Light: Reflection and Refraction

NCERT Exemplar Problems

Long Answer Type Questions

Question 30: Draw ray diagrams showing the image formation by a concave mirror when an object is placed

(a) Between pole and focus of the mirror

Answer:

(b) Between focus and centre of curvature of the mirror

Answer:

Answer:

(d) A little beyond centre of curvature of the mirror

Answer:

(e) At infinity

Answer:

Question 31: Draw ray diagrams showing the image formation by a convex lens when an object is placed

(a) Between optical centre and focus of the lens

Answer:

(b) Between focus and twice the focal length of the lens

Answer:

Answer:

(d) At infinity

Answer:

(e) At the focus of the lens

Answer:

Question 32: Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.

Answer: The angle between incident ray and normal is called Angle of Incidence and it is denoted as ‘i’. The angle between refracted ray and normal is called the Angle of Refraction. Angle of refraction is denoted by ‘r’.

$refraction through glass slab$

Fig: Refraction of Light

Laws of Refraction:

The incident ray, refracted ray and normal to the interface of given two transparent media, all lie in same plane.
The ratio of sine of angle of incidence and sine of angle of refraction is always constant for the light of given colour and for the pair of given media.

The Second Law of Refraction is also known as Snell’s Law of Refraction.

That is, `text(Sin i)/text(Sin r)=text(constant)`

The constant is called refractive index of the second medium in relation to the first medium.

Question 33: Draw ray diagrams showing the image formation by a concave lens when an object is placed

(a) At the focus of the lens

Answer:

(b) Between focus and twice the focal length of the lens

Answer:

Question 34: Draw ray diagrams showing the image formation by a convex mirror when an object is placed

(a) At infinity

Answer:

(b) At finite distance from the mirror

Answer:

object between infinity and pole convex mirror

Question 35: The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?

Answer: Since, image is formed at the other side of lens, thus it is a convex lens.

Given, Distance of image, v = 80cm

Height of image = 3 × height of object

Let height of object = h

Therefore, height of image = 3h

`m=(h')/h`

Or, `m=(3h)/h=+3`

Since image is forming at the other side of lens, thus, it will be inverted, thus, magnification will have negative sign.

Therefore, `m = - 3`

`m=-v/u`

Or, `-3=(80)/u=(80)/(-3)=-26.33` cm

Thus, candle should be placed at 26.33 cm before the lens.

Nature of image is real and inverted.

Question 36: Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Answer: Since, focal length is positive, thus this is a convex mirror.

Focal length, f = 20 cm

Height of image = 1/3rd of height of object.

Let height of object = h

Therefore, height of image = h/3

`m=(h’)/h`

Or, `m=(h/3)/h`

Or, `m=h/(3×h)=1/3`

Now, `m=-v/u`

Or, `1/3=-v/u`

Or, `u=-3v`

Or, `v=-u/3`

Mirror formula: `1/v+1/u=1/f`

Or, `1/(-u/3)+1/u=1/(20)`

Or, `-3/u+1/u=1/(20)`

Or, `(-3+1)/u=1/(20)`

Or, `-2/u=1/20`

Or, `u=-40` cm

Position of image can be calculated as follows:

`m=-v/u`

Or, `1/3=-v/(-40)`

Or, `1/3=v/(40)`

Or, `v=(40)/3=16.66` cm

The positive sign of distance of image shows that image is forming behind the mirror.

Thus, position of image = 40 cm in front of the mirror.

Nature of image – erect and virtual

Question 37: Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of –50 cm. What is the nature of the lens and its power used by each of them?

Answer: Power of lens is the reciprocal of focal length of the lens. The unit of power is dioptre, which is represented by ‘D’.

Case 1: Power of lens when focal length = 50cm

50 cm = 50/100 m = 0.5 m

`P=1/f`

Or, `P=1/(50)=0.02` D

Case 2: Power of lens when focal length = - 50cm

- 50 cm = - 0.5 m

`P=1/(-50)=-0.002` D

In the case of lens having focal length equal to 50cm, the power is equal to +0.02 D and lens is convex.

In the case of lens having focal length equal to - 50cm, the power is equal to - 0.02 D and lens is concave.

Question 38: A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under

Position of candle = 12.0 cm

Position of convex lens = 50.0 cm

Position of the screen = 88.0 cm

What is the focal length of the convex lens?
Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?
What will be the nature of the image formed if he further shifts the candle towards the lens?
Draw a ray diagram to show the formation of the image in case (iii) as said above.

Answer: Let distance of different component is measured from a wall.

Thus, position of candle from lens = position of lens – position of candle = 50 cm – 12 cm = 38 cm

Position of screen from lens = position of screen – position of lens = 88 cm – 50 cm = 38 cm

(i) Since, distance of image and object is equal from lens, thus, magnification is 1, i.e. size of image and object is equal.

In the case of convex lens, when object is placed at 2F, the image of same size, is formed at 2F1 at the other side of lens.

Thus, 2F = distance of object = 38cm

Therefore, F = 38/2 cm= 19 cm

Thus, focal length of the lens = 19cm

(ii) Distance of object = position of lens – distance of object = 50 cm – 31 cm = 19cm

Here, focal length f = 19cm

Since, the distance of object is equal to the focal length, i.e. object is placed at the focal length, thus image will be formed at infinity.

(iii) If the candle is shifted further towards lens, i.e. the distance of object will be less than 19 cm. This means object is placed between focal length and pole.

In this condition an enlarged, virtual and erect image will be formed at the same side of the object will be formed.