Class 10 Science Chapter 9 Board Questions of Light – Reflection and Refraction. This is the collection of main questions asked in last 10 years CBSE Board Papers and all questions of last 5 years CBSE Exams Papers and Sample Paper. The questions are divided into sets one mark, two marks, three marks and five marks questions with answers. Questions are important not only CBSE Board but the UP Board students also. It takes hardly 50 minutes to revise the entire chapter 9 using these board questions. One mark questions are answered in one sentence or one word. Two marks questions also solve according to marks or points mentioned in the questions. 3 or 5 marks questions are described well as per the need of questions, giving the proper reason behind and examples. All the contents on Tiwari Academy website as well as on Apps are free to use. If you are facing any problem to use these contents, please contact us for help.

## Class 10 Science Chapter 9 Board Questions for 2024-25

 Class: 10 Science Chapter: 9 Light – Reflection and Refraction Contents: Previous Years Board Questions Session: CBSE 2024-25

### Why do we see our image in a shining spoon? [CBSE 2004]

The shining surface of spoon behaves like a curved mirror hence we can see our image in it.

### Define real image of an object? [CBSE 2011]

Real image of an object is that where light rays from the object after reflection/refraction from mirror/lens/prism etc., actually converges too.

### What do you mean by the pole of a spherical mirror? [CBSE 2000, 2011]

The center point of the reflecting surface of a spherical mirror is called its pole.

### What is the center of the curvature of a spherical mirror? [CBSE 2004]

The center of the curvature of a spherical mirror is the center of the sphere of which the mirrors forms a part.

### Define principal axis of the spherical mirror? [CBSE 2011, 2012]

A straight light passing through the pole and the center of curvature of a spherical mirror is called its principal axis.

### Class 10 Science Chapter 9 Board Questions for Exams

Get here the answers of Class 10 Science Chapter 9 Board Questions. Questions are answered according to their marks. For example, one mark questions are answered in one word or one sentence. Similarly, the 5 marks questions are answered point wise as well as descriptive format. UP Board students can take the benefits of these questions as in UP Board, students are using NCERT Textbooks for Class 10 Science.

#### The Power of Lens

The power (P) of a lens is a measure of the degree of convergence or divergence of light rays achieved by it. Mathematically, power of a lens is defined as the reciprocal of its focal length (f).
Thus, P = 1/f
SI unit of power is diopter (D) where 1 D = 1m-1. The power of a convex(converging) lens is positive and that of a concave(diverging) lens is negative. If a number of thin lenses having power P1,P2,P3… are placed in contact then the net power (P) of the lens combination is equal to the algebraic sum of the individual powers i.e.,
P= P1 + P2 + P3……

### A ray of light moving along the principal axis is falling on a concave mirror. In which direction is its reflected? [CBSE 2011]

The ray is reflected back along the principal axis of mirror because here
Lr = Li = 0.

### An object is placed in front of concave mirror between the pole and the focus of the mirror, what is the nature of the image formed by the mirror? [CBSE 2011]

Virtual, erect and magnified image formed behind the mirror.

### At what position the object be placed in front of a concave mirror to form the real image of same size? [CBSE 2011]

At the center of curvature (or at 2F).

Zero.

A convex mirror.

### If the image formed by the spherical mirror for all position of the object placed in front of it is always erect and diminished, what type of mirror is it? [CBSE 2015]

The mirror is convex mirror.

### Name the mirror that is used by a dentist in examining teeth? [CBSE 2012, 2015]

Dentists uses a concave mirror to see large image of the teeth of patients.

### What do you mean by the principal focus of a concave mirror? [CBSE 2012]

Principal focus of a concave mirror is a point situated on its principal axis where light rays coming parallel to the principal axis of mirror converges after reflection.

### What is the SI unit of refractive index? [CBSE 2012]

Refractive index is a unit less quantity.

### State the mirror formula of a spherical mirror. [CBSE 2011, 2014]

As per mirror formula 1/v + 1/u = 1/f, Where u = distance of object from the pole, v = distance of image from the pole and f= focal length of given mirror.

###### Center of Curvature of a Spherical Mirror

Center of curvature of a spherical mirror is the center of the sphere the given mirror is the part of which.

We are gradually adding new question answers and improving the contents as per the suggestions received from the uses. All the answers are confined to solutions of Class 10 Science textbooks issued by NCERT.

### Define the principal focus of a concave mirror. [CBSE 2012, 2015]

Principal focus of a concave mirror is the point on its principal axis, where light rays coming parallel to the principal axis actually converge after reflection from the mirror.

### The radius of the curvature of a spherical mirror is 20 cm. what is its focal length? [CBSE 2012]

Give that radius curvature of the mirror R=20cm
Focal length f=R/2 =(20 cm )/2= 10cm.

### Name the mirror that can give an erect and enlarged image of an object? [CBSE 2012, 2015]

Only a concave mirror can give an erect and enlarged image of an object.

### Why do we prefer a convex mirror as a rear view mirror in vehicle? [CBSE 2007, 2012, 2014]

We prefer a convex mirror as a rear view mirror in vehicle because a convex mirror gives an erect and diminished image. As a result, convex mirror helps the driver to have a much wider field of view.

### A ray of light travelling in air enters obliquely into water. Does the light ray bend toward the normal or away from the normal? Why? [CBSE 2006]

The light bend toward the normal on entry into water. It is due to the fact that as compared to air, the water is an optically denser medium.

###### Pole of a Spherical Mirror

The center of the reflecting surface of spherical mirror is called its pole (P).

### 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 3×10⁸ m/s. [CBSE 2005, 2006, 2008, 2011, 2015, 2017]

Given speed of light in vacuum c=3×10⁸ m/s and
The refractive index of glass ng = 1.50
From the relation n = c/v, we have
Speed of light in glass vg = c/ng = 3 ×10⁸/1.50 = 2 × 10⁸ m/s

### The refractive index of diamond is 2.42. What is the meaning of this statement? [CBSE 2012, 2011, 2014, 2016]

When we say that refractive index of diamond is 2.42 it means that the speed of light in diamond
= (Speed of light in vacuum)/2.42
= (3.6 × 10⁸ m/s)/2.42
= 1.24 × 10⁸ m/s

### Define one diopter of power of lens? [CBSE 2011, 2013]

1 diopter of power of lens having focal length of 1 m.

### A convex lens forms a real and inverted image of a needle at a distance of 50cm 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 lens? [CBSE 2012, 2013, 2014]

As real and inverted image of a needle is formed by convex lens at a distance of 50cm from it hence v= +50 cm. As size of image = size of object hence,
h1 = h. But ((h₁))/((h)) = v/u
u= (vh)/h₁ = (50×h)/h = 50 cm
so, the needle is placed 50 cm in front of convex lens as per sign convention u=50 cm.
now,
1/v – 1/u = 1/f
1/f= 1/((50)) – 1/((-50))
= 1/50 + 1/50 = 1/25 cm

Power of lens
P = 1/f = 1/((-2m)) = 0.5D

### What type of lens would you prefer to use while reading small letter found in a dictionary? [CBSE 2011, 2013]

A convex lens of 5 cm is preferred for reading small letter found in a dictionary.

### 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 3×10⁸ m/s. [CBSE 2005, 2006, 2008, 2011, 2015, 2017]

Given speed of light in vacuum c = 3 × 10⁸ m/s and
The refractive index of glass ng = 1.50
From the relation n=c/v, we have
Speed of light in glass vg = c/ng = 3 ×10⁸/1.50 = 2 × 10⁸ m/s

### The refractive index of diamond is 2.42. What is the meaning of this statement? [CBSE 2012, 2011, 2014, 2016]

When we say that refractive index of diamond is 2.42 it means that the speed of light in diamond
= (Speed of light in vacuum)/2.42
= (3.6 × 10^(8) m/s)/2.42
= 1.24 × 108 m/s

### Define one diopter of power of lens? [CBSE 2011, 2013]

1 diopter of power of lens having focal length of 1 m.

### A convex lens forms a real and inverted image of a needle at a distance of 50cm 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 lens? [CBSE 2012, 2013, 2014]

As real and inverted image of a needle is formed by convex lens at a distance of 50cm from it hence v= +50 cm. As size of image = size of object hence,
h1 = h. But ((h1))/((h)) = v/u
u= (vh)/h1 = (50×h)/h = 50 cm
so, the needle is placed 50 cm in front of convex lens as per sign convention u=50 cm.
now,
1/v – 1/u = 1/f
1/f= 1/((50)) – 1/((-50))
= 1/50+1/50= 1/25 cm

Power of lens
P = 1/f= 1/((-2m)) = 0.5D

### What type of lens would you prefer to use while reading small letter found in a dictionary? [CBSE 2011, 2013]

A convex lens of 5 cm is preferred for reading small letter found in a dictionary.

#### Difference between Concave Mirror and Convex Mirror

Concave Mirror Convex Mirror
1. Its reflecting surface is covered inward and its center of curvature and principal focus are situated in front of it.1. Its reflecting surface is curved outward and its center of curvature and principal focus lie behind the mirror.
2. It can be formed real and virtual image of the object depending on the position of the object.2. It can be formed only virtual image of the object irrespective of its position.

### 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. [CBSE 2013, 2014]

Given that the focal length of concave mirror is 15 cm.
To form an erect image object must be placed in front of concave mirror between its pole P and principal focus F. Hence the object AB should be placed in front of given concave mirror at a distance less than 15cm. The image formed AB is virtual and erect. The size of the image is larger than the object.

### 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. [CBSE 2013, 2014]

(a) For headlight of a car we use a concave mirror and headlight is fitted at the principal focus of the mirror. Thus, we get a powerful, parallel beam of light after reflection from the concave mirror.

(b) A convex mirror is used as a side/rear view mirror of a vehicle and enable the driver to see traffic behind him/her and thus facilities safe driving. Convex mirror forms erect and diminished image of vehicles coming behind. Thus it provides a wider field of view to the driver. As a result, the driver can view much larger than would be possible with a plane mirror.

(c) Large size concave mirror are used to concentrate sunlight to produce heat in solar furnaces. Incident sunlight is a parallel beam of light coming from infinity. The concave mirror focuses the light at its principal focus, where temperature rises sharply.

### One half of a concave lenses is covered with a black paper will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observation. [CBSE 2011, 2012, 2013, 2015]

When one half of a convex lens is covered with a black paper this lens produces a complete image of the object. To prove it we perform an experiment as given below:
Take a convex lens and cover half part of it by using a black paper. Place it vertically in a stand. On one side of it place a burning candle on opposite side of the lens fix a white screen. Adjust the position of candle or screen till clear image of burning candle is formed on the screen. We observe that the image is a complete image of the object.
From the experimental observation we find that image formation does not depend upon the size of the lens. A smaller lens can also form complete image of an object placed in front of it. However, brightness of the image decreases when some part of lens is blocked it is because now lesser number of raise pass through the lens.

### The magnification produced by a plane mirror is +1. What does this mean? [CBSE 2012, 2015]

We know that magnification produced by a mirror is given by
m= ( (h₁ ))/( (h))= -v/u
As magnification produced by a plane mirror m=+1, hence we have
+1 =( (h₁ ))/( (h))= -v/u = h₁ = h and v = -u
Thus, the size of image is exactly equal to the size of the object placed in front of it. Moreover, the image is formed as much behind the mirror as the object in front of it.

### Find the focal length of a lens of power -2.0 D. What type of lens is this? [CBSE 2013]

Given that power of lens P = -2.0 D
As power
P= 1/f, Hence focal length of lens F = 1/p = 1/(-2.0D) = 0.5m.
The –ve sign of focal length means that the lens is a concave lens.

### 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? [CBSE 2012]

Given that power of the prescribed lens P = +1.5D and P = 1/f
Focal length of the lens F = 1/p = 1/(+1.5D) = +0.67m
As the power of the lens is +ve, the lens is a converging (concave) lens.

### (a) How can you identify the three type of mirrors without touching them? (b) What do you mean by focal length of a spherical mirror? [CBSE 2012, 2014]

(a) Stand in front of mirror near it and look to your image.
(i)If the image is erect and of same size and size of image does not change even if you move forward or backward, the mirror is a plane mirror.
(ii) If the image is erect and diminished one the mirror is a convex mirror.
(iii) If the image is erect and magnified and became inverted if one moves away from the mirror, the mirror is a concave mirror.
(b) Focal length of a spherical mirror is the distance between principal focus of mirror, thus focal length f = PF.

### Draw the ray diagram and also state the position, the relative size and the nature of image formed by a concave mirror when the object is placed at the center of curvature of mirror. [CBSE 2011]

Let an object AB be placed normally to the principal axis of a concave mirror at its center of curvature C. When we draw a ray diagram to locate the image formed we find that the image A1 B1 is formed at center of curvature C itself.
The image is real inverted image. Size of the image is equal to the size of the object it means the magnification of the image is -1.

### State two position in which a concave mirror produces a magnification of a given image of the object. List the two difference between the two images. [CBSE 2016]

Two positions in which a concave mirror produces a magnified image are:
(i) When an object is placed between focus point F and center of curvature C of the mirror. For this position the image is real, inverted and magnified, and the image is formed beyond center 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.

### State two position in which a concave mirror produces a magnification of a given image of the object. List the two difference between the two images. [CBSE 2016]

It is given that focal length of concave mirror 12cm.
(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 <12cm. (ii) The image is larger than the object. (iii) If object be placed at 24 cm in front of the mirror than it means that the object situated at the center of curvature(there for u=24 cm = 2F = 2×12 cm = R) C of the given mirror. Hence the real, inverted image of same size is formed at center of curvature C itself, 1 (v) = 24 cm.

###### Principal Axis

Principal axis of a spherical mirror is a straight line passing through its pole and center of its curvature.

###### Principal Focus

Principal focus a spherical of a spherical mirror is a point on its principal axis where light rays coming parallel to its principal axis after refection, actually converges to or appears to diverge from.

### To construct a ray diagram, we use to rays of light which are so chosen that it is easy to determine their direction after reflection from the mirror. Chose these two rays and state the path of this rays after reflection from a concave mirror. Use this two rays to find the nature and the position of the image of an object placed at a distance of 15cm from a concave mirror of focal length 10cm. [CBSE 2012, 2015]

The position of the image formed by a spherical mirror can be found by considering following two rays:
(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 center of curvature in a concave mirror after reflection retraces its path. The image formation is shown in the adjacent figure. Here for every five centimeter distance we have one cm in the ray diagram. Here AB is the object and A1B1 is the real and inverted image formed on the basis of above mentioned two rays.
Actual measurement shows that the image is formed 30cm in front the concave mirror that is, V= 30cm.

### If the image formed by a mirror for all position of the object placed in front of it is always erect and diminished. What type of mirror is it? Where and why do we generally use this type of mirror? [CBSE 2015, 2017]

Only a convex mirror always forms an erect and diminished image behind the mirror between its pole and focus point for all positions of the object placed in front of the mirror. 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 auto mobiles because it gives erect and diminished images of vehicle coming from behind. As a result, it helps the driver in having a much wider field of view.

### A student holding a mirror in his hand directed the reflecting surface of the mirror toward the sun. he then directed the reflected light on to a sheet of paper held close to 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. [CBSE 2019]

(a) The student should adjust the distance between the mirror and the paper so that solar rays are sharply focused on the paper.
(b) The mirror is a concave mirror.
(c) The student can find the image approximate focal length by measuring the distance between the paper and the mirror.

### State the laws of reflection 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]

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 sign of the angle of incidence (i) to the sign of angle of refraction (r) is a constant. It is known as Snail’s law. Thus, according to snail’s law.
(Sin i )/(Sin r ) = constant =n
Generally, the constant an is known as the absolute refractive index of given medium. Thus, absolute refractive index of a medium is defining as the ratio of sin of angle of incidence of a light ray in air (or vacuum) to the sin of angle of refraction of the ray in given medium. Absolute refractive index of a medium in a unit less quantity and its value is 1 or greater than 1. In terms of speed of light, the absolute refractive index of a medium is defining as:
Absolute refractive index of a medium (n)
= (Speed of light in vacuum (or air) c)/(Speed of light in given medium v)

### 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, 2019]

A ray diagram showing refraction through a rectangular glass slab. The emergent ray GH is exactly parallel to the incident ray EFNM. It means that angle Lr2=Li1
However, the emergent ray is laterally (sideways) 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.

### A convex lens forms real and inverted image of a needle at the distance of 50cm from it. Where it the needle placed in front of the convex lens if the convex lens if the image is equal to the size of the object? Also find the power of lens. [CBSE 2012, 2013, 2014]

A real image and inverted image of a needle is formed by convex lens at a distance of 50 cm from it hence v= +50 cm. As size of image = size of object, hence h’=h. But h’/h = v/u
u = (vh)/h’
= (50 × h)/h = 50 cm
So, the needle is placed 50 cm in front of convex lens and as per sign convention u=-50 cm.
Now 1/v – 1/u = 1/f
1/f = 1/(50 ) – 1/((-50)) = 1/50+1/50 = 1/25 cm
f = +25 cm = +0.25 m
Power of the lens P = 1/f(m) = 1/0.25m = +4 D

### Find the power of a concave lens of focal length 2m. [CBSE 2012, 2015]

Given that focal length of concave lens f = -2m
Power of concave lens P = 1/f = 1/((-2m)) = -0.5 D

### An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image. [CBSE 2011]

u = -27cm, f = -18 cm (As the mirror is concave) h = 7.0cm.
From mirror formula
1/v + 1/u = 1/f
= 1/v + 1/((-27)) = 1/((-18))
1/v = – 1/18 + 1/27= (-3+2)/54 = (-1)/54
or v= -54cm. The screen must be placed at a distance of 54cm from the mirror in front of it
m = – v/u = -(-54)/(-27) = -2. The image is real, inverted and enlarged.
Size of the image h’ =m.h = (-2) × 7 cm = -14 cm. Thus the image is of 14 cm length and is an inverted image i.e., formed below the principal axis.

### Find the focal length of a lens of power -2.0 D. What type of lens is this? [CBSE 2013]

Given that power of lens P = -2.0 D.
As power P = 1/f, hence focal length of lens f = 1/P = 1/(-2.0 D) = -0.5 m.
The –ve sign of focal length means that the lens is concave lens.

### 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? [CBSE 2012]

Given that power of the prescribed lens P =+1.5 D and P = 1/f.
Focal length of the lens f= 1/P = 1/(+1.5 D) = +0.67 m.
As the power of lens is +ve the lens is a converging (convex) lens.

###### Spherical Mirror

A mirror whose reflecting surface is a part of a sphere is called a spherical mirror. Spherical mirrors are of two type namely (i) Concave mirror (ii) Convex mirror.

### An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image. [CBSE 2017]

Here distance of object u = -15cm, height of object h = +4cm and the focal length of concave mirror f = -10cm.
As per mirror formula
1/v + 1/u = 1/f we have
1/v = 1/f – 1/u
1/v = 1/((-10)) – 1/((-15)) = -1/30
v = -30cm
Magnification m= h’/h =- v/u
h’ = – v/u × h = – ((-30))/((-15)) × 4 = – 8 cm
Thus image is an inverted image of height 8 cm.

### A convex mirror used for rear view on an automobile has radius of curvature of 3.00 m. If a bus is located at 5.0 m from the mirror find the position, nature and size of the image. [CBSE 2012, 2015]

Here R= +3.0 m and u = -5.0 m
1/v = 2/R – 1/u = 2/3.0 – 1/((-5.0) ) = 2/3 + 1/5 = (10+3)/15 = 13/15
v = 15/13 m = + 1.16 m
The image is virtual erect and diminished one and is formed behind the mirror.

### An object 2.0 cm high is placed 20.0 cm in front of a concave mirror of focal length 10.0 cm. Find the distance from the mirror at which a screen should be placed in order to obtain a sharp image. What will be the size and nature of the imaged formed? [CBSE 2005, 2011]

Here h= 2.0 cm, u = -20.0 cm and f = -10.0 cm.
1/v = 1/f – 1/(u ) = 1/((-10)) – 1/((-20)) = (-1)/10 + 1/20 = -1/20
v = -20cm
Again h’/h = – v/u
h’= – v/u h= – ((-20))/((-20))×2.0 = – 2.0 cm
Hence an image of length 2.0 cm is formed in front of mirror at a distance of 20 cm. The –ve sign of v and h’ show that the image is real and inverted image.

### Light enters from air into water which have refractive index of 1.33 [or 4/3] Calculate the speed of light in water. The speed of light in air is c=3×10⁸ m/s [CBSE 2012, 2013, 2014, 2015]

The speed of light in air is c=3×10⁸ m s⁻¹ and refractive index of water nw = 1.33
Speed of light in water nw = c/nw = (3.6 ×10⁸ m/s )/1.33 = 2.25 ×10⁸ m/s

### 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 image formed. [CBSE 2019]

As per question focal length of concave mirror f = -20 cm. height of the object h = +4 cm and distance of image v = 30cm
Following two cases may rise here:
Case I: If the image formed is real then v = -30cm and so far mirror formula 1/v + 1/u = 1/f
We have:
1/u = 1/f – 1/v = 1/((-20)) – 1/((-30)) = (-1)/20 + 1/30 = -1/60
u = -60cm
and h’ = – v/u× h = ((-30))/((-60))× (+4) = -2cm
so the object be placed 60cm in front of mirror and image is inverted image of size 2cm.

Case II: If the image formed is virtual then v = +30cm and now
1/u = 1/f – 1/v = 1/((-20)) – 1/((+30)) = (-1)/20 + 1/30 = -1/12 u = -12cm
and h’ = – v/u× h = ((+30))/((-12))× (+4) = +10cm
so the object is placed at 12cm in front of mirror and image is erect and image of height 10cm.

### List the sign convention for reflection of light by spherical mirrors. Draw a diagram and apply these conventions in the determination of focal length of a spherical mirror which forms a three time magnified real image of an object placed 16cm on front of it. [CBSE 2012]

As per question and using the sign convention followed u=-16 cm and magnification m =-3 (because image is real).
m = – v/u = -3 v = 3u = 3×(-16) = -48 cm
1/v + 1/u = 1/f
1/((-48)) + 1/((-16)) = (1+3)/((-48))
= -4/((-48)) = -1/12
f = -12 cm
The –ve sign of shows that mirror is a concave mirror.

### List the new Cartesian sign convention for reflecting of light by spherical mirrors. Draw a diagram and apply these convention for calculating and nature of a spherical mirror which forms a 1/3 times magnified virtual image of an object placed 18cm in front of it. [CBSE 2012]

As per question and using the sign convention followed u = -18 cm and magnification m=+ 1/3
m = – v/u = + 1/3
v = – u/3 = – ((-18))/3 = +6cm
1/f = 1/v + 1/u = 1/((+6)) + 1/((-18))
= (3-1)/18 = 2/18 = 1/9
f = +9 cm
The +ve sign of significance that the mirror is a convex mirror.

### A student wants to project the image of candle flame on the wall of school laboratory by using a lens: (a) Which type of lens should be used and Why? (b) At which distance in term of focal length f of the lens should he placed the candle flame so as to get (i) a magnified, and (ii)a diminished image respectively on the wall? [CBSE 2014]

(a) the student should use a convex lens because only a convex lens can form real image of the candle flame which can be obtain on the wall of school laboratory.
(b) (i) The candle should be placed at a distance d, where f < d < 2f to obtain a magnified image on the wall. (ii) The candle should be at distance d > 2f to obtain a diminished image of the candle on the wall.

### (a) A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it Stating the position of the object with respect to the lens in each case. (b) An object of height 4cm is placed at distance of 20cm from a concave lens of focal length 10cm. use lens formula to determine the position the image formed. [CBSE 2015]

(a) A convex lens can form a magnified real and inverted image of an object placed in front of it when object is situated somewhere between F1 and 2F2. On the other hand, a convex lens can form a magnified and erect image of an object when the object is situated between optical center o and focus f1.
(b) As per question u= -20 cm, h = 4 cm and f = – 10 cm
Lens formula is 1/f = 1/v -1/u
1/v = 1/u + 1/f = 1/((-20))+1/((-10)) =(-1-2)/20 = (-3)/20
v = -20/3 cm = – 6.67 cm

### (a) List four characteristics of the image formed by the convex lens when an object is placed between its optical center and principal focus. (b) size of the image of an object by a concave lens of focal length 20 cm observed to be reduce to 1/3 rd of its size. Find the distance of the object from the lens. [CBSE 2019]

(a) When an object is placed between the optical center 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 the 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 foe a lens we get
1/3 = v/u v = u/3
Therefore, as per sign convention followed both u and v are –ve
Using lens formula 1/v – 1/u = 1/f we have
1/(-u/3)- 1/((-u)) = 1/((-20))
(-3)/u + 1/u = – 1/20 =- 2/u = (-1)/20 u = 40cm
So the object is placed at a distance of 40cm from the lens.

### List the sign convention for reflection of light by spherical mirrors. Draw a diagram and apply these conventions in the determination of focal length of a spherical mirror which forms a three time magnified real image of an object placed 16cm on front of it. [CBSE 2012]

As per question and using the sign convention followed u=-16 cm and magnification m =-3 (because image is real).
m = – v/u = -3 v = 3u = 3×(-16) = -48 cm
1/v + 1/u = 1/f
1/((-48)) + 1/((-16)) = (1+3)/((-48))
= -4/((-48)) = -1/12
f = -12 cm
The –ve sign of shows that mirror is a concave mirror.

### List the new Cartesian sign convention for reflecting of light by spherical mirrors. Draw a diagram and apply these convention for calculating and nature of a spherical mirror which forms a 1/3 times magnified virtual image of an object placed 18cm in front of it. [CBSE 2012]

As per question and using the sign convention followed u=-18 cm and magnification m=+ 1/3
m = – v/u = + 1/3
v = – u/3 = – ((-18))/3 = +6cm
1/f = 1/v + 1/u = 1/((+6)) + 1/((-18))
= (3-1)/18 = 2/18 = 1/9
f = +9 cm
The +ve sign of significance that the mirror is a convex mirror.

### A student wants to project the image of candle flame on the wall of school laboratory by using a lens: (a) Which type of lens should be used and Why? (b) At which distance in term of focal length f of the lens should he placed the candle flame so as to get (i) a magnified, and (ii)a diminished image respectively on the wall? [CBSE 2014]

(a) the student should use a convex lens because only a convex lens can form real image of the candle flame which can be obtain on the wall of school laboratory.
(b) (i) The candle should be placed at a distance d, where f < d<2f to obtain a magnified image on the wall. (ii) The candle should be at distance d>2f to obtain a diminished image of the candle on the wall.

### (a) A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it Stating the position of the object with respect to the lens in each case. (b) An object of height 4cm is placed at distance of 20cm from a concave lens of focal length 10cm. use lens formula to determine the position the image formed. [CBSE 2015]

(a) A convex lens can form a magnified real and inverted image of an object placed in front of it when object is situated somewhere between F1 and 2F2. On the other hand, a convex lens can form a magnified and erect image of an object when the object is situated between optical center o and focus f1.
(b) As per question u= -20 cm, h = 4 cm and f = – 10 cm
Lens formula is 1/f = 1/v -1/u
1/v = 1/u + 1/f = 1/((-20))+1/((-10)) =(-1-2)/20 = (-3)/20
v = -20/3 cm = – 6.67 cm

### (a) List four characteristics of the image formed by the convex lens when an object is placed between its optical center and principal focus. (b) size of the image of an object by a concave lens of focal length 20 cm observed to be reduce to 1/3 rd of its size. Find the distance of the object from the lens. [CBSE 2019]

(a) When an object is placed between the optical center 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 the 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 foe a lens we get
1/3 =v/u v=u/3
Therefore, as per sign convention followed both u and v are –ve
Using lens formula 1/v – 1/u = 1/f we have
1/(-u/3)- 1/((-u)) = 1/((-20))
(-3)/u + 1/u =- 1/20 =- 2/u = (-1)/20 u=40cm
So the object is placed at a distance of 40cm from the lens.

###### Power of a Converging Lens

The power (P) of a lens is a measure of the degree of convergence of light rays achieved by it. Mathematically, power of a lens is defined as the reciprocal of its focal length (f).
Thus, P = 1/f.

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Last Edited: April 26, 2023