CHAPTER-11 –Light
1-Marks questions
Q.1. What do you mean by lateral inversion in reflection of light?
Ans: The virtual image formed by a mirror produces an alternation of left and right parts of an
object in the image. Such a left-right alternation in the image formation is called lateral inversion.
Q.2. You see your image through a plane mirror at a distance of 1 m form you. What is the distance
of the mirror from you.
Ans: Distance of the mirror=½ x distance of the image.
=½ x1 =½ m.
Q.3. You are looking into a mirror which is at a distance of 50 cm. what is the distance of your
image from you?
Ans: Image distance =2 x object distance
= 2 x distance of mirror
= 2 x 50 cm
=100 cm = 1 m.
Q.4. What is the reason behind for not seeing an image through a brick wall through it reflects light?
Ans: Because, the reflection produced by a wall is not a uniform one.
Q.5. How do we see an object which does not emit light?
Ans: We see an object which does not emit light, by seeing the reflected light from the object, (i.e.
we can not see such object in absence of light).
Q.6. What are spherical mirror?
Ans: Spherical mirror are those mirror having spherical surface i.e. their surface are parts of a
sphere. (there are two types vic. Convex and concave spherical mirror).
Q.7. Name the types of spherical mirrors.
Ans: i) Concave ii) Convex.
Q.8. Define pole of a spherical mirror.
Ans: The centre of a spherical mirror is called its pole [read other _______ also]
Q.9. What is principal section of a spherical mirror?
Ans: It is the principal section of the mirror obtained with a plane pass through the principal axis.
Q.10. How a virtual image differs from a real image?
Ans: A virtual image is formed due to virtual intersection of light rays, hence it cannot be collected
on a screen. But, real image is formed to actual intersection of light rays and can be collected on a
screen.
Q.11. Write the expression of magnification produced by a spherical mirror.
image height,hi
image distance,v
Ans: Magnification, M =object height,he=-object distance ,u
Where new sign convention should be applied.
v
hi
[in case of magnification produced by a lens, M=u = ,he]
Q.12. what is the magnification produce by a concave mirror when the object is placed at centre of
curvature.
Ans: Since, the object is at centre of curvature, the image size equal object since i.e. hi=ho.
.
. .
hi
magnification, M=ho = 1.
Q.13. If the image of a distant object is collected on a screen at a distance of 30 cm in front of a
spherical mirror. What is its focal length?
Ans: Since, the image of a distant object is form near the focus of the concave mirror, the focal
length must be nearly equal to the image distance.
.
. . focal length=30 cm.
Q.14. What is refractive index of a medium?
Ans: The refractive index of a medium is the ratio of speed of light in vacuum (or air) to the speed of
light in that medium.
c
i.e. refractive index, u=v= when c is speed of light in air or vacuum and v is speed light in the
medium.
Q.15. Define refraction of light.
Ans: Refraction of light is a phenomenon of sudden change in the path of a ray of light while
crossing one medium to another.
Q.16. Write an expression for refractive index of a medium.
Ans: [See Q.14 above]
Q.17. What do you mean by relative refractive index of a medium with respect to another medium.
And: The relative refractive index of a medium with respect to another medium is the ratio of the
speed of light in the first medium for which relative refraction index is to be measured.
Q.18. State second law (Snell’s law) of refraction of light.
Ans: It states that during refraction of light the ratio of line of angle of incidence to the line of angle
of refraction for the particular pair of media is always constant and is equal to the refractive index of
the 2nd medium with request to the 1st medium.
Q.19. Define critical angle of a medium.
Ans: It is the angle of incidence of the medium when the angle of refraction became 900 while light
refracts from the medium to another optically rarer medium.
Q.20. Write a relation for critical angle with refractive index of a medium.
1
Ans: For the critical angle, c Sin C=µ where µ is the relative refractive index of the medium with
respect to the reason medium.
Q.21. Relative real depth with apparent depth of an object in a medium.
Ans: Real depth=n x apparent depth. Where n is the refractive index of the medium.
Q.22. if the refraction index of glass is 1.5 what will be the apparent thickness of a glass plate of 1.5
cm thick?
1
Ans: Apparent thickness =n x Real thickness
1
=1.5= 1.5 cm=1 cm
Q.23. How do you differentiate a concave lens from a convex lens?
Ans: Concave lens has thin central region than its outer region whereas a convex lens has central
thicker region.
[OR a concave lens diverges parallel rays where as a convex lens converges parallel rays.]
Q.24. How do you define the power of lens? What is its unit in SI?
Ans: The receiprocal of the focal length of a lens is known as its power. In SI the of power is dioptre
(D) or m -1.
Q.25. A lens can focus an image having same height as the height of the object on the screen 50 cm
away from the lens. What is the focal length of the lens?
Ans: The image size is equal to the object size when the object is placed at twice the focal length of
the lens.
i.e. object distance, u= 2 x f. where u and f are object distance and focal length of the lens.
i.e. 50= 2 f
f=25 cm
[this question may be of 2 marks type]
Q.26. What is least distance of distinct vision? [near point]
Ans: The smallest distance at which the eye can see objects clearly without strain is called the least
distance of distinct vision or near point. [normal vision, LDDV=25 cm].
Q.27. What are the refractive defects of vision?
Ans: a) Myopic. b) Hypermetropia c) Presbyopia. d) Astigmatism are the refractive defects of
vision.
Q.28. Give reason why the violet colour in rainbow is seen at the bottom.
Ans: Violet colour is refracted at maximum deviation than the other colour. So, it is seen at the
bottom of the rainbow.
Q.29. Why is the Sun seen red during early morning and at sun set?
And: Red colour is less scattered than other colours. So, red colour from the sun reaches us while
other colour are scattered during the Sun’s ray passes through thicker atmospheric column during
early morning or during Sun set.
Q.30. Which quantity measures the optical density of medium.
Ans: Refractive index.
Q.31. Why do we see distant object very small?
Ans: Because, the image formed at the retina of our eye is small in comparison to that of the image
of nearer object.
2-Marks Questions
Q.1. Explain how do we see non-luminous objects?
Ans: [See Q.5. of 1-mark question]
Q.2. What are the properties of the image formed by a plane mirror?
Ans:
a) The image size equals object size.
b) The image distance equals object distance.
c) The image is laterally inverted.
Q.3. How do you differentiate a concave mirror from a convex mirror?
Ans:
i) The concave mirror has reflecting surface on the concave side whereas convex mirror has
reflecting surface on the convex side.
ii) A concave mirror can form both real as well as virtual image where as convex mirror can
form virtual image only.
Q.4. What are the two types of images? How are they differed from one another?
Ans: Real and virtual images are the two types of images.
i)
Real image is formed due to actual intersection of rays from a mirror or lens
whereas virtual image is due t virtual intersection of rays.
ii)
Real image can be collected on a screen whereas virtual image cannot be collected
on the screen.
Q.5. A dentist uses a concave mirror of focal length 10 cm to see a small spot on a patient’s teeth.
Find the magnification product if a virtual image is formed at 25 cm behind the mirror.
Ans: We have focal length, f=10 cm; image distance, v=25 cm.
v
By magnification, M=u where u is object distance.
1 1
=-v( f -v)
v
v
25
=- f +1=1- f =1-10=1-25.
= - 1.5
nb
na
Q.6. Prove that the relative refractive index of medium “b” with respect to medium “a”, anb=
where na and nb are the absolute refractive indices of the two media respectively.
v
Ans: We have, anb= v
a
b
where Va and Vb are the speed of light in medium a and b.
VxC
C/vb
=>anb=V xa C = C/va where C is speed of light in air.
n
b
b
=>anb= n hence prove
a
Q.7. Explain why refractive index of medium in different for different colour of light.
Ans: Different colour of light have same speeds in vacuum but they have different speeds in the
some medium. So, some medium have different values of refractive index according to the relation
c
µ=v where c is speed of light in vacuum and v is speed of a particular light in the medium.
1
Q.8. Prove that for a medium its refractive index, n=sin C Where C is its critical angle.
Ans: We have by Snell’s law,
1 sin i
=
n Sin r
where n is the refractive
index of the medium from when the ray starts.
1
sin c
When i=c, r=900, i.e. n = sin 90
1
=> n =sin c [
. . . sin 900=1]
1
=>n=sin c
Q.9. what are the conditions for total internal
reflection?
Ans: i) Refraction must takes place from denser to rare medium.
ii) The in angle incidence must be greater than the critical angle.
Q.10. What are the two principle foci of a lens? Describe them.
Ans: i) First principle focus is a point on the principal axis. The rays diverging from this point in case
of convex lens or the rays converging towards this point are refracted by the lens parallel to the
principal axis.
ii) The second principal focus is a point on the principal axis. The rays incident on the lens
parallel to the principal axis after refraction actually converge to this point or appears to diverge
from the point.
Q.11. A convex mirror is often used as the rear viewer of a vehicle. Give two reasons.
Ans: Reasons: a) The image is erect and b) the field of view is large due to the formation of
diminished virtual image.
Q.12. Explain with the help of diagram, why pencil partly immersed in water appear to be bent at
the water surface.
Ans: When a pencil is immersed partly the tip inside
water will have apparent rise. Similarly, all part of
the pencil from the tip to the part near the surface
in the water will be apparently rise correspondingly.
Thus, the part of below the water surface apparently
bent upward at water surface.
Q.13. What do you mean by power of accommodation of eye? What is the least distance of distinct
vision ( or near point) of a normal eye?
Ans: The ability of the eye lens to adjust its focal length is called power of accommodation.
The least distance of distinct vision of normal eye is about 25 cm from the eye.
Q.14. Describe Prebyopia.
Ans: Sometimes, the may suffer from both myopia and hypermetropia such defect is known as
presbyopia. In such defect, the near point recedes away and far point become close. That is the eye
lens. Short range of vision.
Q.15. When does we see a rainbow?
Ans: Rainbow is caused by dispersion of sunlight by tiny water droplets, present in the atmosphere.
Presence of such droplets is quite common after a shower of rain. To see a rainbow the seen will be
at our backside.
Q.16. Explain why red light is used for warning signal light.
Ans: Of the entire colour light, red colour light suffer less scattered by air and dust particles. So, it
can be seen from a distant place. That is why red light is used for warning signal light.
Q.17. If the earth had no atmosphere, then the sky would looked dark. Why?
Ans: This is because, of the absence particle which scatters the sunlight. It there is atmosphere, the
particles of it will scatters sunlight and the sky look blush as violet and blue component of light are
highly scattered.
Q.18. Describe total internal reflection with a neat diagram.
Ans: Consider a dense medium form where light is refracted. The angle of refraction is greater than
the angle of incidence. Before the angle of
refraction become less than or equal to 900
there will be partial reflection inside the
same medium. But, there will be a total
internal reflection without any refraction
after the incident angle for which the angle of
refraction is 900 is made increase. This
condition is shown in the figure. [This
question can be treated as 3-marks
question].
{**Work out numerical problems in the text
book are also important}
3-Marks questions
Q.1. Which rays are used for making ray diagram of image formation by a lens (or mirror). Describe
the properties of these rays.
Ans: Rays used are:
a) The ray passing through the optical centre O.
(Ray directed towards or passing through centre of curvature for spherical mirror retrace
back after reflection).
It does not suffer any deviation.
b) The ray incident parallel to the principal axis.
It is refracted through the second principal focus F2 in case of convex lens. In the case of
concave lens it is refracted as if it emerges from the second principal focus F2.
(It is reflected to converge to the focus or appear to diverge from the focus for spherical
mirror)
c) The ray incident on the lens through first principal focus F1 in case of convex lens or
incident towards the first principal focus F1 in case of concave lens.
(A ray, incident on the mirror passing through the principal focus F in case of
concave mirror or approaching towards the principal focus F in the case of convex
mirror. It is reflected back parallel to the principal axis)
It is refracted parallel to the principal axis.
Q.2. Write the new Cartesian sign convention.
Ans: New Cartesian sign convention:
i) All measurements of distance are made from the pole of the mirror [from the optical
centre of the lens]
ii) If the measurement is along the direction of incident ray, a +ve sign is assigned to the
value.
iii) If the measurement is against the direction of incident ray, a –ve sign is assigned to the
value.
iv) * Length above the principal axis is assigned +ve and length below the principal axis is
assigned –ve.
Q.3. Why does light suffer refraction? What is the role of refractive index of a medium during
refraction of light through it? Can a material medium has its absolute refraction index less than 1?
Why.
Ans: The seep of light changes its value while crossing one medium to another medium. So, there is
a sudden change in the path of light which we called it as refraction. The refraction index of a
material tells about the optical density of a material medium. The speed of light will be shower in
the medium having larger refractive index.
The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the
speed of light in that medium. As, it is assumed that the speed of light in vacuum is the greatest, the
absolute refractive of material medium must be greater that 1.
Q.4. Derive an expression for apparent depth of an object inside a medium with a suitable diagram.
Ans: Consider a point A at a depth BA in a liquid of refractive index n.
A ray from point A is incident C and refract towards observed eye.
Then A/ will be the apparent
position of A.
Now,
BC
tan i =BA--------------1
r
B
1
& tan r= -------------2
BA
r
/
/
A
i
c
i
n
A
For
the
apparent
depth
measurement B &C is very closed with each
other. So, eqn. (1) & (2) we be written as,
BC
=i------------------(3)
BA
BC
and =r-------------------(4)
.
. .
/
BA
/
BA i Sin i
==
for
BA r Sin r
/
BA 1
=>BA=n
small ‘i’ & ‘r'
. /
. . BA=n1 x BA
1
i.e. Apparent depth=n x real depth.
Q.5. Define total internal reflection? Why is it called total?
Ans: When a ray of light approaches a rarer medium from a denser one such the angle of incident is
increased beyond critical angle, the ray will not suffer refraction. The ray will be reflected back
completely obeying the laws of reflection. The reflection is total because there is no reflection. So,
this reflection is known as total internal reflection.
Partial reflection
reflection
Partial reflection
V
f
Q.6. Using lens formula, derive the relation of magnification M=(1- ) where V is image distance and
f is the focal length of the lens. Define 1 dioptre of power of lens.
Ans: We have the magnification produced is given by,
v
u
M=
= u( f -u) ( . .
1 1
u
f
.
1
v
1 1
+ u= f )
=- + 1
u
=>M= (1- f )
I dioptre of power of lens is the power of lens having a focal length of 1 metre.
Q.7. What is myopia? How is it corrected? Illustrate with suitable diagram?
Ans: It is a defect of eye which makes the eye cannot see clearly the far off object but can see
directly the objects nearly. The far point of myopic eye is less than infinity and the near point is also
less than 25 cm.
This defect is corrected using a concave lens of suitable focal length.
Myopia eye focusing of it∞ and at near point
[Please read
about other
defects also]
Corrected myopic eye
Q.8. Write a note on twinkling of stars.
Ans: Twinkling of star is due to unstable/irregular refraction caused by the atmosphere. There are
always irregular currents of hot air in the atmosphere. This causes the refraction index of the
atmosphere very continually. As a result the position of the star appears to fluctuate to an observe
on earth. If the line of vision get deflected away due the reason, the star is temporarily lost from
view. Thus the stars twinkle.
[Read about mirages or advance sunrise e x delayed sunset].
5-Marks Questions
Q. 1. Discuss how the position of the image of an object will change when it is brought from infinity
to a position near a convex lens [mirror ( concave)]. Illustrate it with suitable ray diagram. Where
will an object be place to obtain a virtual magnified image?
Ans: The following ray diagram illustrates how position of the image of the object at various
positions:
For concave lens:-
F1
F2
F1
F2
i) Object placed at infinity
ii) Object placed beyond a distance equal to
a) A point image is form at the 2nd principal focus.
twice focal length.
It is a real image.
a) A real inverted and diminished in size image
is form at a distance less than twice focal
length behind the lens.
F2
2f
F1
iii) Object placed at a distance equal to
twice focal length.
I
iv) Object placed at a distance less than
twice focal length but greater than focal
length
a) An image which is real, inverted and of same
size as the object is formed at the distance equal a) An image which is real, inverted and greater
to twice focal length behind the lens.
than object size is formed at distance greater
than twice focal length.
o
F2
F1
o
F2
I
F1
o
iii) Object placed at the focal length
iii) Object in between focus and the lens
a) A real and of infinite size is formed at infinity
a) A virtua, errect and enlarged image is
formed beyond the focus on the same size of
the object.
Q.2. Describe myopia and hypermetropia stating the nature, the cause and correction of the defect.
Ans: Myopia or nearsightedness:Nature: Cannot see clearly the far objects but can see distinctly the objects nearby.
Cause:- i) The eyeball is elongated.
Or ii) The eye lens become thicker making focal length shorter
Correction:- Use of concave lens of suitable focal length. The focal length of the concave
lens is equal to the distance of the far point of the defects eye.
Hypermetropia or for sightedness:Nature:- Can see distant objects clearly but cannot see near by objects distinctly.
Causes:- i) Shortening of eye ball length.
Or ii) The eye lens becomes thinner making focal length longer.
Corrections:- Use of convex lens of suitable focal length.
[Numerical problem related to image position, size nature etc. for lens and mirror are also
important. Try the work out example in the text book]
(They will carry 2 or 3 marks)