Lesson: Light-Reflection and Refraction
Define the principal focus of a concave mirror.
It is a point on the principal axis where the light rays that are parallel to the principal axis converge, after reflecting from the mirror.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Radius of curvature, $R=20\text{cm}$
Radius of curvature of the spherical mirror $=2\times \text{focallength}\left(f\right)$
i.e.,
$\begin{array}{l}R=2\text{\hspace{0.17em}}f\\ f=\frac{R}{2}=\frac{20}{2}=10\text{cm}\end{array}$
Name the mirror that can give an erect and enlarged image of an object.
Concave mirror.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Convex mirrors always form a virtual, erect, and diminished image of the objects placed in front of it. This helps in getting a wider view of objects, traffic and people behind the vehicle.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Given:
The radius of curvature, $R=32\text{cm}$.
Radius of curvature,
$\begin{array}{l}R=2\times \text{focallength}\left(f\right)\\ \Rightarrow R=2,\text{}f=\frac{R}{2}=\frac{32}{2}\text{=}16\text{cm}\end{array}$
A concave mirror produces three times magnified (enlarged) real image of object placed at 10 cm in front of it. Where is the image located?
Magnification produced by a spherical mirror, m = height of the image/height of the object.
Let the height of the object be ${h}_{0}$ and the height of the image be h_{1}.
$\Rightarrow m\text{\hspace{0.17em}}\text{=}\frac{{h}_{1}}{{h}_{0}}\text{=}\text{\hspace{0.17em}}-\frac{u}{v}$
The height of the image, ${h}_{1}=\text{\hspace{0.17em}}-3{h}_{0}$ (The image formed is real.)
$\begin{array}{l}\frac{\text{-3}h}{h}\text{=}\frac{\text{-}v}{u}\\ \frac{v}{u}\text{=3}\end{array}$
Object distance, $u=-10\text{cm}$
$v=3\times \left(-10\right)=-30\text{cm}$
$\Rightarrow $ An inverted image is formed at a distance of 30
cm in front of the given concave mirror.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
When a ray of light enters from an optically rarer medium to an optically denser medium, it bends towards the normal. Since water is optically denser than air, a ray of light travelling in water from air will bend towards the normal.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is $3\times {10}^{8}$ ms^{-1}.
Let the refractive index of the glass be n_{g.}
$\Rightarrow {n}_{g}=$ speed of
light in vacuum/speed of light in glass
Speed of light in vacuum, $c=3\times {10}^{8}m{s}^{-1}$
Refractive index of glass, ${n}_{g}=1.50$
Therefore, the speed of light in the glass
= speed of light in vacuum/refractive index of glass
$=\frac{c}{{n}_{g}}=\frac{3\times {10}^{8}}{1.50}=2x{10}^{8}{\text{ms}}^{\text{-1}}$
Find out, from Table, the medium having highest optical density. Also find the medium with lowest optical density.
Material medium |
Refractive index |
Material medium |
Refractive index |
Air |
1.0003 |
Canada Balsam |
1.53 |
Ice |
1.31 |
- |
- |
Water |
1.33 |
Rock salt |
1.54 |
Alcohol |
1.36 |
- |
- |
Kerosene |
1.44 |
Carbon disulphide |
1.63 |
Fused quartz |
1.46 |
Dense flint glass |
1.65 |
Turpentine oil |
1.47 |
Ruby |
1.71 |
Benzene |
1.50 |
Sapphire |
1.77 |
Crown glass |
1.52 |
Diamond |
2.42 |
Diamond has the highest optical density.
Air has the lowest optical density.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table.
Material |
Refractive index |
Material medium |
Refractive index |
Air |
1.0003 |
Canada Balsam |
1.53 |
Ice |
1.31 |
- |
- |
Water |
1.33 |
Rock salt |
1.54 |
Alcohol |
1.36 |
- |
- |
Kerosene |
1.44 |
Carbon disulphide |
1.63 |
Fused |
1.46 |
Dense |
1.65 |
Turpentine oil |
1.47 |
Ruby |
1.71 |
Benzene |
1.50 |
Sapphire |
1.77 |
Crown |
1.52 |
Diamond |
2.42 |
Light travels faster when the refractive index of a medium is less. The speed of light is inversely proportional to the refractive index. Of all three given media, water has the lowest refractive index. So light will travel the fastest in water.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Light travels faster when the refractive index of a medium is less. The speed of light is inversely proportional to the refractive index. The refractive index of air is approximately 1.The refractive index of diamond is 2.42.
This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air.
Define 1 dioptre of power of a lens.
1 dioptre is defined as the power of a lens having a focal length of 1 metre.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Given:
i. Image is real and of same size. It implies that the position of image is at 2F.
ii.
Object
distance, $u=-\text{}50\text{cm}$
Image distance, $v=50\text{cm}$
Focal length $=f$
According to the lens formula,
$\begin{array}{l}\frac{\text{1}}{v}\text{-}\frac{\text{1}}{u}\text{=}\frac{\text{1}}{f}\\ \frac{\text{1}}{f}\text{=}\frac{\text{1}}{\text{50}}\text{-}\frac{\text{1}}{\text{(-50)}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\frac{\text{1}}{\text{50}}\text{+}\frac{\text{1}}{\text{50}}\text{=}\frac{\text{1}}{\text{25}}\\ \Rightarrow f=\text{25}\text{\hspace{0.17em}}\text{cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{0}\text{.25}\text{\hspace{0.17em}}\text{m}\end{array}$
Power of the lens, $P\text{=}\frac{\text{1}}{f\text{(in}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{metres)}}\text{=}\frac{\text{1}}{\text{0}\text{.25}}\text{=+4D}$
Find the power of a concave lens of focal length 2 m.
Let the power of lens be P and the focal length be f
The power of the lens (P) of focal length $f=\frac{1}{f}$
$=\frac{1}{-2}=-0.5\text{}D$