Please refer to Light Reflection and Refraction Class 10 Science Important Questions with answers below. These solved questions for Chapter 10 Light Reflection and Refraction in NCERT Book for Class 10 Science have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these solved problems properly as these will help them to get better marks in your class tests and examinations. You will also be able to understand how to write answers properly. Revise these questions and answers regularly. We have provided Notes for Class 10 Science for all chapters in your textbooks.

Important Questions Class 10 Science Chapter 10 Light Reflection and Refraction

All Light Reflection and Refraction Class 10 Science Important Questions provided below have been prepared by expert teachers of Standard 10 Science. Please learn them and let us know if you have any questions.

Question: Which kind of mirrors are used in the headlights of a motorcar and why?
Answer: A concave mirror is used in the headlight of a car because concave mirror is a converging mirror. In headlights the bulb is kept at the focus of the mirror, due to which the rays emerging from the bulb after reflection from the mirror becomes parallel to the principal axis hence we get the powerful beam of light which can go to farther distances.

Question: The refractive indices of water and glass are and respectively. Write the relation and find the value of refractive index of water with respect to glass and glass with respect to water.
Answer: 

Question: What is the distance between the image and the plane mirror, if the object is at 15 m?
Answer: In plane mirror the distance between object and mirror is same as the distance between the image and mirror.
⇒ Image Distance = Object Distance.
∴ Distance between image and plane mirror is 15 cm.

Question: When do an object and its image coincide in a convergent mirror?
Answer: When the object is kept at the centre of curvature of the concave mirror the image of object coincide with the object, because image is formed at the centre of curvature and is same as the size of the object as shown in the figure below. 

Question: What is the value of 1n2 × 2n1?
Answer:  

Question: What type of lens is an air bubble inside water?
Answer: Due to higher refractive index of water (1.33) than air (1.0003) the air bubble acts as the diverging lens (Concave lens).The figure below illustrates it.

Question: What is the magnification produced by a plane mirror?
Answer: The magnification produced by a plane mirror is +1. Because,   

And the height of image formed by plane mirror is equal to height of object. Therefore, Height of Image = Height of Object.
 Hence Magnification is +1.

Question: How can we find the (rough) focal length of a concave mirror?
Answer: Method We can put the object far away from the concave mirror and place the screen where the image is formed. The place where the image formed is sharpest is the focus of the mirror, measure the distance between the focus and mirror to get the focal length. 

Explanation when an object is kept at the infinity than the image is formed at the focus of the concave mirror.

Question: Define ‘refractive index of a transparent medium.’ What is its unit? Which has a higher refractive index, glass or water?
Answer: Refractive index of transparent medium is defined as the ratio of speed of light in vacuum to the speed of light through medium.
Refractive index (n) = ^C/V
Where;
C = speed of light;
V = speed of light in medium;
It has no unit because it is the ratio of two quantities of same type (speed);
Glass has higher refractive index of 1.52. (Because Refractive index of water = 1.33).

Question: An object 2 cm in size is placed 30cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the nature and the size of the image formed? Draw a ray diagram to show the formation of the image in this case.
Answer: According to the question;
Object distance (u) = -30 cm;
Focal length (f) = -15 cm;
Image distance = v;
By mirror formula; 

Thus, screen should be placed 30cm in front of the mirror (Centre of curvature) to obtain
the real image.
Height of object h₁= 2cm;
Magnification = h₂/h₁=-v/u
Putting values of v and u
Magnification =h₂/2=-30/-30
⇒h₂/2-1;
⇒h₂=2x-1=-2
Height of image is 2 cm.
Negative sign means image is inverted.
Thus real, inverted image of size same as that of object is formed.
Diagram below shows the image formation. 

Question: A convex lens has a focal length of 10 cm. At what distance from the lens should the object be placed so that it forms a real and inverted image 20 cm away from the lens? What would be the size of the image formed if the object is 2 cm high? With the help of a ray diagram show the formation of the image by the lens in this case.
Answer: According to the question;
Object distance = u;
Image distance (v) = 20cm;
Focal length = 10cm
By lens formula; 

⇒ u = -20cm.
Therefore, object is placed at 20 cm in front of lens.
Now; 

Height of image is 2 cm.
Negative sign means image is inverted.
Thus, the image is real, inverted and same as size of object.
Diagram below shows the image formation. 

Question: What is meant by power of a lens? Name and define its S.I. unit. One student uses a lens of focal length +50 cm and another of –50cm. State the nature and find the power of each lens. Which of the two lenses will always give a virtual, erect and diminished image irrespective of the position of the object?
Answer: Power of lens is defined as the efficiency with which a lens can converge or diverge the light ray.
The reciprocal of focal length of the lens is called the power of the lens (P).
P (in dioptre) =1/(in meters)
The S.I unit of power of lens is dioptre denoted by D.
f₁ = 50 cm = 0.5 m (1 m = 100 cm)
⇒ P₁ = 1/f1
P₁ = 1/0.5
P₁ = 2D
Since Power is positive therefore the lens with focal length 50 cm is convex lens.f₂ = – 50cm = -0.5m (1m = 100cm)
⇒ P₂ = 1/f1
P₂ = 1/-0.5
P₂ = -2D
Since Power is negative therefore the lens with focal length -50 cm is concave lens. The image formed by the concave lens is virtual, erect and diminished irrespective of the position of the object.

Question: A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18 cm at a distance of 12 cm from it. Use lens formula to determine the position, size and nature to the image formed.
Answer: According to the question;
Object distance = -12cm;
Image distance = v;
Focal length = 18cm;
By lens formula; 

⇒ v = -36cm.
Therefore, image is formed at 36 cm in front of lens.
Now;
Height of object h1= 5cm;
Magnification = h₂/h₁=v/u
Putting values of v and u 
Magnification = h₂/5=-36/-12
⇒h₂/5=3;
⇒h₂=3×5=15
Height of image is 15 cm.
Thus, the image is virtual, enlarged, and erect and three times the size of object.

Question: A. What is meant by ‘power of a lens’?
Answer: Power of lens is defined as the efficiency with which a lens can converge or diverge the light ray.
P (in dioptre) =1/f(in meters)

Question: B. State and define the S.I. unit of power of a lens.
Answer: The S.I unit of power of lens is dioptre denoted by D.
1 D =1/1m=1m^-1
Thus 1 Dioptre is the power of the lens having focal length of 1m.

Question: C. A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close focal contact with each other. Calculate the lens power of this combination.
Answer: Given;
Focal Length of convex lens f1 = 25cm = 0.25m (1m = 100cm);
Focal Length of concave lens f2 = -10cm = -0.1m (1m = 100cm);
Power of convex lens P₁ = 1/f₁=1/0.25=1/1/4
∴ P1 = 4D.
Power of concave lens P₂ = 1/f₂=1/-0.1=-1/1/10
∴ P2 = -10D.
Power of combination = P₁ + P₂ = 4-10 = 6D.
Thus, power of combination is -6 Dioptre.

Question: A. With the help of a ray diagram, state the meaning of refraction of light. State Snell’s law of refraction of light and also express it mathematically.
Answer: Refraction of light is the phenomenon of bending of light as it travels from one medium to the other. When a ray of light travels from one transparent medium to the other it bends at the surface, this is due to the different optical density of different medium.
The figure below illustrates the refraction of light. 

Snell’s Law states that ratio of sine of angle of incidence and angle of refraction of two medium is always constant.
Mathematically it is given as aμb
aμb = is the refractive index of medium a with respect to medium b

Question: B. The refractive index of water with respect to vacuum is and refractive index of vacuum with respect to glass is . If the speed the speed of light in glass is 2 × 108 ms-1, find the speed of light in (i) vacuum, (ii) water. 
Answer:  Given

Question: A. With the help of a ray diagram explain why a concave lens diverges the rays of a parallel beam of light.
Answer: Concave lens works on the principle of refraction. We can divide the lens into tiny pieces which resembles a triangular prism (shown in figure), thus when the light ray fall it refracts the light either inwards or outwards due to its shape. Concave lens is thicker outside and thinner at edges. So the light rays are refracted outside the figure below illustrates it more clearly. 

Q. B. A 2.0 cm tall object is place perpendicular to the principal axis of a concave lens of focal length 15 cm. At what distance from the lens, should the object be placed so that it forms an image 10 cm from the lens? Also find the nature and the size of image formed.
Answer: According to the question;
Object distance = u;
Image distance (v) = -10cm;
Focal length = -15cm;
By lens formula; 

Question: A. Draw the ray diagram in each case to show the position and nature of the image formed when the object is placed: At the centre of curvature of a concave mirror
 Answer: Position of Image: At the centre of curvature.
Nature of Image: Real and inverted and same as object size
Ray diagram: 

Question: B. Draw the ray diagram in each case to show the position and nature of the image formed when the object is placed: Between the pole P and focus F of a concave mirror
 Answer: Position of Image: Behind the mirror Nature of Image: Virtual, erect and magnified.
Ray diagram:  

Question: C. Draw the ray diagram in each case to show the position and nature of the image formed when the object is placed: In front of a convex mirror 
Answer: Position of Image: Behind the mirror Nature of Image: Virtual, erect and diminished.
Ray diagram:  

Question: D. Draw the ray diagram in each case to show the position and nature of the image formed when the object is placed:
At 2f of a convex lens
Answer: Position of Image: At 2f Behind the lens. Nature of Image: Real, inverted and same as object size.
Ray diagram: 

Question: E. Draw the ray diagram in each case to show the position and nature of the
image formed when the object is placed:
In front of a concave lens
Answer: Position of Image: Behind the lens Nature of Image: Virtual, erect and diminished.
Ray diagram:

Question:  (A) A concave mirror of focal length 10 cm can produce a magnified real as well as virtual image of an  object placed in front of it. Draw ray diagrams to ??ustify this statement.
(B) An object is placed perpendicular to the principal axis of convex mirror of focal length 10 cm. The distance of the object from the pole of the mirror is 10 cm. find the position of the image formed.
Answer: (A) A concave mirror can form real and magnitude image of an object when the object is kept between the centre of curvature and the focus of the mirror 

A concave mirror can form virtual and magnified image of an object when the object is kept between the pole and the focus of the mirror. 

(B) Convex mirror
f = + 10 cm (as principal focus of a convex mirror is behind the mirror)
u = – 10 cm
v = ?
Mirror formula 

The image is formed between P and F behind the mirror.

Question: (A) Draw a labelled ray diagram to show the path of a ray of light incident obliquely on one face of a glass slab.
(B) Calculate the refractive index of the material of a glass slab. Given that the  speed of light through the glass slab is 2 x 10⁸ m/s and in air is 3 ?? 108 m/s.
(C) Calculate the focal length of a lens, if its power is -2.5D. 
Answer: (A) 

(B) Speed of light in glass slab (v) = 2 x 10⁸ m/s
Speed of light in air (c) = 3 x 10⁸ m/s
Refractive index of glass (n) = ?
n=c/v
n=3×10⁸ /2×10⁸ =1.5
Hence, refractive index of material of glass slab is 15.
(C) Power P = – 2.5 D
Total length f = ?
P =1/f
f=1/p or 1/-25 or 100 cm/-25
= – 40 cm Or – 0.4 m
The focal length of a lens of power – 2.5 D
is – 0.4 m Or – 40 cm

Question:  Draw a ray diagram in each of the following cases to show the formation of image, when the object is placed:
(A) between optical centre and principal focus of a convex lens.
(B) anywhere in front of a concave lens.
(C) at 2f of a convex lens.
State the signs and values of magnifications
in the above mentioned cases (A) and (B). 5
Answer: (A) When an object is placed in front of the lens between optical centre and principal focus of a convex lens, the image is formed beyond 2F₁ (on the same side of the object.)

AB is the object and A’B’ is the image. The image formed is enlarged, virtual and erect so the value of magnification will be greater than 1 and its sign will be positive (t).
(B) When an object a placed anywhere infront of a concave lens.
When we place an object between infinity and optical centre (0) of the concave lens,the image will be formed between focus (F₁) and optical centre (0) on the same side of the lens.

The image formed is real, inverted and of the same size.

Question:  An object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of focal length 15.0 cm.
(A) At what distance from the mirror should a screen be placed in order to obtain a sharp image?
(B) ??ind the size of the image.
(C) Draw a ray diagram to show the formation of image this case. 
Answer: (A) Height of object h₁ = + 4 cm
object distance h = –25.0 cm
focal length of a concave mirror of = –15.0 cm using the mirror formula 

∴ Image distance v = – 375. cm
A screen should be placed at 37.5 cm infront of the mirror. (3)
(B) Size of image : using the expression for linear magnification.

The height of image is 6 cm i.e. it is magnified. The minus sign shows that the image is below the principal axis. Therefore, it is real and inverted.

(C) Scale is given as :
5 cm = 1 cm
FP = 3 cm
FC = ?? cm
AB = 0.8 cm
A’B’ = 1.2 cm
AB’ = 7.5 cm 

Question:  While tracing the path of a ray of light passing through a rectangular glass slab a
student tabulated his observations. If in his experiment for two readings he takes two
values of ∠i as 40⁰ and 50⁰, what should be the correct values of ??e and ??r in each case?
Answer: In the experiment on refraction of light by a glass slab, the correct values of e and r are shown below: 

Question:  A child has a spherical mirror of focal length 1?? cm and he observes an erect and magnified image of his face in the mirror. State the type of mirror used by him and nature of the image he observes. What should be the range of distance of the mirror from his face? Draw ray diagram to justify your answer.
Answer: The mirror used by the child is cancave mirror. The nature of image is virtual and magnified.The range of face from mirror is between focal length and pole.

Question:  (A) Name and state the law of refraction of light that explains the refractive index of one medium with respect to the other. Express it mathematically. How
is refractive index related to velocity of light in one medium with respect to
the second? Write an expression for the absolute refractive index of a medium.
(B) Refractive index of water is 1.33 and that of turpentine oil is 1.47. Draw a ray diagram to show the path of a ray passing from turpentine oil into water.
Answer: (A) Snell’s law: It states that the ratio of sine of angle of incidence and sine of angle of refraction is constant.
sin i/sinr = μ Here is refractive index.
refractive index is the ratio of speed of light in one medium and speed of light in second medium.
μ₂₁ =Speed of light inmedium 1/Speed of light inmedium 2 V1/V2
Expression for absolute refractive index of a medium μm= Speed of light in air/Speed of light inmedium = C/V
(B) Refractive index of water = 1.33
Refractive index of turpentine oil = 1.47 

Explanation: The speed of light is higher in a rarer medium than in denser medium. Thus, a ray of light traveling from a rarer medium to a denser medium slows down and bends towards the normal. When it travels from a denser medium to a rarer medium, it speeds up and bends away from the normal. Turpentine oil has 1.47 refractive Index so it is optically denser than water.

Question:  (A) Explain with the help of diagrams the following terms in the context of spherical mirrors:
(i) Pole
(ii) Centre of curvature
(iii) Principal axis
(B) Consider the following diagram. In this M is a mirror, P is an object and Q is the image of the object formed by the mirror:

(i) State the type of mirror M.
(ii) If the numeric value of focal length of the mirror is 1?? cm, what must be the
range of distance of the object from the mirror?
(iii) Draw a ray diagram to show the formation of image in this case.
Answer: (A) Diagrams showing the pole, Centre of Curvature and Principal axis of the spherical mirrors, concave mirror and convex mirror are drawn below: 

(1) Pole: The centre of the reflecting surface of the spherical mirror is a point called the pole.
(2) Centre of curvature: It is the centre of the holl ow sphere of glass of which the reflecting surface of the spherical mirror is a part.
(3) Principal axis: It is the straight imaginary line passing through the centre of curvature and pole of a spherical mirror.
(B) (1) The type of mirror is a concave mirror as image formed is magnified, virtual and erect, and behind the mirror. This is possible when object lies between pole and principal focus of a concave mirror.
A convex mirror always forms a virtual erect but diminished image of the object and a plane mirror forms a virtual, erect image of same size as the object.
(2) The object distance u should be less than the focal length, 15 cm. Range of object distance is 0 cm < u < 15 cm.
(3) Ray diagram showing the formation of image. 

Here AB represents the object ‘P’ and A’B’ represents its image ‘Q’ given in question.

Question:  State Snell’s law of refraction of light. Express it mathematically. Write the relationship between absolute refractive index of a medium and speed of light in vacuum. 
Answer: Snell’s law of refraction of light states that the ratio of sine of angle of incidence (i) to the angle of refraction (r) is a constant for a given pair of media and for light of a given colour.
Snell’s law can be stated mathematically as  sin i /sin r , where i is the angle of incidence and r is the angle of refraction.
The absolute refractive index ‘n’ of a medium is the ratio of speed of light in vacuum (c) to the speed of light in that medium (v).
n = Speed of light in vaccum (c)/Speed of light in the medium (v)

Question: (A) A person suffering from myopia (near-sightedness) was advised to wear corrective lens of power ?? 2.5 D. A spherical lens of same focal length was taken in the laboratory. At what distance should a student place an object from this lens so that it forms an image at a distance of 10 cm from the lens?
(B) Draw a ray diagram to show the position and nature of the image formed in the above case.
Answer: (A) P = –2.5 D
f = ?
P = 1/f Or f =1/p=1/-2.5D
= – 0.4 m Or – 40 cm
f = – 40 cm
v = – 10 cm
u = ?

1/f=1/v-1/u

1/u=1/v-1/f
 
=1/-10- 1/- 40 = 1/-10+1/40

-4+1/40 -3/40

u=-40/3 or -13.33 cm

The object should be placed – 13.33 cm from the lens so that A forms an image at a distance of 10 cm from the lens.
(B)

Question:  Analyse the following observation table showing variation of image distance (v) with object distance (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 convexlens? Give reason in support of your answer.
(B) Write the serial number of that observation which is not correct. How did you arrive at this conclusion?
(C) Take an appropriate scale to draw ray diagram for the observation at S. No.4 and find the approximate value of magnification.
Answer: (A) The focal length of the convex lens = 15 cm.
We observe from observation at S.No. 3 that object distance (–30 cm) and image
distance (+30cm) are both equal. We know that for a convex lens, when object is at 2F,image is also formed behind the lens at 2F.
Focal length = f = 30/2 = 15 cm

(B) Observation at S. No. 6 is not correct as object distance = – 10 cm, that is object lies between Optical centre and focus of lens. Image should be formed in front of the lens beyond 2F. Sign of the image distance
should have been “–” instead of “+”.
(C) Taking f = + 15 cm, u = – 20 cm, v = + 60

cm, and height of object = 1?? cm, the ray diagram has been drawn with scale 10 small divisions = 10 cm.
Change in Ray diagram 

We measure height of image and find that it is equal to 45 cm.
Therefore, magnification = 3

Question:  Define the term magnification with reference to spherical mirrors. If a concave mirror forms a real image 40 cm from the mirror, when the object is placed at a distance of 20 cm from its pole, find the focal length of the mirror.
Answer: Magnification by a spherical mirror is defined as the ratio of the size of the image formed to the size of the object.
Thus, magnification, m = – v/u = h₂/h₁

where h₁= size of the object, and h2 = size of the image.
To find the focal length of the concave mirror, we use the mirror formula,
1/f = 1/v +1/u .

where f = focal length of the mirror, u = object distance = – 20 cm, and v = – 40 cm (the minussign is used as a real image is formed, which is in front of the mirror).
    ∴      1/f= -1/40- 1/20= -1-2/40= -3/40
 
      ⇒     f=-40/3cm=-13.33 

Question:  (A) Name the lens which can be used as a magnifying glass. ??or which position of the object a convex lens form:
(i) a real and inverted image of the same  size as that of the object?
(ii) a virtual and erect image?Draw ray diagram to justify your answer in each case.
(B) One half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Draw ray diagram to justify your answer.
Answer: (A) Convex lens can be used as a magnifying glass Position of the object:
(i) When an object is placed at centre of curvature, a real, inverted and equal size image is obtained. 

(ii) When an object is placed between the focus and the optical centre of a convex lens, a virtual and erect image of the object is formed. 

(B) Yes, even when one half of the convex lens is covered with a black paper, the complete image of the object will be formed. When the upper half of the lens is covered: In this situation, rays of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.
When the lower half of the lens is covered:
In this situation, rays of light coming from the object will be refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.
We will get a sharp image but the brightness of the image will be less now.
Ray diagram:
(a)

Question:  (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 mirror produces 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?
Answer: (A) Rays which are chosen for constructing a ray diagram in case of a concave mirror are given as under:(1) A ray parallel to the principal axis will pass through the principal focus, afterreflection.

(2) A ray passing through the principal focus will emerge parallel to the principal axis after reflection.

(3) A ray passing through the centre of curvature will be reflected back along the same path.

(4) A ray incident obliquely to the principal axis, towards a point P (pole of the mirror) will be reflected obliquely.

Ray diagram to locate the image of an object placed between pole and focus of a concave mirror using the two rays at (iii) and (iv) is given below:

(B) In the given question, image is obtained on a screen which means image is a real image.Magnification = – 3 (3 times magnified real and inverted image).
Here, u = – 20 cm. To find image distance v, we use the relation m =- v/u
∴ -3=-v/20 ⇒ v =-60 cm
The screen is placed ??0 cm in front of the concave mirror.
Distance of screen from the object = 40 cm

Question:  15.8. Suppose you have three concave mirrors A, B and C of focal lengths 10 cm, 15 cm and 20 cm.
For each concave mirror you perform the experiment of image formation for three values of object distance of 10 cm, 20 cm and 30 cm. Giving reason answer the following:
(A)For the three object distances, identify the mirror??mirrors which will form an image of magnification -1.
(B) Out of the three mirrors identify the mirror which would be preferred to be used for shaving purposes??makeup.
(C)For the mirror B draw ray diagram for image formation for object distances 10 cm and 20 c.m
Answer: (A) A real, inverted and same size image as that of object formed by the concave mirror will form an image
of magnification –1. It is possible only when the object is placed at C (R = 2 f). Hence for the object distances of 20 cm and 30 cm, concave mirrors ‘A’ and ‘B’ will form the real, inverted and same size images as that of the object therefore, the mirrors ‘A’ and ‘B’ will form an image of magnification –1.
Explanation:

(B) Concave mirror ‘C’ of focal length 20 cm will be preferred to be used for shaving purpose / make up. This is because when we bring our face within its focal length it forms a virtual, erect and enlarged image of our face.
(C) Ray diagram for image formation by mirror ‘B’
(i) For object distance 10 cm. 

(ii) For object distance 20 cm. 

Question. (a) Define radius of curvature and focal length of a spherical mirror and show it on a figure.
(b) Write relation between radius of curvature and focal length of a spherical mirror.
Answer : 
a. Distance between pole and centre of curvature is known as radius of curvature or the radius of the sphere of which mirror is a part is called radius of curvature.   (Ques 57)

PC =radius of curvature R ^ h b. Focal length of a mirror is half of the radius of curvature    f = R/2

Question. An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm, determine the position, nature and size of the image formed using the lens formula.
Answer : 
Given, object height h =+ 5 cm for a concave lens.
u =- 20 cm, f =- 10 cm
v = ?  (Que. 58)

Question. If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror?
Answer :
Only in convex mirror, for all positions of the object placed in front of it is always virtual, erected and diminished. Hence this mirror is convex mirror.   

Convex mirrors are used in automobiles as a rear view mirror because of wider field of view and formation of erect image.

Question. The nature, size and position of image of an object produced by a lens or mirror are as shown below.
Identify the lens/ mirror (X) used in each case and draw the corresponding complete ray diagram, (size of the object about half of the image).    

Answer :     

Question. (a) Calculate the distance at which an object should be placed in front of a convex lens of focal length 10 cm to obtain a virtual image of double its size.
(b) In the above given case, find the magnification, if image formed is real. Express it in terms of relation between v and u
Answer : Given f =+ 10 cm, u = ?
For virtual image
m =+ 2

Question. One half of a convex lens is covered with a black paper.
a. Show the formation of image of an object placed at 2Fp of such covered lens with the help of ray diagram. Mention the position and nature of
image.
b. Draw the fay diagram for same object at same position in front of the same lens, but now uncovered. Will there be any difference in the image obtained in the two cases? Give reason for your answer.
Answer :
a. If the lower half of the lens is covered even then it will form a complete real, inverted image of samesize at C2(2F2) with reduced intensity of image.       

b. There will be no change in the nature and position of the object except in later case the image will be brighter.

Question. 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.
Answer : 
Power of lens is the ability of a lens to converge or diverge light rays passing through it. It is the reciprocal of the focal length.
S.I. Unit: If focal length is measured in metre then unit of power of a lens is Dioptre (D)
p = 1/f(D)

Power of first lens: Focal length=+40 cm. Focal length is positive hence it is a convex lens.
P1 = + 100 /f(cm) = + 100 / 40 cm = + 2.πD

Power of second lens
Focal length =- 20 cm 

Its focal length is negative hence it is a concave lens. 

Question. 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.
Answer : 
Laws of refraction:
a. The incident ray, refracted ray and normal to the point of incidence, all lie in the same plane.
b. The ratio of sine of incident angle and sine of angle of refraction for a given pair of medium is constant.

sin i /sin r = Constant

Absolute refractive index of a medium is the ratio of speed of light in air or vacuum and speed of light in the medium.
Absolute refractive index   = speed of life in air vacuum / speed of light in medium

Question. An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre ‘O’ of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre ‘O’ and principal focus ‘F’ on the diagram. Also find the approximate ratio of size of the image to the size of the object.
Answer : Ray diagram: Position of O and F 

Question. (a) Define real image of an object.
(b) Name the mirror that:
– can give real as well as virtual image of an object.
– will always give virtual image of same size of an object.
– will always give virtual and diminished image of an object.
– is used by a doctor in examining teeth,
(b) With the help of a ray diagram explain the use of concave mirror as solar concentrators.
Answer : 
a. When the reflected rays after reflection actually meet at a point then real image is formed.
b. Concave mirror
Plane mirror
Convex mirror
Concave mirror
c. The rays coming from the sun are parallel to principal axis and will concentrate at focus after reflection. 

Question. Name the type of mirror used in the following:
a. Solar furnace
b. Side/rear – view mirror of a vehicle.
Draw a labelled ray diagram to show the formation of image in each of the above two cases.
Which of these mirrors could also form a magnified and virtual image of an object? Illustrate with the help of a ray diagram.
Answer :  
a. Concave mirror 

b. Convex mirror  

Concave mirror form magnified virtual image of an object. 

Question. (a) A thin converging lens forms a
– Real magnified image.
– Virtual magnified image of an object placed in front of it.
Write the positions of the objects in each case.
(b) Draw labelled ray diagrams to show the image formation in each case.
(c) How will the following be affected on cutting this lens into two halves along the principal axis?
– Focal length
– Intensity of the image formed by half lens.
Answer : 
a. When object is placed in between F and 2F(C) of a converging lens it will form a real magnified image. 

When object is placed in between F1 and optical centre O of a converging lens, it will form a virtual magnified image of the object.  

c. When lens is cut along the principal axis its focal length remains same but intensity is reduced.

Question. (a) Define principal focus of a spherical mirror.
(b) For what position of the object does a concave mirror form a real, inverted and diminished image of the object? Draw the ray diagram.
(c) An object 4 cm high is placed at a distance of 6
cm in front of a concave mirror of focal length 12 cm. Find the position of the image formed.
Answer : 
a. The point on the principal axis at which the light rays parallel to principal axis after reflection from a concave mirror actually meet or appear to come from in convex mirror on the principal axis iscalled principal focus.
b. In case of a concave mirror, when the object is placed beyond 2F(C) then image formed is real, inverted and diminished.  

Question. (a) Define optical centre of a spherical lens.
(b) You are given a convex lens of focal length 30 cm.
Where would you place an object to get a real, inverted and highly enlarged image of the object?
Draw a ray diagram showing the image formation, (c) A concave lens has a focal length of 20 cm. At what distance an object should be placed so that  it forms an image at 15 cm away from the lens?
Answer : 
a. Mid point of a lens through which a ray of light passes undeviated.
b. At f =+ 30 cm, because when object is kept at fin case of a convex lens then its real, inverted and highly enlarged image is formed