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**Short Answer Type Questions****Q.1. State laws of reflection of light. List four characteristics of the image formed by a plane mirror. [CBSE 2011,2015,2019] Ans. **The laws of reflection of light state that:

- The incident ray, the reflected ray and the normal to the surface at the point of incidence all lie in the same plane.
- The angle of incidence (i) is always equal to the angle of reflection (∠r) i.e. ∠i = ∠r

Characteristics of the image formed by a plane mirror

- Image formed in a plane mirror is virtual and erect.
- It is the same size as the object.
- The image formed is laterally inverted.
- The image formed is at same distance behind the mirror as object is in front of mirror.

**Q.2. A student, holding a mirror in his hand, directed the reflecting surface of the mirror towards the sun. He then directed the reflected light on to a sheet of paper held close to the mirror. **

**(a) What should he do to burn the paper? **

**(b) Which type of mirror does he use? **

**(c) Will he be able to determine the approximate value of focal length of this mirror from this activity ? Give reason and draw ray diagram to justify your answer in this case. [CBSE 2019]**

**Ans. **(a) The student should adjust the distance between the mirror and the paper so that solar rays are sharply focussed on the paper.

(b) The mirror is a concave mirror.

(c) The student can find the approximate focal length by measuring the distance between the paper and the mirror.

As shown in Fig. 10.29, parallel rays from the sun are focussed on the paper at point A' in focal plane of mirror such that PB' = f.

**Q.3. A concave mirror has a focal length of 20 cm. At what distance from the mirror should a 4 cm tall object be placed so that it forms an image at a distance of 30 cm from the mirror ? Also calculate the size of the image formed. [CBSE 2019]**

**Ans. **As per question, focal length of concave mirror f = — 20 cm, height of object h = + 4 cm and distance of image v = 30 cm.

Following two cases may arise here :

**Case I: **I f the image form ed is real then v - — 30 cm and so from mirror formula

we have

and

So the object be placed 60 cm in front of mirror and image is an inverted image of size 2 cm.

**Case II: **If the image formed is virtual then v - + 30 cm and now

and

So the object is placed at 12 cm in front of mirror and image is an erect image of height 10 cm.

**Q.4. A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 18 cm. Find the nature, position and size of the image. [CBSE 2019]**

**Ans. **As per question f = + 12 cm, u = - 18 cm and h = + 10 cm

As per lens form

we have

The image is formed on opposite side of lens at a distance of 36 cm from it. The image is a real and inverted image.

Moreover, magnification

So, the size of image is 20 cm tall and is formed below the principal axis.

**Q.5. A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the (i) position, (ii) nature, and (iii) size of the image formed. [CBSE 2004, 2006, 2014, 2019]**

**Ans. **Here h - + 5 cm, f= + 20 cm, u = - 30 cm

(i) Using lens formula

we have

(ii) +ve sign of v means that image is being formed on the other side of lens i.e., the image is a real image.

**Q.6. A real image, 2/3 rd of the size of an object, is formed by a convex lens when the 3 object is at a distance of 12 cm from it. Find the focal length of the lens. [CBSE 2019]**

**Ans. **Here distance of the object from the lens u = - 12 cm and magnification of real image m = -2/3

As per relation m = v/u , we have v = mu = (-2/3) x (-12) = +8 cm

So as per lens formula

we have

**Q.7. Draw a ray diagram to show refraction through a rectangular glass slab. How is the emergent ray related to incident ray ? What is its lateral displacement ? [CBSE 2010, 2011, 2013, 2014, 2019]**

**Ans. **A ray diagram showing refraction through a rectangular glass slab has been shown in adjoining Fig. 10.32.

The emergent ray GH is exactly parallel to the incident ray EFNM. It means that ∠r_{2} = ∠i_{1}.

However, the emergent ray is laterally (side ways) displaced as compared to the original path of light ray. In ray diagram, the lateral displacement is GN. Its value increases on increasing the width of glass slab.

**Q.8. State the laws of refraction of light.**

**Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in vacuum. [CBSE 2018]****Ans.** Two basic laws of refraction of light are :

(i) The incident ray, the refracted ray and the normal to the separating surface at the point of incidence, all lie in the same plane.

(ii) The ratio of sine of the angle of incidence (i) to the sine of angle of refraction (r) is a constant. It is known as Snell’s law. Thus, according to Snell’s law

Sin i/sin r = a constant = n

Generally, the constant n is known as the absolute refractive index of given medium. Thus, absolute refractive index of a medium is defined as the ratio of sine of angle of incidence of a light ray in air (or vaccum) to the sine of angle of refraction of the ray in given medium.

Absolute refractive index of a medium is a unitless quantity and its value is one or greater than one In terms of speed of fight, the absolute refractive index of a medium is defined as:

Absolute refractive index of a medium (re) = Speed of lieht in vacuum (or air) ‘ c / Speed of light in given medium ‘ v’**Q.9. What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of - 20 cm. Write the nature and power of each lens. [CBSE 2018]****Ans. **The power (P) of a lens is defined as the reciprocal of its focal length (f) expressed in metre.

Thus,

SI unit of power is dioptre (D), where 1 D = 1 m^{-1}.

The focal length of lens used by one student, f_{1} = + 40 cm = + 0.4 m

∴ The lens is a convex lens and the power of lens

Again the focal length o f lens used by another student, f_{2} = - 20 cm = - 0.2 m

∴ The lens is a concave lens and the power of lens **Q.10. What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of -20 cm. Write the nature and power of each lens. [CBSE 2018]****Ans. Power of a Lens: **The ability of a lens to converge or diverge the ray of light after refraction, is called power (P) of the lens. It is defined as the reciprocal of the focal length, i.e. P = 1/f.

The SI unit of power of a lens is ‘dioptre’. A lens of focal length 100 cm has a power of 1 dioptre = 1 m^{-1}.

Given: f_{A} = + 40 cm = 0.4 m, f_{B} = -20 cm

Hence, the nature of lens A is convex with power + 2.5D and lens B is concave with power-5D.**Q.11. What is mean by power of lens? You have three lenses L _{1}, L_{2} and L_{3} of power +10 D, +5 D and - 10 D respectively. State the nature and focal length of each lens. Explain which of the three lenss will from a virtual and magnified image of an object placed at 15 cm from the lens. Draw the ray diagram in support of your answer. [CBSE Sample Paper 2018]**

= + 20 cm (convex lens)

When object is placed between focus and optical centre of convex lens, virtual erect and magnified image is formed on the same side of the lens.

Hence, for the object distance of 15 cm, lens L

(a) Name the type of lens that should be used.

(b) What should be the focal length of the lens and at what distance from the candle flame te lens be placed.

(c) Draw a labelled diagram to show the image formation in this case. [CBSC 2018C]

(b) 2 F = 4 ⇒ f = 2 m

Distance of candle flame from the lens = 4 m.

A ray diagram showing the image formation of an object AB is shown here.

A convex mirror is used as a rear view mirror in automobiles because it gives erect and diminished images of vehicles coming from behind. As a result, it helps the driver in having a much wider field of view.

As per mirror formula

we have

Thus, a screen be placed infront of mirror at a distance of 30 cm from it.

∴

Thus, image is an inverted image of height 8 cm.

Therefore, m = v/u = -1

or u = -v, v = -u, - u = v = — 25

Using lens formula,

Thus, the positive focal length shows that the given lens is a convex lens of focal length 12.5 cm. If the object is now displaced 15 cm towards the optical centre of the lens i.e, object is now placed at a distance of 25 - 15 = 10 cm from the optical centre.

Therefore u = - 10 cm , and f = +12.5 cm.

Using lens formula again,

or

v = -50

So, in this case, virtual image is formed on the same side of the object at a distance of 50 cm from the optical centre o f the lens as shown in figure

Or

f = -10cm.

Given: For concave mirror u = - 20 cm,

v = - 40 cm,

Using mirror equation,

or

⇒

Mathematically, sin i/ sin r = constant = n

The constant n

Absolute refractive index of the medium is given by

**Fig. 10.11****Q.23. Complete the following diagram [Fig. 10.12]: [CBSE 2011,2013,2016]****Ans.** The completed diagram is as given above in Fig. 10.13.**Q.24. ****Define optical centre of a spherical lens. [CBSE 2016]****Ans. **Optical centre of a spherical lens is a point on its principal axis, a ray of light passing through which goes undeviated along its path after refraction.**Q.25. ****Define a lens. [CBSE 2015,2016]****Ans. **See Point Number 30 under the heading “Chapter At A Glance”**Q.26. ****Draw the following diagram [Fig. 10.22], in which a ray of light is incident on a concave/convex mirror, on your answer sheet. Show the path of this ray, after reflection, in each case. [CBSE 2011, 2012, 2014,2016]****Ans. **The diagrams have been drawn and path of rays after reflection have been shown:**Q.27. ****State two positions in which a concave mirror produces a magnified image of a given object. List two differences between the two images. [CBSE 2016]****Ans. **Two positions in which a concave mirror produces a magnified image are :

(i) When an object is placed between focus point F and centre of curvature C of the mirror. For this position the image is real, inverted and magnified, and the image is formed beyond centre of curvature of the mirror.

(ii) When an object is placed between pole P and focus point F of the mirror. For this position the image is virtual, erect and magnified and the image is formed behind the mirror.**Q.28. ****It is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm. (i) What should be the range of distance of an object placed in front o f the mirror? (ii) Will the image be smaller or larger than the object ? Draw ray diagram to show the formation of image in this case. (iii) Where will the image of this object be, if it is placed 24 cm in front of the mirror? Draw ray diagram for this situation also to justify your answer. [CBSE 2016]**

(i) To obtain an erect image of an object by this mirror, the object should be placed in front of the mirror between its pole and focus point, that is |u| < 1 2 cm.

(ii) The image is larger than the object. The ray diagram has been shown in Fig. 10.25.

(iii) If object be placed at 24 cm in front of the mirror then it means that the object is situated at the centre of curvature [ ∵ | u | = 24 cm = 2f - 2 x 12 cm = R ] C of the given mirror. Hence as shown in Fig. 10.24 the real, inverted image of same size is formed at centre of curvature C itself [ |v| - 24 cm].

(b) In the above ray diagram mark the object-distance (u) and the imagedistance (v) with their proper signs (+ve or -ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f) of the convex lens in this case. [CBSE 2016]

(b) The object distance 'u', the image distance 'v' and the focal length 'f' of the convex lens are correlated as per relation :

While applying this formula we must specify +ve or -ve signs of u, v and f as per new Cartesian sign convention being followed.

∵ m = v/u, hence v = mu =(-1) x (-20) = + 20 cm

∴

⇒ f= 10 cm = 0.1 m

∴ Power of convex lens P =

As per lens formul

we have

∴

⇒ u = -20 cm

Thus, the object is placed at a distance 20 cm from the lens.

Moreover

A ray diagram to show the formation of image is given here in Fig. 10.39.

distance, v = -40 cm

Magnification, m = -v/u

⇒

Therefore, the object is placed at a distance of 40 cm in front of the spherical mirror.

Case I: when u = -40 cm and v = -40 cm,

Using mirror formula, we get

⇒

⇒

Hence the focal length of the mirror is 20 cm, and the negative focal length shows that it is a concave mirror.

The new position of the object when it moves 20 cm towards the concave mirror, u’ = - ( 4 0 - 20) = -20 cm.

Case II: u’ = - 20 Cm, f = - 20 cm, v =?

From mirror formula,

⇒

⇒

⇒

Thus, the image is formed at infinity.

Hence when the object is moved 20 cm towards the mirror, a real, inverted and highly enlarged image is formed at infinity.

A convex mirror is commonly used as a rear-view mirror in vehicles because it always produces a virtual and erect image whose size is smaller than the object. Therefore, it enables the driver to see a wide field of view of the traffic behind the vehicle in a small mirror.

(i) When an object is placed at infinity.

(ii) When an object is placed between F

Thus, from the above figures, it is clear that whatever be the position of the object in front of a concave lens, the image formed is always virtual, erect and diminished.

The power of the given lens is calculated as

⇒ R = - 5 D

(i) The ray incident parallel to the principal axis, after reflection, passes through the principal focus of a concave mirror.

(ii) A ray passing through the centre of curvature in a concave mirror after reflection, retraces its path.

The image formation is shown in the adjacent figure. Here for every 5 cm distance wc have 1 cm in the ray diagram. Here AB is the object and ATT is the real and inverted image formed on the basis of above mentioned two rays.

Actual measurement shows that the image is formed 30 cm in front o f the concave m irror i.e., v = 30 cm.

(b) Define principal focus and focal length of a lens. Draw ray diagram to show the position of principal focus of a lens. [CBSE2015

In Fig. 10.33, the point O is the optical centre.

(b) Principal focus of a lens is a point where a light beam incident parallel to the principal axis of the lens, after refraction, actually converges to (in case of a convex lens) or diverges from (in case of a concave lens). Since, a lens has two refracting surfaces, a lens has two principal foci. F

The distance of principal focus of a lens from its optical centre is called its focal length f.

Thus, f = OF_{1}= OF_{2}.**Q.45.**** ****What is the magnification of the images formed by plane mirrors and why?****Ans. **The magnification of the images formed by plane mirrors is 1 as the size of the image is equal to the size of object.**Q.46.**** ****A student wants to project the image of a candle flame on a screen 80 cm in front of a mirror by keeping the candle flame at a distance of 20 cm from its pole.****(i) Which type of mirror should the student use? ****(ii) Find the magnification of the image produced. ****(iii) Find the distance between the object and its image. ****(iv) Draw a ray diagram to show the image formation in this case and mark the distance between the object and its image. [Foreign 2015]****Ans. **(i) Concave mirror, as it forms a real image on the same side of the mirror.

(ii) Magnification, The negative sign in magnification shows that the image formed is real and inverted.

(iii) Distance between the object and its image = 80 - 20 = 60 cm.

(iv) The focal length of the concave mirror is given by

∴ f= - 16 cm, R = 2f = - 32 cm

Since u = - 20 cm, it implies that the object lies between F and C, so image is formed beyond the centre of curvature as shown below:

The image is real, inverted and enlarged.**Q.47.**** ****Draw a ray diagram to show the path of the reflected ray in each of the following cases. A ray of light incident on a convex mirror. (a) strikes at its pole making an angle θ from the principal axis.(b) is directed towards its principal focus.(c) is parallel to its principal axis. [Foreign 2015]**

v =?, h

Using lens formula,

⇒

∴

So, the image is formed on the same side of the object at a distance of 6.67 cm. The negative sign indicates that the image is virtual.

Also |u|< |u| so the image is diminished.

∵

or

or

So, the image is virtual, erect, diminished and of size 1.66 cm.

**Q.1.** An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. **(i) Use lens formula to find the distance of the image from the lens.****(ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case.****(iii) Draw ray diagram to justify your answer of part (ii). [Delhi 2019]****Ans. **Given u = - 60 cm, f = - 30 cm

(i) Using lens formula,

∴

∴ v = -20 cm

(ii) Nature of image : Virtual

Position of image : Between optical centre and focus Of concave lens.

Size of image: Smaller than the object using

So size of image is is one third of the object.

Erect/inverted : Erect image

(iii)**Q.2.** (a) List four characteristics of the image formed by a convex lens when an object is placed between its optical centre and principal focus. **(b) Size of the image of an object by a concave lens of focal length 20 cm is observed to be reduced to 1/3 rd of its size. Find the distance of the object from the lens. [CBSE2019]****Ans.** (a) When an object is placed between the optical centre and principal focus of a convex lens, the image formed is virtual, erect and enlarged. Moreover, the image is formed on the same side of lens behind the object.

(b) Here magnification of given concave lens m = +1/3 and focal length of lens f = - 20 cm.

As per relation m = v/u for a lens, we get

Therefore, as per sign convention followed, both u and v are -ve.

Using lens formula we have

⇒

So the object is placed at a distance of 40 cm from the lens.**Q.3.** (a) What is the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of a virtual image by a concave mirror. **(b) The linear magnification produced by a spherical mirror is+3. Analyse this value and state the (i) type o f mirror and (ii) position of the object with respect to the polerof the mirror. Draw ray diagram to show the formation of image in this cased****(c) An object is placed at a distance of 30 cm in front of a convex mirror of focal length 15 cm. Write four characteristics of the image formed by the mirror. [Delhi 2017]****Ans. **(a) Two rays are required.

(b) The linear magnification produced by a spherical mirror is +3. It shows that the size of image is three times the size of object, image is virtual and erect and formed behind the mirror. Hence

(i) the mirror is a concave mirror, and

(ii) the object is placed betw een the pole and the focus of a concave mirror.

(c) The four characteristics of the image formed by the convex mirror are virtual, erect, diminished and laterally inverted.**Q.4.** (a) To construct a ray diagram we use two rays which are so chosen that it is easy to know their directions after reflection from the mirror. List two such rays and state the path of these rays after reflection in case of concave mirrors. Use these two rays and draw ray diagram to locate the image of an object placed between pole and focus of a concave mirror. **(b) A concave m irror p roduces three times magnified image on a screen. If the object is placed 20 cm in front of the mirror, how far is the screen from the object? [Delhi 2017] ****Ans. **(a) Rays which are chosen to construct a ray diagram for reflection are: O') A ray parallel to the principal axis and 00 A ray passing through the centre of curvature of a concave mirror.

Path of these light rays after reflection:

(i) It will pass through the principal focus of a concave mirror

(ii) It gets reflected back along the same path. When an object is placed between the pole and the principal focus of a concave mirror, a virtual, erect and enlarged image is formed behind the concave mirror as shown in the figure.

(b) Given: u = — 20 cm and m = 3

Magnification, m, is given by

∴ v = - m x u

= - (- 3) (- 20 cm) = - 60 cm

Distance between the object and the screen is

= - 60 cm - (- 20 cm)

= - 40 cm.**Q.5.** **Analyse the following observation table showing variation of image-distance (v) with objectdistance (u) in case of a convex lens and answer the questions that follow, without doing any calculations:****(a) What is the focal length of the convex lens? Give reason to justify your answer. (b) Write the serial number of the observation which is not correct. On what basis have you arrived at this conclusion? (c) Select an appropriate scale and draw a ray diagram for the observation at S.No. 2. Also find the approximate value of magnification. [AI 2017]**

Reason: Objects at S.No. 3 indicates that u - -40 cm, v = +40 cm

Thus, object is at 2F.

Therefore, 2f - 40 cm ⇒ f= 20 cm

(b) Observation at S.No. 6 is not correct.

The value, w = - 15 cm, indicates that the object is in between the optical centre and the focus (i.e., less than the focal length) of the lens and hence, the image should be on the same side as the object. Accordingly, the image distance should be negative and cannot be positive (+120 cm) as shown in table.

(c) Ray diagram for the observation at S.No. 2 : Given: u = - 60 cm; v =+30 cm; f= 20 cm

view mirror.

Reason: (i) It always produces a virtual and erect image.

(ii) The size of image formed is smaller than the object.

Therefore, it enables the driver to see a wide field view of the traffic behind the vehicle in a small mirror.

(b) Radius of Curvature: The separation between the pole and the centre of curvature or the radius of the hollow sphere, of which the mirror is a part, is called radius of curvature (R), i.e., PC = R.

Since focal length of the mirror is +24 cm. It indicates that nature of the given spherical mirror is convex/diverging mirror.

As R = 2f = 24 cm

Therefore, f = +12 cm**Q.7. (a) Draw labelled ray diagrams for each of the following cases to show the position, nature and size of the image formed by a convex lens when the object is placed. (i) between its optical centre (O) and principal focus (F) ****(ii) between F and 2 F ****(b) How will the nature and size of the image formed in the above two cases, (i) and (ii) change, if the convex lens is replaced by a concave lens of same focal length? [Delhi 2017C]**** Ans. **A convex lens of focal length 'f' can form

(i) a magnified and erect image only when the object is placed between its focus 'F' and optical centre ‘O’ of the lens.

(ii) a magnified and inverted image when an object is placed in the following positions: Between F

(b) Whatever be the position o f object as given in case (i) and (ii),the image formed by the concave lens is always virtual, erect and diminished.

(i) The angle of incidence is equal to the angle of reflection.

(ii) The incident ray, the normal to the reflecting surface at the point of incidence and the reflected ray from that point, all lie in the same plane.

(i) Image formed is behind the mirror between pole (P) and focus (F).

(ii) Virtual, erect and diminished image is formed.

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