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Question-1

State the basic laws which determine the reflection of light by mirrors.

Solution:
The reflection of light from mirrors takes place according to the following laws of reflection of light:   

(a) the angle of incidence (i) is equal to the angle of refection (r), and

(b) the incident ray, the normal to the mirror at the point of incidence, and the reflected ray, all lie in the same plane.

Question-2

State Snell's law. Give its uses.

Solution:
Snell's law states that the ratio of sine of angle of incidence (i) to the sine of angle of refraction (r) is a constant. This constant is called the refractive index of the material of the prism.

Thus, angle of incidence (i) and the angle of refraction (r) are related as
(sin i /sin r) = n21
where n21 is a constant, and is called the refractive index of the second medium with respect to the first medium.

Uses  of
Snell's law.
In the following two examples, use Snell's law, the sine button on your calculator, a protractor, and the index of refraction values to complete the following diagrams. Measure , calculate , and draw in the refracted ray with the calculated angle of refraction.
In each of these two example problems, the angle of refraction is the variable to be determined. The indices of refraction (ni and nr) are given and the angle of incidence can be measured. With three of the four variables known, substitution into Snell's law followed by algebraic manipulation will lead to the answer.

Question-3

Can the laws for plane refracting surfaces be applied to spherical refracting surfaces also? State the laws of refraction of light.

Solution:
Yes, the laws that are valid for plane refracting surfaces can also be applied for spherical refracting surfaces. The laws of refraction are as follows:

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

(ii) The ratio of sine of angle of incidence 'i' to the sine of angle of refraction 'r' is a constant. This constant is called the refractive index of the material of the prism. Thus, angle of incidence 'i' and the angle of refraction 'r' are related as       

(sin i /sin r) = n21

where n21 is a constant, and is called the refractive index of the second medium with respect to the first medium. 

Question-4

What does the positive sign associated with the virtual image and the negative sign associated with the real image denote?

Solution:
The positive sign associated with the virtual image indicates the presence of an erect image (above the principal axis of the lens) and the negative sign associated with the real image indicates the presence of an inverted image (below the principal axis of the lens).

Question-5

State the condition at which total internal reflection occurs.

Solution:
Total internal reflection occurs if the angle of incidence 'i' exceeds the critical angle 'ic'.

Question-6

Define the term "star light flux".

Solution:
Star light flux is the luminous energy emitted per second by the star. 

Question-7

What impact does atmospheric refraction have on sunrise and sunset?

Solution:
The Sun is visible to us about two minutes before the actual sunrise, and about two minutes after the actual sunset. This is because of atmospheric refraction, which is due to the apparent flattening of the Sun's disc at sunrise and sunset.

Question-8

What happens when a ray of light falls normally on the surface of a mirror?

Solution:
If a ray of light is incident normally on a mirror, the angle of incident is zero. Hence the angle of reflection is also zero. This means that a ray of light, which is incident normally on a mirror, is reflected back along the same path.

Question-9

What is lateral inversion? Explain with a suitable example.

Solution:
If an object is placed in front of a plane mirror, then the right side of the object appears to be on the left side of the image and the left side of the object appears to be on the right side of its image. This change of sides of an object and its mirror image is called lateral inversion.

For example when we sit in front of a plane mirror and write with our right hand, it appears in the mirror that we are writing with the left hand.

Question-10

Define principal focus of concave and convex mirror.

Solution:
Principal focus of a concave mirror
(i) The principal focus of a concave mirror is a point on its principal axis to which all the light rays, which are parallel and close to the axis, converge after reflection from the concave mirror.

(ii) A concave mirror has real focus.

Principal focus of a convex mirror
(i) The principal focus of a convex mirror is a point on its principal axis from which
a beam of light rays, initially parallel to the axis, appear to diverge after being
reflected from the convex mirror.

(ii) A convex mirror has virtual focus.

Question-11

What is the nature, size and position of the image formed of an object at infinity? Explain how it is used as a doctor’s mirror.

Solution:
When an object is placed at infinity from a concave mirror, the image formed is

(i) real and inverted

(ii) highly diminished and

(iii) at the focus

When the object is at infinity, the image is formed at the focus of the concave mirror, hence the mirror can concentrate all the parallel rays of the light to its focus. This explains the use of concave mirror as a doctor’s head-mirror. A concave mirror is used as a head-mirror by the doctors to concentrate light on the body parts to be examined.

Question-12

How will you distinguish between a plane mirror, a concave mirror and a convex mirror without touching them?

Solution:
We can distinguish these mirrors by identifying the image of our face produced by them when we just look into those mirrors.

(i) A plane mirror will produce an image of the same size as our face.

(ii) A concave mirror will produce a magnified image.

(iii) A convex mirror will produce a diminished image.

Question-13

Explain the path of the light traveling in air incident on a rectangular glass block and emerging out into the air from the opposite face. Draw a ray diagram to show the complete path of this ray of light.

Solution:
A ray of light AO traveling in air is incident on a rectangular glass block at point O. On entering the glass block, it gets refracted and bends towards the normal ON1. A second change of direction takes place when the ray of light OO¢ traveling in the glass block emerges into air at point O . Since the ray of light OO¢ now goes from a denser medium ‘glass’ into the rarer medium ‘air’, it bends away from the normal ON 2 and takes the direction OB, which is called emergent ray. The incident ray AO and the emergent ray OB are parallel to each other.

Question-14

An empty test tube is placed slanting in water and viewed from above, what will you observe? What difference will it make when the tube is partially filled with water?

Solution:
When an empty test tube is placed slanting in water, the portion of tube within water will appear to be silvery due to total internal reflection. If the tube is partially filled with water, only the portion CE (Fig. 2) will appear silvery. The lower portion AD (Fig 2) will not be silvery since total internal reflection takes place only if there is an optically rarer medium also.

Question-15

Why do stars twinkle on a clear night?

Solution:
When the light coming from a star enters the earth’s atmosphere, it undergoes refraction due to the varying optical densities of air at various altitudes. The atmosphere is continuously changing. The continuous changing atmosphere refracts the light from the stars by different amounts from one moment to the next. When the atmosphere refracts more starlight towards us, the star appears to be bright and when the atmosphere refracts less starlight, then the star appears to be dim. In this way, the starlight reaching our eyes increases and decreases continuously due to atmospheric refraction and the star appears to twinkle at night.

Question-16

A beam of red light is allowed to fall in turn on

(a) a black cloth

(b) red cloth

(c) white cloth

(d) blue cloth. Describe and explain the appearance in each case.

Solution:
(a) When red light falls on a black cloth it still appears black, because it does not reflect the red colour instead, it absorbs the light falling on it.

(b) When red light falls on a red cloth it still appears red, because it reflects only that colour.

(c) When red light falls on a white cloth it changes to a red coloured cloth, because the white cloth reflects the red colour of the light.

(d) When red light falls on a blue cloth it changes to black coloured cloth, because there is no blue colour in the red light which can be reflected by the cloth and since the blue cloth absorbs the red light falling on it, it appears black.

Question-17

An object is placed between the pole and focus of a concave mirror. What is the nature and size of image?

Solution:
The image will be virtual and highly magnified.

Question-18

State the condition at which total internal reflection occurs.

Solution:
Total internal reflection occurs if the angle of incidence 'i' exceeds the critical angle.

Question-19

A stamp collector uses a lens (convex) of focal length 7 cm, to examine a stamp. What is the power of the lens?

Solution:
Power = 1/f = 1/7 = 0.1428 cm.

Question-20

What is critical angle? Draw a ray diagram showing light incident at the critical angle.

Solution:
When a ray of light passes from denser to rarer medium, the angle for which the refracted ray traces the horizontal path of the denser medium is called the critical angle.

Question-21

Draw the colour circle showing primary and composite colours.

Solution:
Red, blue and green are the primary colours. The other colours which are obtained by using these primary colours are called secondary colours.


Question-22

What is meant by dispersion of light?

Solution:
The splitting of white light into different constituent colours is called dispersion of light.

Question-23

How are focal length and radius of curvature connected?

Solution:
The radius of curvature is twice the focal length of the lens.

Question-24

An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagrams and find the position, size and the nature of the image formed.

Solution:
Object distance u = -25 cm
Image distance v = ?
Focal length f = 10 cm

Using the lens formula


     =
 
v = 16.67 cm

Positive sign shows that the image formed is to the right of the lens and is at 16.67 cm from the lens, i.e., the image is formed between F and 2f. Since only the real and inverted image is formed on the right hand side of a convex lens, therefore, the image formed is real and inverted.

Magnification, m = = -0.67.

Image formed is two times magnified and is inverted.
Again m =

where h1 and h2 are the sizes of objects and images

i.e., -0.67 = (h1 is always positive)
h2 = - 3.35 cm
Size of image is 3.35 cms, hence smaller than the object and is always inverted.

Question-25

A concave lens of focal length 15 cm, forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Solution:
Focal length of the lens f = -15 cm (since concave lens are negative)
Since the image is formed on the same side, v = -10 cm.
u = ?

Using lens formula


=
Therefore u = -30 cm

Question-26

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Solution:
Object distance, u = -10 cm
Image distance, v = ?
Focal length, f = -15cm

Using the mirror formula



+

= -2 +3 /30 = 1/30

V= 30 cm

The image is at a distance of 6 cm behind the mirror. It is virtual, erect and magnified.

Question-27

What is the angle of incidence when the ray reflected back, when it passes from denser to rarer medium.

Solution:
The angle of incidence is greater than the critical angle, and then the ray, which passes from denser to rarer medium is reflected back.

Question-28

Draw ray diagrams to show image formation.

Object at C – concave mirror

Object beyond 2f – convex lens

Object at infinity – concave lens
. Write the nature of the image in each case.

Solution:
Object at C – concave mirror

Object within 2f and beyond 2f – convex lens

Object at infinity – concave lens

Question-29

Light, of wavelength of 500 nm in air, enters a glass plate of refractive index 1.5. 
Find   

(i) speed    

(ii) frequency and     

(iii) wavelength of light in glass. 

Assume that the frequency of light remains the same in both media.?

Solution:
Wavelength λ = 500 nm = 500 × 10-9 m
Refractive index of the glass n = 1.5

(i) We know that n =
     vg = = 2 × 108 m/s

(ii) We know that c = ν λ
     or ν =
           = 6 × 1014 Hz

(iii) If λ g is the wavelength of light in glass, then
      vg = ν λ g
     λ g =
         = 3.333 × 10-7 m = 333.3 nm.

Question-30

Why do we use blue colour solution after washing the clothes?

Solution:
After washing clothes we use blue color for brightening the clothes As the white clothes are deficient in ultra-violet wavelengths, this can be rectified by using violet or blue colors solutions.

Question-31

How is critical angle related to the refractive index of the medium?

Solution:
If the refractive index of denser medium with respect to rarer medium is known, the critical angle for the given pair of media can be calculated.

μ = 1/sin ic.

Question-32

The refractive index of dense flint glass is 1.65, and for alcohol it is 1.36, with respect to air. What is the refractive index of dense flint glass with respect to alcohol?

Solution:
Refractive index of dense flint glass w.r.t. air, n1 = 1.65
Refractive index of alcohol w.r.t. air, n2 = 1.36
Refractive index of dense flint glass with respect to alcohol is

= = 1.21.

The refractive index of dense flint glass with respect to alcohol = 1.21.

Question-33

A concave mirror produces 3 times magnified real image of an object placed at 10 cm in front of it. Where is the image located?

Solution:
The image is real and highly magnified and it is obtained at a distance which is beyond 2f, where f is the focal length of the mirror.

Question-34

With the help of labelled ray diagram only show the formation of an image in a compound microscope.

Solution:

Question-35

Two thin lenses of power +3.5 D and –2.5 D are placed in contact. Find the power and focal length of the lens combination.

Solution:
Two thin lenses of powers,

P1 = +3.5 D

P2 = -2.5 D

Power of combination, P = P1 + P2 = 3.5 + (- 2.5) = 1 D

Power of combination, P = 1 D.
Thus the combination acts as a convex lens of power 1D (positive sign)

Focal length of combination, f = = 1 m.

Question-36

How can you distinguish between a real image and a virtual image?

Solution:
The image which can be observed through naked eye is called real image and the image which cannot observed through naked eye is called virtual image.

Question-37

State Snell’s law

Solution:
Snell's Law relates the indices of refraction of the two media to the directions of propagation in terms of the angles to the normal.

Question-38

How image can be located which is formed by a plane mirror?

Solution:
The image obtained by using a plane mirror is virtual and a distance which is equal to the object distance.

Question-39

What is the relation between Speed of light in a medium and refractive index. 

Solution:
Speed of the light is inversely proportional to the refractive index of the medium.

Question-40

What happens to light when it enters a rarer medium from a denser medium?

Solution:
When a ray of light enters into the rarer medium from a denser medium, the velocity of the light increases to the maximum.

Question-41

Define 1 dioptre.

Solution:
One dioptre is equal to the reciprocal of the focal length measured in metres (that is, 1/metres).

Question-42

Why are reflecting prisms used in periscopes?

Solution:
In periscopes, reflecting prisms are used in which the incoming light is reflected perpendicularly and this reflected light is again reflected by another prism to get the image.

Question-43

Do frequency and wavelength change when light travels from one medium to another?

Solution:
Yes, the frequency and wavelength changes when a ray of light travels from one medium to another medium, because of the change in velocity.

Question-44

How a virtual image bigger than the object can be produced?

Solution:
When an object is placed at the focus of any lens, then the image obtained is bigger than the object and it is virtual.

Question-45

Define refractive index in terms of velocity and mass density.

Solution:
Refractive index of any medium is the amount of density of the medium which resists the incident light and the ratio of change in velocity when it passes from rarer to denser medium.

Question-46

Give one application for convex mirror and convex lens.

Solution:
(i) It is used as a shaving mirror.

(ii) It is used in camera.

Question-47

Define power of a lens. For what type of lens is it positive?

Solution:
Power of a lens is the inverse of its focal length.

Question-48

Distinguish between supplementary and complementary colours. What is composite colour?

Solution:
The output of the primary colours are called secondary or supplementary colours and the inverse of these colours are called complimentary colours. The mixing of different colours to get a new colour is called composite colour.

Question-49

What is centre of curvature?

Solution:
A lens has two centers of curvature. These are the centers of the two spherical surfaces which combine to form a lens.

Question-50

Define focus of a convex lens. Draw a ray diagram showing the focus.

Solution:
The distance between the convex lens and the screen where the clear image is obtained, when the object is placed at infinity is called focus and the length is called focal length.

Question-51

What is a simple microscope?

Solution:
Simple microscope is an instrument which is used to magnify small objects. It is used by a convex lens for which the object placed within the focus, then the image obtained is highly magnified.

Question-52

Light of wavelength 500nm in air enters a glass plate of refractive index 1.5. find its wavelength in glass.

Solution:
Refractive index = λ a/λ m = 1.5

Wavelength in glass medium (λ m) = 500/1.5 = 333.33 nm.

Question-53

What is the consequence of making angle of incidence of light, at an interface, greater than the critical angle?

Solution:
When a ray of light passes from denser to rarer medium, the ray of light reflected back into the denser medium provided that the angle of incidence of light, which passes from denser medium to rarer medium has an angle which is greater than the critical angle.

Question-54

With the help of a labeled ray diagram show the formation of image of an object by a compound microscope.

Solution:

 

This shows how the enlarged image is formed in compound microscope.

Question-55

What is meant by refraction?

Solution:
It is a phenomenon in which the path of a ray of light changes when it moves from one medium to another.

Question-56

Express the refractive index of the medium.

Solution:
It is the ratio of velocity of light in vacuum to the velocity of light in medium.

Question-57

If a ray of light passes from medium I to medium II without any change of direction, what can be said about the refractive indices of these media.

Solution:
The two media have the same refractive index.

Question-58

Mention two properties of a wave: one property which varies and the other which remains constant when the wave passes from one medium to another.

Solution:
Wavelength and frequency.

Question-59

Name one factor on which the direction of bending of a ray of light depends.

Solution:
It depends on the different speed of light in the two media through which a ray of light travels.

Question-60

A ray of light while traveling from a denser to a rarer medium, is incident at critical angle on the interface. What is the angle of refraction in the rarer medium?

Solution:
90 degree.

Question-61

A glass slab is placed over a page in which letters are printed in different colours. Will the image of all the letters lie in the same plane?

Solution:
Since the refractive index of glass is different for different colours, the images of different colours will not be raised up equally.

Question-62

Does the apparent depth of a tank of water change, normally.

Solution:
No, when viewed normally, Snell’s law fails and hence no refraction takes place. Therefore the depth remains same.

Question-63

For which colour of white light is the refractive index of a transparent medium is maximum and minimum.

Solution:
It is maximum for violet light and minimum for red light.

Question-64

What is total reflecting prism?

Solution:
A right angled isosceles prism is called a total reflecting prism. It is used in a periscope and binocular.

Question-65

State three actions that a total reflecting prism can produce.

Solution:
It can deviate light through an angle of 180 degree. It can deviate light through an angle of 90 degree. It can invert the rays of light.

Question-66

Mention the types of lenses.

Solution:
There are two types of lenses

(i) convex lens

(ii) concave lens.

Question-67

What is a convex lens?

Solution:
A convex lens is thinner at the edges and thicker in the middle.

Question-68

Mention the types convex lens.

Solution:
Double convex lens, plano-convex lens, convex meniscus.

Question-69

What is radius of curvature?

Solution:
It is the radius of the spherical surfaces which form the lens.

Question-70

What is optical centre?

Solution:
The central point in the lens is called the optical centre. If a ray is incident towards the optical centre, it passes undeviated through the lens.

Question-71

What is pupil?

Solution:
The small hole in the iris is the pupil.

Question-72

What is aqueous houmour?

Solution:
A clear liquid region between the cornea and the lens.

Question-73

What is vitreous humour?

Solution:
The space between eye lens and retina is filled with another liquid called vitreous humour.

Question-74

What is far point?

Solution:
The farthest point up to, which the eye can see clearly.

Question-75

What is cornea?

Solution:
The transparent spherical membrane covering the front of the eye.

Question-76

What is retina?

Solution:
The back surface of the eye is called retina.

Question-77

What is blind spot?

Solution:
The point at which the optic nerve leaves the eye. An image formed at this point is not sent to the brain

Question-78

What is eye lens?

Solution:
It is a transparent lens made of jelly like material.

Question-79

What is iris?

Solution:
The coloured diaphragm between the cornea and lens.

Question-80

Define magnification power of a simple microscope.

Solution:
Magnification power of a simple microscope is given by
m = 1+D/f

Question-81

How to increase the magnification and sharpness of the image in optical instruments?

Solution:
Many optical instruments consist of a number of lenses. They are combined to increase the magnification and sharpness of the image. The net power (P) of the lenses placed in contact is given by the algebraic sum of the individual powers P1, P2, P3,... as P = P1+P2+P3.

Question-82

Optical density. Explain

Solution:
The ability of a medium to refract light is also expressed in terms of optical density. Optical density has a definite connotation. It is not the same as mass density. We have been using the terms ‘rarer medium’ and ‘denser medium’. It actually means ‘optically rarer medium’ and ‘optically denser medium’, respectively. The one with the larger refractive index is optically denser medium than the other. The speed of the light is higher in a rarer medium than in a denser medium.

Question-83

Explain diffraction of light.

Solution:
The bending of light at the edges of the obstacle (opaque objects).

Question-84

Give some of the uses of convex mirrors.

Solution:
Convex mirrors are commonly used as rear-view (wing) mirrors in vehicles. These mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic behind him/her to facilitate safe driving. Convex mirrors are preferred because they always give an erect, though diminished, image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror.

Question-85

How far will a light beam travel? If there was an extremely strong light beam that could be seen for miles, where would the light stop? If a flashlight has low batteries, how far will its light beam travel?

Solution:
Each little bit of light travels in a straight line, to the end of the universe. That's why we can see the stars. But how far away can you still see a dim flashlight? Now there are three issues:

*No light beam is perfectly collimated -- that is, to some extent it contains light that is traveling in different directions. Then as we get farther and farther away, the beam is spread out over a larger area, and is correspondingly dimmer. The stars are as bright as the sun when we are close to them, but very distant stars are hard to see. A laser makes a pretty good beam, but the best laser beam we know is a mile wide by the time it arrives at the moon. A man on the moon would not be able to see it -- it would be too dim.
 
*If we try to do the experiment on earth, we have to deal with the atmosphere, which is not completely transparent. A little bit of haze (small droplets of water or something), dust, bugs, or even temperature variations in the air can affect the light beam. If the light is being scattered into other directions, it is ruining the collimation. If the light is being absorbed, it's just gone.
 
*Finally, when we say we can't see the light, what we usually mean is that we see something else. We can't see through fog because it is bright -- it is scattering light into our eyes which blots out the light that is trying to go directly through the fog. The result is that on a foggy day, a distant building can barely be seen -- there is only a slightly different shade of gray against the sky behind it. The gray is the light the fog is sending; the slight extra brightness of the sky is the light that has made it through the fog to show us the scene beyond. This is also why most people can't see the stars at night: they are hidden behind the glow in the sky caused by the lights of the city they live in.

Putting all this together, the brightness of the light source and how well it is collimated are important, but the real answer to this question lies in factors outside the light source. The light beam doesn't stop -- it just gets harder and harder to see, and eventually disappears into the background. This is also true for flashlight having low batteries.

Question-86

Give the applications of plane mirror.

Solution:
Plane mirrors are

(i) Used in periscope.

(ii)Used in galvanometer scale.

(iii) Used for removing parallax in compass boxes.

Question-87

Give the expression for refractive index in terms of speed of light.

Solution:
In terms of speed of light, refractive index is defined as the ratio of that speed of light in vacuum to the speed of light in medium. n = c/v, where ‘n’ is the refractive index of the denser medium with respect to air, ‘c’ is the velocity of light in air and ‘v’ is the velocity of light in denser medium.




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