- What is the nature of lens A?
- What is the nature of lens B?
- Convex lens (since image is real, inverted and diminished).
- Concave lens (since image is virtual, erect and diminished).
| | 1 | 2 | 3 | 4 | 5 |
| Medium: | Water | Crown glass | Rock salt | Ruby | Diamond |
| Refractive index: | 1.33 | 1.52 | 1.54 | 1.71 | 2.42 |
It is a convex lens since its focal length is positive.
$\text{P}=\frac{1}{\text{f}}=\frac{1}{0.15}=+6.66\text{D}$

Give reasons for your choice.



| Medium | Refractive index |
| A B C D | 1.33 1.44 1.52 1.65 |
$\text{Refractive index}=\frac{\text{Speed of light in vacuum}}{\text{speed of light in a mediums}}$
Since the speed of light in vacuum is a constant, the refractive index becomes inversely proportional to the speed of light in a medium.refractive index of glass with respect to air = nGA = absolute refractive index of glass,
refractive index of diamond with respect to glass = nDG
Given,
NDG = 1.6 and nGA = 1.5
Now, $\text{n}_\text{DG}=\frac{\text{n}_{\text{DA}}}{\text{n}_{\text{GA}}}$
⇒ nDA = nDG × nGA
⇒ nDA = 1.6 × 1.5 = 2.4


Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-0.50}=\frac{1}{0.20}$
$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.50}$
$\text{v}=0.33\text{m}$
Image is formed 0.33m behind the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{0.33}{-0.50}=-0.66$
Image is real and inverted.
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-0.25}=\frac{1}{0.20}$
$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.25}$
$\text{v}=1\text{m}$
Image is formed 1m behind the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{1}{-0.25}=-4$
Image is real and inverted.
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-0.15}=\frac{1}{0.20}$
$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.15}$
$\text{v}=-0.60\text{m}$
Image is formula 0.60m in front of the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{-0.6}{-0.15}=+4$
Image is virtual and erect.
Camera, Case (i)
Magnifying glass, Case (iii)

$\text{Refractive index of water}=\frac{\text{Speed of light in air}}{\text{Speed of light in water}}$
$\text{Refractive index of water}=\frac{300\text{ million m}/ \text{ s}}{225\text{million m}/\text{s}}=1.33$
$2.42=\frac{3\times10^8}{\text{Speed of light diamond}}$
Speed of light diamond = 1.239 × 108m/ s

$\text{m}=\frac{\text{h}_2}{\text{h}_1}$
$2=\frac{\text{h}_2}{1}$
$\text{h}_2= 2\text{m}$

$\Rightarrow1.36=\frac{3\times10^8}{\text{speed of light in alcohol}}$
$\Rightarrow\text{speed of light in alcohol}=\frac{3\times10^8}{1.36}$
$\therefore$ speed of light in alcohol = 2.21 × 108m/s
speed of light in alcohol with respect to air is 2.21 x 108ms-1The object is placed at B such that PB = 8cm. This means that the object lies between the pole and focus of the concave mirror. The image formed is virtual, erect and magnified.
Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.

Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.




