# Light

## NCERT Exercise

### Part 1

Question 1: Which one of the following materials cannot be used to make a lens?

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

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: (d) Between the pole of the mirror and its principal focus

## Chapter List

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: (b) At twice the focal length

Question 4: A spherical mirror and a thin spherical lens have each 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

Question 5: No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

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

Answer: (a) either plane or convex

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: (c) A convex lens of focal length 5 cm

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: When object is between principal focus and pole of a concave mirror, an erect, enlarged and virtual image is formed. So, we need to keep the object at a distance which is less than 15 cm (the given focal length). Question 8: Name the type of mirror used in the following situations.

Answer: When a source of light is kept at focus of a concave mirror, the reflected rays form a parallel beam of light and go to infinity. Hence, concave mirror is used as reflector in the headlights. This helps in getting a parallel beam of light.

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

Answer: A convex mirror can show image of a wider area, because of its wide field of view. This enables the driver to see more of traffic coming from behind. Hence, convex mirror is used as rear-view mirror in vehicles.

(c) Solar furnace.

Answer: The light rays coming from infinity converge at the focus after reflection from a concave mirror. Hence, a concave mirror is used in solar furnace because it helps in concentrating the solar energy at a point.

Question 9: One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer: Covering the half portion of a convex lens will not affect the image making ability of the lens. The following two figures illustrate this.

Condition 1: When upper half of the lens is covered: Condition 2: When the lower half of the lens is covered 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 the nature of the image formed.

Answer: Given, height of object = 5cm

Position of object, u = - 25cm

Focal length of the lens, f = 10 cm

Hence, position of image, v =?

We know that,

1/v-1/u=1/f

Or, 1/v+(1)/(25)=(1)/(10)

Or, 1/v=(1)/(10)-(1)/(25)

Or, 1/v=(5-2)/(50)

Or, 1/v=(3)/(50)

Or, v=(50)/(3)=16.66 cm

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

Now, magnification = v/u

Or, m=(16.66 cm)/(-25 cm)=-0.66

m=(h_i)/(h_o)

Or, -0.66=(h_i)/(5 cm)

Or, h_i=-0.66xx5 cm=-3.3 cm

The negative sign of height of image shows that an inverted image is formed. Thus, position of image = At 16.66 cm on opposite side of lens

Size of image = - 3.3 cm at the opposite side of lens

Nature of image – 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: Given, focal length, f = - 15 cm

Distance of image, v = - 10 cm

Distance of object, v =?

Object distance can be calculated using the lens formula

1/v-1/u=1/f

Or, (1)/(-10)-1/u=(1)/(-15)

Or, -1/u=-(1)/(15)+(1)/(10)

Or, -1/u=(-2+3)/(30)=(1)/(30)

Or, u=-30 cm

Negative sign shows that object is at 30cm in front of the lens.