Science Answer the questions.[Phy-3M]

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

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

$=\frac{-1}{18}+\frac{1}{27}-\frac{-1}{54}$

$\text{f}=\frac{\text{R}}{2}=+15 \ \text{cm}$

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

$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{(-10)}$

$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{15}+\frac{1}{10}$

$\Rightarrow \ \frac{1}{\text{v}}=\frac{2+3}{30}$

$\Rightarrow \ \frac{1}{\text{v}}=\frac{5}{30}$

$\Rightarrow \ \text{v}=6 \ \text{cm}$

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

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

$\frac{1}{\text{u}}=\frac{1}{-10}-\frac{1}{-15}=\frac{-5}{150}$

$\text{u}=\frac{150}{-5}=-30 \ \text{cm}$

1. Concave mirror.
2. $\text{Linear magnification} = -\frac{\text{Image distance}}{\text{Object distance}}$

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

Object distance, u = -15 (u is always negative)

Image distance, v = -90 (-ve sign as the image is formed in front of the mirror on the screen)

$m = -\frac{(- 90)}{(-15)} \ \ \ \ \ \ \ \ \ \ \therefore m = -6$

3. The distance between the object and its image is 75 cm.
4. Ray diagram:

1. He should use a concave mirror, as it forms a real image on the same side of the mirror.
2. Object distance, u = -12 cm.

Image distance, v = - 48 cm.

Magnification, $m = -\frac{v}{u} = - \frac{(-48)}{(-12)} = -4$

The minus sign in magnification shows that the image formed is real and inverted.

3. The image is formed at a distance of 36 cm from the object.
4.  

1. Focal length, f =?

$f = \text{ + 21 cm,}$

Object position is at 29 cm

$\Rightarrow u = - \text{21 cm}$

$v = ?$

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

$\Rightarrow \frac{1}{v} = \frac{1}{21} + \frac{1}{-21}$

$\Rightarrow\frac{1}{v} = \frac{1 - 1}{21}$

$\Rightarrow \frac{1}{v} = \frac{0}{21}$

$\therefore v = \frac{21}{0}$

$\Rightarrow v = \frac{21}{0} = \infty$

So image is formed at infinity.

3. If the object is further shifted towards the lens then a virtual, erect and magnified image will be formed behind the object.

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

$\text{Or}\text{ } \frac{1}{v}=\frac{1}{+15{\text{cm}}}+\frac{1}{-10{\text{cm}}}=\frac{-1}{30{\text{cm}}}$

1. (m = - 1, i.e. Image is real, inverted and same size as the object.

and, Object distance = image distance = 2f = 35 cm.

2. Nature of lens: Convex/ Converging.
3. As 2f = 35 cm

$\therefore\text{f}=\frac{35}{2}=+17.5\text{cm}$

4. On displacing the object 20 cm towards the lens, the object distance becomes 15c m (35 cm – 20 cm) i.e. it lies between F and O of the lens. Image formed now is virtual/on the same side of lens as the object.​

• (m = -1, means that the image is real, inverted and of the same size as the object).

$\therefore$ Object distance = image distance = 2f = 25 cm.

$\therefore\text{f}=\frac{25}{2}=12.5\text{cm}$

• Nature of the lens is convex/converging.
• On displacing the object distance by 15 cm, towards the lens, the object distance becomes 10 cm which is less than the focal length. Image formed now is virtual/same side of lens as the object.

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

$\therefore \frac{1}{\text{f}}=\frac{1}{60}+\frac{1}{-30}$

$\Rightarrow \frac{1}{\text{f}}=\frac{1}{60}+\frac{1}{30}$

$\Rightarrow \frac{1}{\text{f}}=\frac{(1+2)}{60}$

$\Rightarrow \frac{1}{\text{f}}=\frac{3}{60}$

f = +20 cm

$\Rightarrow \text{m}=\frac{60}{(-30)}$

$\Rightarrow \text{m} = (-2)$

Marking $\angle\text{r}$ and x

2. $_\text{a}\text{n}_\text{g}=\frac{3}{2}$

$\therefore\ _{\text{a}}\text{n}_\text{g}=\frac{1}{_\text{a}\text{n}_\text{g}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$

Alternately, $\frac{\text{C}_\text{air}}{\text{C}_\text{glass}}=\frac{3}{2}$

$\therefore\ \frac{\text{C}_\text{glass}}{\text{C}_\text{air}}=\frac{2}{3}$

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

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

$\frac{1}{\text{v}}=\frac{1}{\text{-12cm}}-\frac{1}{\text{-18cm}}$

$\therefore\text{v}=-36\text{cm}$

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

$ \because\text{h}_2=\text{-h}_1\times\frac{\text{v}}{\text{u}}$

$=\text{-3cm}-\frac{-36\text{cm}}{-18\text{cm}}$

$=-6\text{cm}$

1. The incident, the refracted ray and normal at the point of incidence, all lie in the same plane.
2. The second law of refraction is called snell’s law of refraction. According to snell’s law, “The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of mediums”.

$\frac{\sin{\text{i}}}{\sin{\text{r}}} = n$ (constant)

$\frac{\text{Speed of light in vaccum}}{\text{Speed of light in a medium}} = \text{Refractive index of the medium}$

$\Rightarrow \frac{3 \times 10^{8} \text{m/s}}{x} = 1.5$

$\Rightarrow x = \frac{3 \times 10^{8} }{1.5} = 2 \times 10^{8} \text{ m/s}$

$\therefore \text{Speed of light in the medium} = 2 \times10^{8} \text{ m/s}$

1. The mirror is concave.
2. The image formed is real and inverted and of the same size as that of the object.
3. In this case, the object is placed at the centre of curvature of the mirror, so the object distance is equal to the image distance. Here, the screen is placed 40 cm from the mirror.

$\therefore$ Image distance = Object distance = 40 cm.

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

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

$\frac{1}{\text{v}}=\frac{1}{\text{-10cm}}-\frac{1}{\text{-15cm}}$

$\therefore\text{v}=-30\text{cm}$

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

$\because\text{h}_2=-\frac{\text{v}}{\text{u}}\times\text{h}_1$

$=-\frac{-30\text{cm}}{-15\text{cm}}\times4\text{cm}$

$=-8\text{cm}$

$\therefore v = - 50$

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

$\Rightarrow - =1 = \frac{(-50)}{-u}$

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

1. Since image is formed on the screen therefore the mirror formed real image which is formed by concave mirror only.
2. Image distance = 50cm in front of the mirror.
3. $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

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

$\frac{1}{f} = -\frac{1}{50} - \frac{1}{50}$

$\Rightarrow \frac{-2}{50} = \frac{-1}{25}$

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

4.  

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

$\frac{1}{f} =\frac{1}{(\text{+ 90)}}-\frac{1}{(\text{- 45)}}$

$ f\text{ =} + 30 \text{ cm}$

$m= \frac{h'}{h} = \frac{v}{u}$

$m= \frac{h'}{2} = \frac{90}{-45}$

$\therefore h' = - \text{ 4} \text{ cm}$