CBSE Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments Important Question Answer
Very Short Type Question – 1 Mark
1. A person is standing before a concave mirror cannot see his image, unless he is beyond the centre of curvature? Why?
Ans. When man stands beyond focus is i.e. between focus and centre of curvature, his real and inverted image is formed beyond C is beyond him and thus he cannot see the image. But when he stands beyond C, image is formed between focus and centre of curvature is in front of him and thus he is able to see his image.
2. For what angle of incidence, the lateral shift produced by a parallel sided glass plate is maximum?
Ans. We know
D =t
Lateral shift is maximum
3. You read a newspaper, because of the light if reflects. Then why do you not see even a faint image of yourself in the newspaper?
Ans. The image is produced due to regular reflection of light but when we read a newspaper, because of diffused (irregular) reflection of light we are not able to see even a faint image.
4. A substance has critical angle of 
for yellow light what is its refractive index?
Ans.
5. An object is placed between the pole and focus of a concave mirror produces a virtual and enlarged image. Justify using mirror formula?
Ans. 
For a concave mirror
Given V < f so 
is positive, hence image is virtual.
Now magnification 
Hence enlarged image is produce
6. A converging and diverging lens of equal focal lengths are placed coaxially in contact. Find the focal length and power of the combination?
Ans. 
For converging lens 
For diverging lens 
P=0
Hence
7. The refractive index of a material of a convex lens is n1 it is immersed in a medium of refractive index
. A parallel beam of light is incident on thelens. Trace the path of the emergent rays when
.
Ans. When then the convex lens behaves as a concave
8. In a telescope the focal length of the objective and the eye piece are 60cm and 5cm respectively. What is? (1) Its magnification power (2) Tube length
Ans. magnification 
Tube length 
9. Show the variation of u and 
in case of a convex mirror?
Ans.
10. Two lenses having focal length 
are placed coaxially at a distance x from each other. What is the focal length of the combination?
Ans. 
:
Short type question answers – 2 Marks
1. What are optical fibres? Give their one use?
Ans. Optical fibres consist of thin and long strands of fine quality glass or quartz coated with a thin layer of material of refractive index less than the refractive index of strands. They work on the principle of total internal reflection so they do not suffer any loss.
Uses – The optical fibres are used in medical investigations i.e. one can examine the inside view of stomach and intestine by a method called endoscopy.
2. How the focal lengths of a lens change with increase in the wavelength of the light?
Ans. 
i.e. when wavelength increases 
decreases and according to
focal length increases.
3. Show with a ray diagram, how an image is produced in a total reflecting prism?
Ans. the two rays from the object PQ undergoes total internal reflection firstly at the face AB and then at BC forming the find image 
(real and inverted image)
4. The radii of the curvature of the two spherical surfaces which is a lens of required focal length are not same. It forms image of an object. The surfaces of the lens facing the object and the image are inter-changed. Will the position of the image change?
Ans. As we know
When 
gets interchanged focal length of the lens remains the same hence position of the image will not change.
5. A thin converging lens has focal length (f) when illuminated by violet light. State with reason how the focal length of the lens will change if violet light is replaced by red light
Ans. Since
n for violet is more that n for red colour hence focal length of the lens will decreases when violet light is replaced by red light.
6. Thin prism of angle 
gives a deviation of 
. What is the refractive index of material of the prism?
Ans.
n =1.41
7. Although the surfaces of a goggle lens are curved it does not have any power. Why?
Ans. The two surface of the goggle lens are parallel i.e. one surface convex and the other concave thus the power of the two surfaces and equal but of opposite sign.
8. A ray of light in incident normally on one face of the prism of apex angle and refractive index
. Find the angle of deviation for the ray of light?
Ans. When ray PQ falls normally on AB then if goes straight at QR (no refraction)
Applying Snell’s law for face AC
r = 45o
Now angle of deviation
9. Following data was recorded for values of object distance and corresponding values of image distance in the experiment on study of real image formation by a convex lens of power +5 D. one of three observation is incorrect. Identify and give reason?
S. No.(u)123456
Object distance
253035455055
Image distance976137353230
Ans. 
Observation (3) is incorrect because both object and the image here lies between f and 2f.
10. A bird flying high in the air appears to be higher than in reality. Explain why?
Ans. Bird flying in air is in the rarer medium and if we see it from denser medium than light form bird refract towards the normal thus appears to come from the higher point. i.e. Apparent height > Real height (BIRD APPEARS TO BE)
11. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Ans. Focal length of the convex lens, 
= 30 cm
Focal length of the concave lens, 
= – 20 cm
Focal length of the system of lenses = f
The equivalent focal length of a system of two lenses in contact is given as:
Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.
12. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Ans. Distance between the object and the image, d= 3 m
Maximum focal length of the convex lens =
For real images, the maximum focal length is given as:
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
13. A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Ans. Distance between the image (screen) and the object, D= 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
Therefore, the focal length of the convex lens is 21.39 cm.
14. A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.
Ans. The power of the spectacles used by the myopic person, P= – 1.0 D
Focal length of the spectacles,
Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.
During old age, the person uses reading glasses of power,
The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.
15. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?
Ans. Focal length of the objective lens,
= 140 cm
Focal length of the eyepiece, 
= 5 cm
Least distance of distinct vision, d = 25 cm
(a) When the telescope is in normal adjustment, its magnifying power is given as:
(b) When the final image is formed at d, the magnifying power of the telescope is given as:
16. Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5°of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
Ans. Angle of deflection, 
Distance of the screen from the mirror, D= 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 
The displacement (d) of the reflected spot of light on the screen is given as:
Hence, the displacement of the reflected spot of light is 18.4 cm.
Short type question answers – 3 Marks
1. Find the radius of curvature of the convex surface of a plane convex lens, whose focal length is 0.3m and the refractive index of the material of the lens is 1.5?
Ans. 
For plane convex lens
R = 0.15 m
2. Show that the limiting value of the angle of prism is twice its critical angle? Hence define critical angle?
Ans. Angle of the prism 
For limiting 
(Maximum)
Value of angle of prism for 
means 
But when = C
Amax = 2C
The angle of incidence for which angle of refraction is 90o is called critical angle.
3. Draw a labeled diagram of telescope when the image is formed at the least distance of distinct vision? Hence derive the expression for its magnifying power?
Ans. magnifying power
=
(since angles are very small)
—-(i)
For eye piece
Multiply by D
Substituting in e.g. (i)
4. Drive the expression for the angle of deviation for a ray of light passing through an equilateral prism of refracting angle A?
Ans. At the surface AB
At the surface AC
—-(1)
In quadrilateral AQOR
—-(2)
Now in 
Or
—-(3)
From (2) and (3)
—-(4)
Substituting equation (4) in equation (1)
= i + e – A
Or
A + = i + e
5. Draw a ray diagram to illustrate image formation by a Newtonian type reflecting telescope? Hence state two advantages of it over refracting type telescopes?
Ans. Advantages
(1) The image formed in reflecting type telescope is free from chromatic aberrations
(2) The image formed is very bright due to its large light gathering power.
NEWTONIAN TELESCOPE (REFLECTING TYPE)
6. The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eye piece is 101cm. calculate the focal length of the objective and the eye piece.
Ans. 
—-(1)
fo = 100 fe —-(2)
Substituting equation. (2) in equation (1)
fe + 100fe = 101
101fe = 101
fe = 1cm
Substituting fe in equation (2)
7. A convex lens made up of refractive index n1 is kept in a medium of refractive index 
. Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if
(1) 
(2) 
(3) 
Ans. (1) When the lens behaves as a convex lens.
(2) When the lens behaves as a plane plate so no refraction takes place
(3) When the lens behave as a convex lens
8. Derive the relation 
Where 
are focal lengths of two thin lenses and F is the focal length of the combination in contact.
Ans. Consider two thin lenses in contact having focal length For the first lens
For the second lens 
acts as an object which forms the final image I
Adding equation (1) & (2)
Using lens formula
For n no. of thin lenses is contact
9. A convex lens has a focal length 0.2m and made of glass is immersed in water 
find the change in focal length of the lens?
Ans. Fair = 0.2m, a
g = 1.50
—-(1)
Now 
Where too is the focal length of the lens when immersed in water?
10. A reflecting type telescope has a concave reflector of radius of curvature 120cm. calculate the focal length of eye piece to achieve a magnification of 20?
Ans. M = 20
R = 120cm (fro concave reflector)
fe = 3cm
11. Show that a convex lens produces an N time magnified image, when the object distances from the lens have magnitude 
. Here f is the magnitude of the focal length of the lens. Hence find two values of object distance, for which value of u convex lens of power 2.5 D will produce an image that is four times as large as the object?
Ans. Magnifying power
For real image m = -N
Or
—-(1)
For virtual image m =N
Or
—-(2)
From equation (1) & (2) we can say that magnification produced by a lens can be N if u
=
Now power of a lens = 2.5 D
m equation (1)
u = 30cm or -50cm
12. Define total infernal reflection of light? Hence write two advantages of total reflecting prisms over a plane mirror?
Ans. The phenomenon of reflection of light when a ray of light traveling from a denser medium is sent back to the same denser medium provided the angle of incidence is greater than the angle called critical angle is called total internal reflection.
Advantages
1. It does require silvering
2. Multiple reflections do not take place in a reflecting prism due to this; only one image is formed, which is very bright.
13. An equi–convex lens of radius of curvature R is cut into two equal parts by a vertical plane, so it becomes a plano-convex lens. If f is the focal length of equi–convex lens, then what will be focal length of the plano –convex lens?
Ans. We know
For equi –convex lens 
—-(1)
For plano convex lens
—-(2)
From (1) & (2)
Or
f’’ = 2f
14. A converging lens of focal length 6.25cm is used as a magnifying glass if near point of the observer is 25cm from the eye and the lens is held close to the eye. Calculate (1) Distance of object from the lens. (2) Angular magnification and (3) Angular magnification when final image is formed at infinity.
Ans. (1) 
u = -5cm
(2) 
m = 5cm
m = 4
15. Draw a graph to show that variation of angle of deviation Dm with that of angle of incidence i for a monochromatic ray of light passing through a glass prism of refracting angle A. hence deduce the relation?
Ans. For the minimum deviation position
(Say)
We know =
Also 
= A
Or 2r = A
Applying minimum deviation condition is equation. (1)
2i = 
Applying Snell’s law
Or
16. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Ans. Size of the object, 
= 3 cm
Object distance, u= – 14 cm
Focal length of the concave lens, f = – 21 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
The magnification of the image is given as:
Hence, the height of the image is 1.8 cm.
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.
17. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Ans.In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12 cm
(a) Focal length of the convex lens, f= 20 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed 7.5 cm away from the lens, toward its right.
(b) Focal length of the concave lens, f= –16 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed 48 cm away from the lens, toward its right.
18. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?
Ans. Refractive index of glass,
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens =
Radius of curvature of the other face of the lens = 
Radius of curvature of the double-convex lens =R
The value of R can be calculated as:
Hence, the radius of curvature of the double-convex lens is 22 cm.
19. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Ans. Focal length of the objective lens, 
= 144 cm
Focal length of the eyepiece, 
= 6.0 cm
The magnifying power of the telescope is given as:
The separation between the objective lens and the eyepiece is calculated as:
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
20. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 
, and the radius of lunar orbit is
.
Ans. Focal length of the objective lens,
= 15 m =
cm
Focal length of the eyepiece,
= 1.0 cm
(a) The angular magnification of a telescope is given as:
Hence, the angular magnification of the given refracting telescope is 1500.
(b) Diameter of the moon,
d= 
Radius of the lunar orbit,
r0=
m
Let 
be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm
22. A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
Ans. (a) Focal length of the magnifying glass, f = 5 cm
Least distance of distance vision, d = 25 cm
Closest object distance = u
Image distance, v= – d= – 25 cm
According to the lens formula, we have:
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distant (u‘), the image distance 
According to the lens formula, we have:
Hence, the farthest distance at which the person can read the book is
5 cm.
(b) Maximum angular magnification is given by the relation:
Minimum angular magnification is given by the relation:
23. A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Ans. (a) Area of each square, A= 1 mm2
Object distance, u= – 9 cm
Focal length of a converging lens, f = 10 cm
For image distance v, the lens formula can be written as:
Magnification, 
∴Area of each square in the virtual image = (10)2A
= 102 ×1 = 100 
= 1 
(b) Magnifying power of the lens 
(c) The magnification in (a) is not the same as the magnifying power in (b).
The magnification magnitude is 
and the magnifying power is
.
The two quantities will be equal when the image is formed at the near point (25 cm).
24. (a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
Ans. (a) The maximum possible magnification is obtained when the image is formed at the near point (d= 25 cm).
Image distance, v= – d= – 25 cm
Focal length, f= 10 cm
Object distance = u
According to the lens formula, we have:
Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.
(b) Magnification =
(c) Magnifying power =
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
25. What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Ans. Area of the virtual image of each square, 
Area of each square, 
Hence, the linear magnification of the object can be calculated as:
But m =
…(1)
Focal length of the magnifying glass, f= 10 cm
According to the lens formula, we have the relation:
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
26. (a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
Ans. Focal length of the objective lens, = 140 cm
Focal length of the eyepiece, = 5 cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece 
(b) Height of the tower, h1= 100 m
Distance of the tower (object) from the telescope, u= 3 km = 3000 m
The angle subtended by the tower at the telescope is given as:
The angle subtended by the image produced by the objective lens is given as:
Where,
= Height of the image of the tower formed by the objective lens
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
(c) Image is formed at a distance, d= 25 cm. The magnification of the eyepiece is given by the relation:
Height of the final image
Hence, the height of the final image of the tower is 28.2 cm.
Long term question answer – 5 Marks
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. 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:
14. (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
15.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.
16.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 
17.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.
18.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.
19.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.
20.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.
21.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:
