Light - Reflection and Refraction

Intext Questions 1

Question 1

Define the principal focus of a concave mirror.

Answer

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.

Question 2

The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer

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.

Question 3

Name the mirror that can give an erect and enlarged image of an object.

Answer

Concave Mirror

Question 4

Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer

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.

Intext Questions 2

Question 1

Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer

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.

Question 2

A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer

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.

Intext Questions 3

Question 1

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?

Answer

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.

Question 2

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 x 10⁸ ms^-1.

Answer

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

Question 3

Find out, from Table 9.3, the medium having the highest optical density. Also, find the medium with the lowest optical density.

Answer

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)

Question 4

You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.

Answer

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.

Question 5

The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer

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.

Intext Questions 4

Question 1

Define 1 dioptre of power of a lens.

Answer

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

Question 2

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.

Answer

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.

Question 3

Find the power of a concave lens of focal length 2 m.

Answer

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

Exercises

Question 1

Which one of the following materials cannot be used to make a lens?

1. Water
2. Glass
3. Plastic
4. Clay

Answer

Clay

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

Question 2

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

1. Between the principal focus and the centre of curvature
2. At the centre of curvature
3. Beyond the centre of curvature
4. Between the pole of the mirror and its principal focus.

Answer

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.

Question 3

Where should an object be placed in front of a convex lens to get a real image of the size of the object?

1. At the principal focus of the lens
2. At twice the focal length
3. At infinity
4. Between the optical centre of the lens and its principal focus.

Answer

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.

Question 4

A spherical mirror and a thin spherical lens have a focal length of -15 cm. The mirror and the lens are likely to be

1. both concave
2. both convex
3. the mirror is concave, and the lens is convex
4. the mirror is convex, but the lens is concave

Answer

both concave

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

Question 5

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

1. only plane
2. only concave
3. only convex
4. either plane or convex

Answer

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.

Question 6

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

1. A convex lens of focal length 50 cm
2. A concave lens of focal length 50 cm
3. A convex lens of focal length 5 cm
4. A concave lens of focal length 5 cm

Answer

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.

Question 7

We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer

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:

Question 8

Name the type of mirror used in the following situations.

(a) Headlights of a car

(b) Side/rear-view mirror of a vehicle

(c) Solar furnace

Support your answer with a reason.

Answer

(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.

Question 9

One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer

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.

Question 10

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed.

Answer

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.

Question 11

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer

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.

Question 12

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer

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.

Question 13

The magnification produced by a plane mirror is +1. What does this mean?

Answer

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

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.

Question 14

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer

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

Question 15

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer

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.

Question 16

Find the focal length of a lens of power -2.0 D. What type of lens is this?

Answer

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.

Question 17

A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer

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.