Q1(NCERT): What is meant by power of
accommodation of the eye?
Answer: Power of accommodation is the ability
of the eye lens to focus near and far objects
clearly on the retina by adjusting its focal length.
Power of accommodation of the eye is limited. It
implies the focal length of the eye lens cannot be
reduced beyond certain minimum limit.
Q2(CBSE 2011): A person cannot see the
objects distinctly, when placed at a distance
less than 50 cm.
(a) Identify the defect of vision.
(b) Give two reasons for this defect.
(c) Calculate the power and nature of the lens he
should be using to see clearly the object placed at
a distance of 25 cm from his eyes.
(d) Draw the ray diagrams for the defective and the
corrected eye.
Answer: (a) Defect of vision = Hypermetropia or Long-
sightedness
(b) Reasons for the defect are:
Curvature of lens or retina becomes less than normal
increases focal length.
Ciliary muscles become stiff
Shortening of eye ball due to which distance between
retina and and lens reduces and image is formed
beyond retina.
(c) Given, u = -25 cm, v = -50cm, f = ?
Applying lens formula, 1/f = 1/v - 1/u
⇒1/f = -1/50 - (-1/25) = -1/50 + 1/25 = (-1+2) /50 =
1/50
⇒ f = 50cm = 0.5m
∴ Power of the lens = 1/f = 1/0.5m = + 2D
∴ A convex lens having power of 2D can be used to
correct the vision.
(d) ray diagrams for the defective and the corrected
eye.
Q3(NCERT): A person with a myopic eye cannot see
objects beyond 1.2 m distinctly. What should be
the type of the corrective lens used to restore
proper vision?
Answer: The distance of far point x = 1.2m
To view distant objects, concave lens of focal length
-1.2 should be used.
i.e. f = -x = -1.2m
Power of lens = 1/f = 1/(-1.2m) = -0.83D
Q4(CBSE 2011): Name the part of the human eye
that helps in changing the focal length of the eye
lens.
Answer: Ciliary muscles.
(Note: Pupil regulates and controls the amount of light
entering the eye.)
Q5(NCERT MCQ): The human eye can focus objects
at different distances by adjusting the focal length
of the eye lens. This is due to
(a) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.
Answer: (b) accommodation.
Q6(NCERT MCQ): The human eye forms the image
of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.
Answer: (d) retina.
Q7(NCERT MCQ): The least distance of distinct
vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm.
(c) 25 cm.
(d) 2.5 m.
Answer: 25cm
Q8(NCERT MCQ): The change in focal length of an
eye lens is caused by the action of the
(a) pupil.
(b) retina.
(c) ciliary muscles.
(d) iris.
Answer: (c) ciliary muscles.
Q9(NCERT): What is the far point and near point of
the human eye with normal vision?
Answer: The near point of the eye is the minimum
distance of the object from the eye, which
can be seen clearly. For a normal human eye, this
distance is 25cm .
The far point of the eye is the maximum distance upto
which the eye can see the objects clearly without strain.
The far point of the normal human eye is infinity.
Q10(CBSE 2011): A boy uses spectacles of focal
length – 60 cm. Name the defect of vision he is
suffering from. Which lens is used for the
correction of this defect? Compute the power of
this lens.
Answer: Given f = -60 cm =-0.60m
The -ve sign of focal length indicates that the lens is
concave. ∴ the boy suffers from myopic (short
sightedness) vision.
Power of the lens = 1/f(in m) = 1/(-0.60) = -1.67 D
Q11(NCERT): A student has difficulty reading the
blackboard while sitting in the last row. What could
be the defect the child is suffering from? How can
it be corrected?
Answer: The student has difficulty in seeing distant
objects. He might be suffering from myopia. The defect
can be corrected using a concave lens of appropriate
focal length.
Q12(CBSE): Why do we have two eyes instead of
one?
Answer: Having two eyes instead of one is advantageous
for the following:
field of view is more with two eyes than one eye.
two eyes gives us three dimensional vision (stereo
vision) of an object.
Q13(NCERT): A person needs a lens of power –5.5
dioptres for correcting his distant vision. For
correcting his near vision he needs a lens of power
+1.5 dioptre. What is the focal length of the lens
required for correcting (i) distant vision, and (ii)
near vision?
Answer:
(i) Distant vision: Given focal length (f) = ?, Power (P) =
-5.5D
P = 1/f ⇒ f = 1/P = 1/(-5.5)m = -100/55cm = -18.2
cm
(ii) For near vision, P = +1.5 D
⇒ f = 1/P = 1/1.5m = 1000/15cm = 66.7 cm
Q14(CBSE 2012): Define dispersion of white light
and name the colours of white light in order?
Answer: The splitting of light into its component colours
is called dispersion. White light splits into seven colours
(VIBGYOR) i.e. violet, indigo, blue, green, yellow, orange
and red. Isaac Newton was the first to use a glass prism
to obtain the spectrum of sunlight.
Q15(NCERT): The far point of a myopic person is 80
cm in front of the eye. What is the nature and
power of the lens required to correct the problem?
Answer: Given, distance of far point (x) = 80 cm, P = ?
To view distant objects correctly, focal length of the
corrective lens = f = -x = -80cm
Power of the lens (P = 1/(f in metres) = 1/(-0.80) =
100/-80 = -1.25D
The lens is concave of powed -1.25D
Q16(NCERT): Make a diagram to show how
hypermetropia is corrected. The near point of a
hypermetropic eye is 1 m. What is the power of the
lens required to correct this defect? Assume that
the near point of the normal eye is 25 cm.
Answer: For diagram see question 2(b) above.
Given v = -1m = -100cm
u = -25cm
Using lens formula, 1/v -/1u = 1/f
⇒ -1/100 + 1/25 = 1/f
⇒ f = 33.3 cm - 0.333m
Power of the lens = 1/f = 100/33.3 = 3D
Q17(NCERT): Why is a normal eye not able to see
clearly the objects placed closer than 25 cm?
Answer: Since the focal length of eye lens cannot be
decreased beyond certain minimum length.
Q18(Old CBSE Curr. 2006): List four common
defects which can be corrected with the use of
spectacles.
Answer: Four common defects that can be corrected by
using spectacles:
1. Hypermetropia
2. Myopia
3. Astigmatism
4. Presbyopia
Q19(NCERT): What happens to the image distance
in the eye when we increase the distance of an
object from the eye?
Answer: The size of the eye can change, so the image
distance is fixed. When we increase the distance of an
object from the eye, the image distance in the eye does
not change. Due to power of accommodation of the
eye, focal length of the eye lens is changed which
compensates the increase in object distance. Hence
image distance remains fixed and image is formed on
the retina of the eye.
Q20(NCERT): Why do stars twinkle?
OR
Q(CBSE 2012): Explain why do stars twinkle and
planets do not?
Answer: The twinkling of a star is
due to the earth's atmospheric
refraction. The earth's atmosphere
is moving and it consists of
pockets of warm and cold air. Thus
the atmosphere has variations in
refractive indices of air. Stars are
very far away from the earth and
emit their own light. Being very far
from the earth, these are
considered point sources. When a
star light enters the earth's
atmosphere, it undergoes multiple
refractions and bends
continuously towards the normal
till it enters our eyes. Its apparent
position appears higher than the
normal one. Due to mobility of air
and variation is temperature, this
apparent position of star is not
steady and moves continuously giving a twinkling effect.
When in space (outside the earth's atmosphere) stars
do not appear twinkle. Planets as compared to stars are
closer to the earth and appear bigger (we cannot
consider them point sized like stars). It nullifies the
twinkling effect.
Q21(NCERT): Explain why the planets do not
twinkle.
Answer: The distanced between planets and the earth is
less as compared to stars. Planets cannot be considered
as point sources. The apparent shift in their position
due to their position cannot be observed because they
subtend greater angle at the eye. Comparatively larger
size, their brightness do not change and do not give
twinkle effect.
Q22: What is Mirage? How does it occur?
Answer: A mirage is an optical illusion which is generally
observed in deserts or over hot surfaces like a coal
tarred road during hot summers. In summers, the layer
near to earth surface is hotter and lighter than upper
layers of the air. Due to variation of optical densities of
air layers, light from an object (e.g. tree) undergoes
series of internal reflections and bend upwards. It gives
an inverted virtual image below the seen object. To a
distant traveller it gives shinning water effect called
mirage.
Q23: What are the causes of myopia?
Answer: Causes of myopia are:
Increase in the length of the eye ball leads to increase in
distance of the retina from the eye.
Decrease in focal length of eye lens.
Q24: What is Tyndall effect?
Answer: The scattering of light by colloidal particles is
called Tyndall effect. When a beam of light strikes the
minute particles of earth’s atmosphere which are
suspended particles of dust and molecule of air the path
of beam become visible. The phenomenon of scattering
of light by the colloidal particle is called Tyndall Effect.
It can be observed when sunlight passes through a
canopy of a dense forest.
Q25: How does colour of scattered light depend on
colloidal particles?
Answer: The colour of the scattered light depends on
the size of the scattering particles
accommodation of the eye?
Answer: Power of accommodation is the ability
of the eye lens to focus near and far objects
clearly on the retina by adjusting its focal length.
Power of accommodation of the eye is limited. It
implies the focal length of the eye lens cannot be
reduced beyond certain minimum limit.
Q2(CBSE 2011): A person cannot see the
objects distinctly, when placed at a distance
less than 50 cm.
(a) Identify the defect of vision.
(b) Give two reasons for this defect.
(c) Calculate the power and nature of the lens he
should be using to see clearly the object placed at
a distance of 25 cm from his eyes.
(d) Draw the ray diagrams for the defective and the
corrected eye.
Answer: (a) Defect of vision = Hypermetropia or Long-
sightedness
(b) Reasons for the defect are:
Curvature of lens or retina becomes less than normal
increases focal length.
Ciliary muscles become stiff
Shortening of eye ball due to which distance between
retina and and lens reduces and image is formed
beyond retina.
(c) Given, u = -25 cm, v = -50cm, f = ?
Applying lens formula, 1/f = 1/v - 1/u
⇒1/f = -1/50 - (-1/25) = -1/50 + 1/25 = (-1+2) /50 =
1/50
⇒ f = 50cm = 0.5m
∴ Power of the lens = 1/f = 1/0.5m = + 2D
∴ A convex lens having power of 2D can be used to
correct the vision.
(d) ray diagrams for the defective and the corrected
eye.
Q3(NCERT): A person with a myopic eye cannot see
objects beyond 1.2 m distinctly. What should be
the type of the corrective lens used to restore
proper vision?
Answer: The distance of far point x = 1.2m
To view distant objects, concave lens of focal length
-1.2 should be used.
i.e. f = -x = -1.2m
Power of lens = 1/f = 1/(-1.2m) = -0.83D
Q4(CBSE 2011): Name the part of the human eye
that helps in changing the focal length of the eye
lens.
Answer: Ciliary muscles.
(Note: Pupil regulates and controls the amount of light
entering the eye.)
Q5(NCERT MCQ): The human eye can focus objects
at different distances by adjusting the focal length
of the eye lens. This is due to
(a) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.
Answer: (b) accommodation.
Q6(NCERT MCQ): The human eye forms the image
of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.
Answer: (d) retina.
Q7(NCERT MCQ): The least distance of distinct
vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm.
(c) 25 cm.
(d) 2.5 m.
Answer: 25cm
Q8(NCERT MCQ): The change in focal length of an
eye lens is caused by the action of the
(a) pupil.
(b) retina.
(c) ciliary muscles.
(d) iris.
Answer: (c) ciliary muscles.
Q9(NCERT): What is the far point and near point of
the human eye with normal vision?
Answer: The near point of the eye is the minimum
distance of the object from the eye, which
can be seen clearly. For a normal human eye, this
distance is 25cm .
The far point of the eye is the maximum distance upto
which the eye can see the objects clearly without strain.
The far point of the normal human eye is infinity.
Q10(CBSE 2011): A boy uses spectacles of focal
length – 60 cm. Name the defect of vision he is
suffering from. Which lens is used for the
correction of this defect? Compute the power of
this lens.
Answer: Given f = -60 cm =-0.60m
The -ve sign of focal length indicates that the lens is
concave. ∴ the boy suffers from myopic (short
sightedness) vision.
Power of the lens = 1/f(in m) = 1/(-0.60) = -1.67 D
Q11(NCERT): A student has difficulty reading the
blackboard while sitting in the last row. What could
be the defect the child is suffering from? How can
it be corrected?
Answer: The student has difficulty in seeing distant
objects. He might be suffering from myopia. The defect
can be corrected using a concave lens of appropriate
focal length.
Q12(CBSE): Why do we have two eyes instead of
one?
Answer: Having two eyes instead of one is advantageous
for the following:
field of view is more with two eyes than one eye.
two eyes gives us three dimensional vision (stereo
vision) of an object.
Q13(NCERT): A person needs a lens of power –5.5
dioptres for correcting his distant vision. For
correcting his near vision he needs a lens of power
+1.5 dioptre. What is the focal length of the lens
required for correcting (i) distant vision, and (ii)
near vision?
Answer:
(i) Distant vision: Given focal length (f) = ?, Power (P) =
-5.5D
P = 1/f ⇒ f = 1/P = 1/(-5.5)m = -100/55cm = -18.2
cm
(ii) For near vision, P = +1.5 D
⇒ f = 1/P = 1/1.5m = 1000/15cm = 66.7 cm
Q14(CBSE 2012): Define dispersion of white light
and name the colours of white light in order?
Answer: The splitting of light into its component colours
is called dispersion. White light splits into seven colours
(VIBGYOR) i.e. violet, indigo, blue, green, yellow, orange
and red. Isaac Newton was the first to use a glass prism
to obtain the spectrum of sunlight.
Q15(NCERT): The far point of a myopic person is 80
cm in front of the eye. What is the nature and
power of the lens required to correct the problem?
Answer: Given, distance of far point (x) = 80 cm, P = ?
To view distant objects correctly, focal length of the
corrective lens = f = -x = -80cm
Power of the lens (P = 1/(f in metres) = 1/(-0.80) =
100/-80 = -1.25D
The lens is concave of powed -1.25D
Q16(NCERT): Make a diagram to show how
hypermetropia is corrected. The near point of a
hypermetropic eye is 1 m. What is the power of the
lens required to correct this defect? Assume that
the near point of the normal eye is 25 cm.
Answer: For diagram see question 2(b) above.
Given v = -1m = -100cm
u = -25cm
Using lens formula, 1/v -/1u = 1/f
⇒ -1/100 + 1/25 = 1/f
⇒ f = 33.3 cm - 0.333m
Power of the lens = 1/f = 100/33.3 = 3D
Q17(NCERT): Why is a normal eye not able to see
clearly the objects placed closer than 25 cm?
Answer: Since the focal length of eye lens cannot be
decreased beyond certain minimum length.
Q18(Old CBSE Curr. 2006): List four common
defects which can be corrected with the use of
spectacles.
Answer: Four common defects that can be corrected by
using spectacles:
1. Hypermetropia
2. Myopia
3. Astigmatism
4. Presbyopia
Q19(NCERT): What happens to the image distance
in the eye when we increase the distance of an
object from the eye?
Answer: The size of the eye can change, so the image
distance is fixed. When we increase the distance of an
object from the eye, the image distance in the eye does
not change. Due to power of accommodation of the
eye, focal length of the eye lens is changed which
compensates the increase in object distance. Hence
image distance remains fixed and image is formed on
the retina of the eye.
Q20(NCERT): Why do stars twinkle?
OR
Q(CBSE 2012): Explain why do stars twinkle and
planets do not?
Answer: The twinkling of a star is
due to the earth's atmospheric
refraction. The earth's atmosphere
is moving and it consists of
pockets of warm and cold air. Thus
the atmosphere has variations in
refractive indices of air. Stars are
very far away from the earth and
emit their own light. Being very far
from the earth, these are
considered point sources. When a
star light enters the earth's
atmosphere, it undergoes multiple
refractions and bends
continuously towards the normal
till it enters our eyes. Its apparent
position appears higher than the
normal one. Due to mobility of air
and variation is temperature, this
apparent position of star is not
steady and moves continuously giving a twinkling effect.
When in space (outside the earth's atmosphere) stars
do not appear twinkle. Planets as compared to stars are
closer to the earth and appear bigger (we cannot
consider them point sized like stars). It nullifies the
twinkling effect.
Q21(NCERT): Explain why the planets do not
twinkle.
Answer: The distanced between planets and the earth is
less as compared to stars. Planets cannot be considered
as point sources. The apparent shift in their position
due to their position cannot be observed because they
subtend greater angle at the eye. Comparatively larger
size, their brightness do not change and do not give
twinkle effect.
Q22: What is Mirage? How does it occur?
Answer: A mirage is an optical illusion which is generally
observed in deserts or over hot surfaces like a coal
tarred road during hot summers. In summers, the layer
near to earth surface is hotter and lighter than upper
layers of the air. Due to variation of optical densities of
air layers, light from an object (e.g. tree) undergoes
series of internal reflections and bend upwards. It gives
an inverted virtual image below the seen object. To a
distant traveller it gives shinning water effect called
mirage.
Q23: What are the causes of myopia?
Answer: Causes of myopia are:
Increase in the length of the eye ball leads to increase in
distance of the retina from the eye.
Decrease in focal length of eye lens.
Q24: What is Tyndall effect?
Answer: The scattering of light by colloidal particles is
called Tyndall effect. When a beam of light strikes the
minute particles of earth’s atmosphere which are
suspended particles of dust and molecule of air the path
of beam become visible. The phenomenon of scattering
of light by the colloidal particle is called Tyndall Effect.
It can be observed when sunlight passes through a
canopy of a dense forest.
Q25: How does colour of scattered light depend on
colloidal particles?
Answer: The colour of the scattered light depends on
the size of the scattering particles
