Class 10 Science Chapter 9 Board Questions

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

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

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

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

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

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

Virtual, erect and magnified image formed behind the mirror.

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

Zero.

A convex mirror.

The mirror is convex mirror.

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

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.

Refractive index is a unit less quantity.

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.

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.

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

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

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.

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.

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

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

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

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

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

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

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

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

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

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

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.

(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.

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.

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.

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.

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) 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.

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.

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.

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.

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.

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) 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.

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)

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 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

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

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.

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.

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.

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.

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.

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.

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

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.

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.

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) 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 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) 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.

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.

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) 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 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) 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.