0.6\text{m}=-60\text{cm}$

2. Lens Y is a concave lens since it has negative focal length.

Here, image distance, v = +80cm

Magnification, m = -3

Object distance, u = ?

Now, magnification, $\text{m}=\frac{\text{v}}{\text{u}}\Rightarrow\ -3=\frac{80}{\text{u}}\ \text{or u }=\frac{-80}{3}\text{cm}$

Nature of image will be real, magnified and inverted. The image will be formed beyond 2F.

By using lens formula,

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

$\Rightarrow\ \frac{1}{\text{f}}=\frac{1}{80}-\frac{3}{-80}=\frac{1}{20}\text{ or f }=20\text{cm}$

Focal length is positive so, the lens is convex.

$\text{m}=\frac{\text{v}}{\text{u}}=-3$

$\text{v}=-3\text{u}$

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

$\frac{1}{\text{v}}-\frac{1}{-150}=\frac{1}{100}$

$\frac{1}{\text{v}}=\frac{1}{100}-\frac{1}{150}$

$\frac{1}{\text{v}}=\frac{1}{300}$

$\text{v}=+300\text{mm}$

1. h₁ = 3cm

u = -24cm

f = 8cm

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

$\frac{1}{\text{v}}-\frac{1}{-24}=\frac{1}{8}$

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

$\text{v}=12\text{cm}$

Image is formed 12cm behind the lens.

$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$

$\frac{12}{-24}=\frac{\text{h}_2}{3}$

$\text{h}_2=-1.5\text{cm}$

Image is 1.5cm hight, real and inverted.

2. u = -3cm

h1 = 3cm

f = 8cm

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

$\frac{1}{\text{v}}-\frac{1}{-3}=\frac{1}{8}$

$\frac{1}{\text{v}}=-\frac{5}{24}$

${\text{v}}=-4.8\text{cm}$

Image is formed 4.8cm in front of the lens.

$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$

$\frac{-4.8}{-3}=\frac{\text{h}_2}{3}$

$\text{h}_2=+4.8\text{cm}$

Image is 4.8cm high, virtual and erect.

3. Case (b).