CBSE Class 12 Physics Chapter-9 Important Questions – Free PDF Download
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CBSE Class 12 Physics Chapter-9 – Ray Optics and Optical Instruments
CBSE Class 12 Physics Important Questions Chapter 9 – Ray Optics and Optical Instruments
5 Marks Questions
1. Prove that 
When refraction occurs of a convex spherical refracting surface and the ray travels
from rarer to denser medium.
Ans. From 
—-(i)
Similarly form 
—-(ii)
From 
Same aperture of the spherical surface is small so point N lies close to P and since angles 
are very small 
and 
and 
Applying sign conventional
and 
Substituting these values in e.g. (1) & (2)
According to snell’s law
(Since angles are very small)
2. A lens forms a real image of an object. The distance of the object. From the lens is U cm and the distance of the image from the lens is 
cm. The given graph shows the variation of and U
(a) What is the nature of the lens?
(b) Using the graph find the focal length of the lens?
(c) Draw a ray diagram to show the formation of image of same size as that of object in case of converging lens hence derive lens equation?
Ans. (a) convex lens
(b) 
i.e 
In the given graph f = 10cm.
(c) 
and 
are similar
—-(2) 
Combining equation (1) & (2)
Using sign conventions
Divide by Uf
Or
Hence derived
3. By stating sign conventions and assumptions used derive the relation between 
in case of a concave mirror?
Ans. Sign conventions
(1) All distances are measured from the pole of the mirror.
(2) Distance measured in the direction of incident light is positive and those measured in the direction opposite to the incident light are negative.
(3) Height measured upwards is positive and height measured downwards is negative.
Assumptions
(1) Aperture of the spherical mirror is considered to be very small.
Remaining angles equal
—-(1) (
Dlies closetol AB + MD)
Similarly 
and are also similar
—-(2)
Combining equation (1) & (2) and using sign conventions
Divide by we 
get
4. (a) A person looking at a mesh of crossed wires is able to see the vertical lines more distinctly than the horizontal wires. What is the effect due to? How is such a defect of vision corrected?
(b) A man with normal near point (25cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5cm.
(i) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass ?
(ii) What is the maximum and minimum angular magnificent (magnifying power) possible using the above simple microscope?
Ans. (a) It is due to the defect called astigmatisms and is caused due to irregular surface of cornea and curvature of the eye lens is different in different planes. This type of defect can be corrected using cylindrical lens.
(b) (i) Here f = 5cm 
For closest point 
and thus 
u = -4.2cm
A for farthest point f = 5cm 
u = -5cm
(b) (ii) Angular magnification
Maximum angular magnification
Minimum angular magnification
5. Four double convex lens with following specification are available.
| LensFocal | lengthAperture | |
| A | 100cm | 10cm |
| B | 100cm | 5cm |
| C | 10cm | 2cm |
| D | 5cm | 2cm |
(a) Which of the given four lenses should be selected as objective and eyepiece to construct an astronomical telescope and why? What will be the magnifying power and length of the tube of this telescope?
(b) An object is seen with the help of a simple microscope, firstly in red light and then is blue light. Will the magnification be same in both the cases? Why?
Ans. (a) objective of the telescope should be of large aperture as it has to gather maximum light and should be of large focal length to have maximum magnification.
Hence lens A is selected as objective and lens D as eyepiece of small aperture and small focal length.
M.P. = 20
(b) L = fo + fe
L = 100 + 5
L = 10.5 cm
When red light as replaced by blue light, magnifying power increases.
6. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Ans. Size of the candle, h= 2.5 cm
Image size = h‘
Object distance, u= – 27 cm
Radius of curvature of the concave mirror, R= – 36 cm
Focal length of the concave mirror,
Image distance = v
The image distance can be obtained using the mirror formula:
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.
7. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Ans. Height of the needle, 
= 4.5 cm
Object distance, u = – 12 cm
Focal length of the convex mirror, f= 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror. The image size is given by the magnification formula:
Hence, magnification of the image, 
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.
8. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Ans. Actual depth of the needle in water, = 12.5 cm
Apparent depth of the needle in water, 
= 9.4 cm
Refractive index of water = 
The value of can be obtained as follows:
Hence, the refractive index of water is about 1.33.
Water is replaced by a liquid of refractive index, 
The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the needle. Hence, we can write the relation:
Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up.
∴Distance by which the microscope should be moved up = 9.4 – 7.67
= 1.73 cm
9. Figures 9.34 (a) and (b) show refraction of a ray in air incident at 
with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 
with the normal to a water-glass interface [Fig. 9.34(c)].
Ans. As per the given figure, for the glass – air interface:
Angle of incidence, i =
Angle of refraction, r= 
The relative refractive index of glass with respect to air is given by Snell’s law as:
…(1)
…(i)
As per the given figure, for the air – water interface:
Angle of incidence, i=
Angle of refraction, r= 
The relative refractive index of water with respect to air is given by Snell’s law as:
…(2)
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
The following figure shows the situation involving the glass – water interface.
Angle of incidence, i= 
Angle of refraction = r
From Snell’s law, rcan be calculated as:
Hence, the angle of refraction at the water – glass interface is 
10. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Ans.
Actual depth of the bulb in water, 
= 80 cm = 0.8 m
Refractive index of water, 
The given situation is shown in the following figure:
Where,
i= Angle of incidence
r= Angle of refraction = 
Since the bulb is a point source, the emergent light can be considered as a circle of radius,
Using Snell’ law, we can write the relation for the refractive index of water as:
Using the given figure, we have the relation:
∴ Area of the surface of water
= 
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61
.
11. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be
. What is the refractive index of the material of the prism? The refracting angle of the prism is
. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Ans.Angle of minimum deviation, 
=
Angle of the prism, A =
Refractive index of water, µ= 1.33
Refractive index of the material of the prism = 
The angle of deviation is related to refractive index
as:
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let 
be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is given by the relation:
Hence, the new minimum angle of deviation is
.
12. A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Ans. Focal length of the objective lens, 
= 2.0 cm
Focal length of the eyepiece, 
= 6.25 cm
Distance between the objective lens and the eyepiece, d= 15 cm
(a) Least distance of distinct vision, 
∴Image distance for the eyepiece, 
= – 25 cm
Object distance for the eyepiece = 
According to the lens formula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens =
According to the lens formula, we have the relation:
Magnitude of the object distance, 
= 2.5 cm
The magnifying power of a compound microscope is given by the relation:
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
∴Image distance for the eyepiece,
Object distance for the eyepiece =
According to the lens formula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens = 
According to the lens formula, we have the relation:
Magnitude of the object distance, = 2.59 cm
The magnifying power of a compound microscope is given by the relation:
Hence, the magnifying power of the microscope is 13.51.
13. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,
Ans. Focal length of the objective lens, 
= 8 mm = 0.8 cm
Focal length of the eyepiece, 
= 2.5 cm
Object distance for the objective lens, 
= – 9.0 mm = – 0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, 
= – d= – 25 cm
Object distance for the eyepiece = 
Using the lens formula, we can obtain the value of as:
We can also obtain the value of the image distance for the objective lens
using the lens formula.
The distance between the objective lens and the eyepiece
The magnifying power of the microscope is calculated as:
Hence, the magnifying power of the microscope is 88.
14. Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of
images that one obtains from explicit ray diagrams.]
Ans. (a) For a concave mirror, the focal length (f)is negative.
∴f < 0
When the object is placed on the left side of the mirror, the object distance (u)is negative.
∴u< 0
For image distance v, we can write the lens formula as:
…(i)
The object lies between f and 2f.
(
u and f are negative)
…(2)
Using equation (1), we get:
∴
is negative, i.e., vis negative.
Therefore, the image lies beyond 2f.
(b) For a convex mirror, the focal length (f) is positive.
∴ f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u < 0
For image distance v, we have the mirror formula:
Using equation (2), we can conclude that:
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
(c) For a convex mirror, the focal length (f) is positive.
When the object is placed on the left side of the mirror, the object distance (u) is negative,
For image distance v, we have the mirror formula:
But we have u <0
Hence, the image formed is diminished and is located between the focus (f) and the pole.
(d) For a concave mirror, the focal length (f) is negative.
When the object is placed on the left side of the mirror, the object distance (u) is negative.
It is placed between the focus (f) and the pole.
For image distance v, we have the mirror formula:
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:
15. (a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68.
The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?
Ans.(a) Refractive index of the glass fibre, 
Refractive index of the outer covering of the pipe, 
= 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i‘
The refractive index 
of the inner core – outer core interface is given as:
For the critical angle, total internal reflection (TIR) takes place only when
, i.e., 
Maximum angle of reflection,
Let, 
be the maximum angle of incidence.
The refractive index at the air – glass interface,
We have the relation for the maximum angles of incidence and reflection as:
Thus, all the rays incident at angles lying in the range 
will suffer total internal reflection.
(b) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe,
= Refractive index of air =1
For the angle of incidence
, we can write Snell’s law at the air – pipe interface as:
Since 
> r, All incident rays will suffer total internal reflection
16.Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen.
Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
Ans.(a) Yes
Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.
(b) No
A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.
(c) The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.
(d) Yes; Decrease
The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.
(e) Yes
The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.
17.(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Ans.Focal length of the convex lens, 
= 30 cm
Focal length of the concave lens, 
= – 20 cm
Distance between the two lenses, d= 8.0 cm
(a) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
Where,
= Object distance = ∞
v1= Image distance
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
Where,
= Object distance
= (30 – d) = 30 – 8 = 22 cm
= Image distance
The parallel incident beam appears to diverge from a point that is
from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula, we have:
Where,
= Object distance = 
= Image distance
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
Where,
= Object distance
= – (20 + d) = – (20 + 8) = – 28 cm
= Image distance
Hence, the parallel incident beam appear to diverge
18.At what angle should a ray of light be incident on the face of a prism of refracting angle 
so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
Ans.The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure.
Angle of prism, 
Refractive index of the prism, µ= 1.524
= Incident angle
= Refracted angle
= Angle of incidence at the face AC
e= Emergent angle = 
According to Snell’s law, for face AC, we can have:
It is clear from the figure that angle
According to Snell’s law, we have the relation:
Hence, the angle of incidence is 
19.For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Ans.Least distance of distinct vision, d= 25 cm
Far point of a normal eye, 
Converging power of the cornea, 
Least converging power of the eye-lens, 
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, 
= 40 + 20 = 60 D
Power of the eye-lens is given as:
To focus an object at the near point, object distance (u)= – d= – 25 cm
Focal length of the eye-lens = Distance between the cornea and the retina
= Image distance
Hence, image distance, 
According to the lens formula, we can write:
Where,
= Focal length
∴Power of the eye-lens = 64 – 40 = 24 D
Hence, the range of accommodation of the eye-lens is from 20 D to 24 D.
20.Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Ans.(a)Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
(b) Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.
(d) The angular magnification produced by the eyepiece of a compound microscope is 
Where,
= Focal length of the eyepiece
It can be inferred that if is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as 
Where,
= Object distance for the objective lens
= Focal length of the objective
The magnification is large when >. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since is small, will be even smaller. Therefore, and are both small in the given condition.
(e)When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.
21.An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Ans.Focal length of the objective lens,= 1.25 cm
Focal length of the eyepiece, = 5 cm
Least distance of distinct vision, d= 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m= 30
The angular magnification of the eyepiece is given by the relation:
The angular magnification of the objective lens (mo) is related to meas:
= m
We also have the relation:
Applying the lens formula for the objective lens:
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
Where,
= Image distance for the eyepiece = – d= – 25 cm
= Object distance for the eyepiece
Separation between the objective lens and the eyepiece 
=4.17+7.5
=11.67 cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
22.An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Ans.Focal length of the objective lens,= 1.25 cm
Focal length of the eyepiece, = 5 cm
Least distance of distinct vision, d= 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m= 30
The angular magnification of the eyepiece is given by the relation:
The angular magnification of the objective lens (mo) is related to meas:
= mm
Applying the lens formula for the objective lens:
And 
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
Where,
= Image distance for the eyepiece = – d= – 25 cm
= Object distance for the eyepiece
Separation between the objective lens and the eyepiece 
=4.17+7.5
=11.67 cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
23.A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
Ans.The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.
Distance between the objective mirror and the secondary mirror, d= 20 mm
Radius of curvature of the objective mirror, 
Hence, focal length of the objective mirror, 
Radius of curvature of the secondary mirror, R1 = 140 mm
Hence, focal length of the secondary mirror, 
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror,
=110-20
=90 mm
Applying the mirror formula for the secondary mirror, we can calculate image distance (v)as:
Hence, the final image will be formed 315 mm away from the secondary mirror.
24.Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
Ans.Focal length of the convex lens,
= 30 cm
The liquid acts as a mirror. Focal length of the liquid =
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
Let the refractive index of the lens be
and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is – R.
R can be obtained using the relation:
Let 
be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror =
Radius of curvature of the liquid on the side of the lens, R= – 30 cm
The value of can be calculated using the relation:
