Light - Reflection and Refraction

Clay

Reason — Clay cannot be used to make a lens as it is opaque and will not allow light to pass through it.

Between the pole of the mirror and its principal focus

Reason — In order to get a virtual, erect and larger image of an object with the help of a concave mirror, the object should be placed between the pole of the mirror and its principal focus.

At twice the focal length

Reason — In order to get a real image of the same size as the object with the help of a convex lens, the object should be placed at twice the focal length.

both concave

Reason — According to the sign convention, the focal length for both a concave mirror and a concave lens is negative.

either plane or convex

Reason — Both plane mirrors and concave mirrors form erect images irrespective of the position of the object. Concave mirror only forms erect image when the object is placed between it's pole and focus.

A convex lens of focal length 5 cm

Reason — A convex lens gives a magnified image and shorter the focal length, more is the magnification, hence, a convex lens of focal length 5 cm will be preferred.

Concave mirror gives a virtual and erect image only when the object is placed between its pole and focus.
Hence, the range of distance of the object should be less than 15 cm from the mirror.

Image is virtual.

Image is larger than the object.

Ray diagram to obtain an erect image using a concave mirror is shown below:

(a) Concave mirrors because they produce powerful parallel beam of light when the light source is placed at its principal focus.

(b) Convex Mirror because they always give an erect image with a wider field of view as they are curved outward, enabling the driver to view much larger area than would be possible with a plane mirror.

(c) Concave Mirror because it concentrates the parallel rays of the sun at a principal focus.

Yes, it will produce a complete image of the object. However, the intensity or brightness of the image will be reduced.

Experiment : To show that a half-covered convex lens can produce a complete image of the object.

Apparatus needed : A convex lens, a candle, and two black sheets of paper.

Procedure :

1. Place a convex lens on the table. Light a candle with a match stick and place it near one side of the lens on the table (turn off the lights of the room in order to get dark).

2. Take the sheet of paper and place it on the other side of the lens as the candle.

3. Now focus the image formed on the sheet of paper by moving the sheet only.

4. Observe the image carefully.

5. Mark the positions of candle, lens and screen.

6. Now, turn off the light. Take the convex lens and cover half of it with the second sheet of black paper.

7. Mount the lens on the table in the original position and repeat the above experiment.

Observations : A complete image of the object is obtained in both the cases. However, the image formed with half of the convex lens covered is less bright.

Conclusion : The half covered lens will also produce a complete image of the object. However, the image formed will be less bright.

Given,

Object height (h_o) = 5 cm

Object distance (u) = -25 cm

focal length (f) = 10 cm

Ray diagram is shown below:

Image distance (v) = ?

Image height (h_i) = ?

According to the Lens formula,

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

Substituting the values we get,

$\dfrac{1}{v} - \dfrac{1}{-25} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{25} = \dfrac{1}{10} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} - \dfrac{1}{25} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{5-2}{50} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{3}{50} \\[0.5em] \Rightarrow v = 16.66 \text{ cm}$

Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.

Magnification (m) = $\dfrac{\text{v}}{\text{u}}$ = $\dfrac{\text{height of image}}{\text{height of object}}$

Substituting the values we get,

$\dfrac{16.66}{-25}$ = $\dfrac{\text{height of image}}{5}$

height of image = $\dfrac{16.66 \times 5}{-25}$ = -3.3

Negative sign of the height of image means that an inverted image is formed.

The image is reduced, real and inverted.

f = -15 cm [∵ focal length of concave lens is negative]

v = -10 cm [∵ image distance of concave lens is negative]

u = ?

According to the lens formula,

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

Substituting the values we get,

$\dfrac{1}{- 10} - \dfrac{1}{u} = \dfrac{1}{- 15} \\[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{1}{- 10} + \dfrac{1}{15} \\[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{-3+2}{30} \\[0.5em] \Rightarrow \dfrac{1}{u} = - \dfrac{1}{30} \\[0.5em] \Rightarrow u = -30\text{ cm}$

The object is placed 30 cm in front of the concave lens.

Given,

f = +15 cm

u = -10 cm

v = ?

According to the mirror formula,

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

Substituting the values we get,

$\dfrac{1}{v} + \dfrac{1}{-10} = \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} + \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{3+2}{30} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{5}{30} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{6} \\[1em] \Rightarrow v = 6\text{ cm}$

Therefore, image is formed 6 cm behind the mirror. Image is virtual and erect.

$\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[1em]$

Given,

$+1 = \dfrac{\text{height of image}}{\text{height of object}} \\[1em] ∴ \text{height of image} = \text{height of object}$

Hence, it means that the image formed by a plane mirror will be virtual and erect and it will be of the same size as the object.

Given,

Object distance (u) = -20 cm

Object height (h) = 5 cm

Image distance (v) = ?

Image height (h) = ?

Radius of curvature (R) = 30 cm

We know, R = 2f or f = $\dfrac{R}{2}$

Hence, f = $\dfrac{30}{2}$ = 15 cm

According to the mirror formula,

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

Substituting the values we get,

$\dfrac{1}{v} + \dfrac{1}{-20} = \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 20} + \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{3+4}{60} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{7}{60} \\[1em] \Rightarrow v = \dfrac{60}{7} = 8.57\text{ cm}$

Image is virtual, erect and formed 8.57 cm behind the mirror.

$\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[0.5em] \dfrac{-8.57}{-20} = \dfrac{\text{height of image}}{5} \\[0.5em] \Rightarrow \text{height of image} = \dfrac{-8.57 \times 5}{-20} \\[0.5em] \Rightarrow \text{height of image} = 2.14 \text{ cm}$

Height of image is 2.14 cm

Given,

Object distance (u) = -27 cm [∵ object is placed infront of mirror so object distance will be negative]

Object height (h) = 7 cm

Image distance (v) = ?

Image height (h) = ?

f = -18 cm [∵ focal length of concave mirror is negative by convention]

According to the mirror formula,

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

Substituting the values we get,

$\dfrac{1}{v} + \dfrac{1}{-27} = \dfrac{1}{-18} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 27} - \dfrac{1}{18} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{2-3}{54} \\[1em] \Rightarrow \dfrac{1}{v} = - \dfrac{1}{54} \\[1em] \Rightarrow v = - 54\text{ cm}$

∴ The screen should be placed at a distance of 54 cm in front of the mirror on the object side.

$\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[1em] -\dfrac{-54}{-27} = \dfrac{\text{height of image}}{7} \\[1em] \Rightarrow \text{height of image} = -\dfrac{54 \times 7}{27} \\[1em] \Rightarrow \text{height of image} = -14 \text{ cm}$

∴ Height of image is 14 cm and image is real, inverted and magnified.

Given,

P = -2.0 D

f = ?

We know,

Power of lens (P) = $\dfrac{1}{\text{focal length}}$

As the given power is negative hence, we can say that the lens used is concave in nature.

Substituting the value of power in formula we get,

-2 = $\dfrac{1}{\text{f}}$

f = - $\dfrac{1}{\text{2}}$

f = -0.5 m

Therefore, focal length is -0.5 m and it is a concave lens.

Given,

P = +1.5 D

f = ?

We know,

Power of lens (P) = $\dfrac{1}{\text{focal length}}$

As the given power is positive hence, we can say that the lens used is convex (converging) in nature.

Substituting the value of power in formula we get,

+1.5 = $\dfrac{1}{\text{f}}$

f = $\dfrac{1}{\text{1.5}}$

f = 0.67 m

Therefore, focal length is 0.66 m and it is a converging lens.

The principal focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to the principal axis, pass after reflection from the mirror.

Given,

Radius of curvature (R) = 20 cm

Focal length (f) = ?

We know, R = 2f or f = $\dfrac{R}{2}$

Hence, f = $\dfrac{20}{2}$ = 10 cm

∴ Focal length of spherical mirror is 10 cm.

Concave Mirror

A convex mirror is preferred as a rear-view mirror in vehicles because they always give an erect though diminished image. They have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror.

Given,

Radius of curvature (R) = 32 cm

Focal length (f) = ?

We know, R = 2f or f = $\dfrac{R}{2}$

Hence, f = $\dfrac{32}{2}$ = 16 cm

Therefore, the focal length of the convex mirror is 16 cm.

Given, object distance (u) = -10 cm

Magnification = -3 (as the image is real and inverted)

image distance (v) = ?

We know,

$\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[1em]$

Substituting we get,

$-3 = \dfrac{-v}{-10} \\[1em] \Rightarrow -v = (-3) \times (-10) \\[1em] \Rightarrow v = -30 \text{ cm}$

The negative sign indicates that an inverted image is formed in front of the given concave mirror at a distance of 30 cm.

The light ray bends towards the normal because when a light ray enters from an optically rarer medium (which has a low refractive index i.e., air) to an optically denser medium (which has a high refractive index i.e., water), its speed slows down and it bends towards the normal.

Refractive index of a medium (n_m) = $\dfrac{\text{Speed of light in vacuum}}{\text{Speed of light in the medium}}$

Speed of light in vacuum (c) = 3 x 10⁸ ms^-1

Refractive index of glass (n_g) = 1.50

Substituting the values we get,

$1.5 = \dfrac{3 \times 10^8}{\text{Speed of light in glass}} \\ [0.5em] \Rightarrow \text{Speed of light in glass} = \dfrac{3 \times 10^8}{1.5} \\ [0.5em] \Rightarrow \text{Speed of light in glass} = 2 \times 10^8$

Hence, Speed of light in glass = 2 x 10⁸ ms^-1

The optical density of a medium is directly proportional to its refractive index.

Hence, in Table 9.3:

Medium with highest optical density = Diamond (Refractive Index 2.42)

Medium with lowest optical density = Air (Refractive Index 1.0003)

Light travels fastest in water as compared to kerosene & turpentine, because, refractive index of water is lower than that of kerosene and turpentine and speed of light is inversely proportional to the refractive index.

The statement "refractive index of diamond is 2.42" means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air/vacuum i.e., the speed of light in diamond will be $\dfrac{1}{2.42}$ times the speed of light in air/vacuum.

1 dioptre is the the power of a lens whose focal length is 1 metre.
1D = 1m^-1

Given,

Image distance (v) = 50 cm

Object height (h_o) = Image height (h_i)

magnification = -1 [∵ image is real & inverted]

Object distance (u) = ?

Power (P) = ?

We know, magnification (m) = $\dfrac{\text{v}}{\text{u}}$

Substituting the values we get,

-1 = $\dfrac{50}{\text{u}}$

⇒ u = -50 cm

Therefore, the needle is placed at a distance of 50 cm in front of the convex lens.

According to the Lens formula,

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

Substituting the values we get,

$\dfrac{1}{50} - \dfrac{1}{-50} = \dfrac{1}{f} \\[0.5em] \Rightarrow \dfrac{1}{f} = \dfrac{1}{ 50} + \dfrac{1}{50} \\[0.5em] \Rightarrow \dfrac{1}{f} = \dfrac{2}{50} \\[0.5em] \Rightarrow \dfrac{1}{f} = \dfrac{1}{25} \\[0.5em] \Rightarrow f = 25\text{ cm} \\[0.5em] \Rightarrow f = 0.25\text{ m}$

Power of lens (P) = $\dfrac{1}{f \text{(in metres)}}$

⇒ P = $\dfrac{1}{\text{0.25}} = \dfrac{100}{\text{25}}$

⇒ P = 4D

Therefore, power of lens = 4D.

Given,

focal length (f) = -2 m [∵ focal length of concave lens is negative]

Power of lens (P) = $\dfrac{1}{f \text{(in metres)}}$

Hence, P = $\dfrac{1}{\text{f}}$ = $\dfrac{1}{\text{-2}}$ = -0.5 D