# Class 12 Physics NCERT Solutions Chapter 9 Ray Optics

Class 12 Physics Chapter 9 “Ray Optics and Optical Instrumentation ” NCERT Solutions are available here for students and teachers for education purpose.. These Class 12 Physics NCERT Solutions Chapter 9 are as per the latest syllabus as prescribed by CBSE

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## Class 12 Physics Chapter 9 “Ray Optics and Optical Instrumentation”

Class 12 physics Chapter 9 is “Optics”. The ninth chapter of Class 12 Physics is very useful for exam point of view. This chapter consists of the topics – Optics, Reflection, Refraction, Lens Formula, Double Lens Formula, etc

### Class 12 Physics NCERT Solutions Chapter 9 “Ray Optics and Optical Instrumentation”

Question 9.1 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?

According to Question
Size of candle h1 = 2.5cm
Object Distance u = -27 cm
R = -36 cm
f = R/2 = -36/2 = – 18
Image distance = ?
As we know
1/u + 1/v = 1/f
therefore putting values.
We get ,
v = -54 cm

m = -h2/h1 = v/u
-h2/2.5 = -54/ -27 = 2
Size of Image = h2 = -5cm
Image is real and inverted

When the candle is moved closer to mirror , Screen has to move away from the mirror.

Class 12 Physics NCERT Solutions Chapter 9 Question 9.2
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.

According to question ,
h1 =4.5 cm
u = -12 cm

f = 15 cm
v =? m =?
We know that
1/u + 1/v = 1/f
v = 6.7 cm

m = h2/h1 = -v/u
h2/4.5 =-6.7/12
h2 = 2.5 cm
Image is virtual and erect.
When needle is moved away from the mirror, then the image moves away from the mirror till focus (F) , and its size keeps on decreasing

Question 9.3 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?

According to question ,
real depth of the tank = 12.5 cm
apparent depth of the tank = 9.4 cm
μ = ?
We know ,
μ = real depth / apparent depth
μ = 12.5/9.4 = 1.33

When water is replaced by the liquid of refractive index 1.63 ,then
1.63 = 12.5 /y
y = 12.5 /1.63 = 7.67

Question 9.4 Figures 9.31(a) and (b) show refraction of a ray in air incident at 60° 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 45° with the normal to a water-glass interface [Fig. 9.31(c)].

In fig (a) , i = 60 and r = 35
aμg = sin i / sin r = sin 60 / sin 35 = 0.8660/0.5736 = 1.51

In fig(b) , i = 60 and r = 41,
aμw = sin i / sin r = sin 60 / sin 41 = 0.8660/0.7314 = 1.18

In fig (c) , i=45 and r = ?
aμw = aμg / aμw = sin i / sin r
1.51 /1.32 = sin 45 / sin r = 0.7071 /sin r
sin r = 0.6181
r = 38.2

Question 9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. 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.)

Area of the surface of water through which light from the bub can emerge is the area of the circle of radius r
μ = 1 /Sin i
i = sin-1 1/μ = 48.6
tan i = r / h
r = 0.8 * 1.1345 = 0.907 m

Question 9.6 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 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

According to Question,
δm =40°
μ = ? , A = 60°
Refractive Index of Prism ( μ ) = [Sin ( A + δm )/2 ] / Sin A/2 = Sin 50° / Sin 30°
= 0.766/1.22 = 1.532

When the given prism is placed in water ,
wμg = [sin(A + δm )/2] / Sin A/2
On calculating , we get
δm =10° 20′

Question 9.7 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?

We know that Meu = 1.55
and R1 = R and R2 = R
Also f = 20cm
We have to calculate the value of R .
So using the formula,
1/f = (Meu -1) (1/R1 – 1/R2 )
Putting the values we get ,
R = 20*1.1 = 22 cm

Class 12 Physics NCERT Solutions Chapter 9 Question 9.8
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20cm, and
(b) a concave lens of focal length 16cm?

Since the point ‘P’ is on the right side of the lens then it act as a virtual object, therefore u = 12 cm , v= ?
(a) Convex lens of focal length 20 cm
i.e. f = 20 cm
Using lens formulas ,
1/v – 1/u =1/f,
Putting all values we get ,
v = 7.5 cm
(b) Concave lens of focal length 16 cm
i.e. f = – 16cm
u =12 cm
Using the lens formula , we get
1/v – 1/u = 1/f
Putting the value we get ,
v = 48cm

Question 9.9 An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

According to question,
h1 = 3 cm
u = -14 cm
f = – 21 cm
v = ?
Using the lens formulas,
1/v – 1/u = 1/f ,
Putting all the values we get ,
v = -8.4 cm
Also
m = h2/h1 = v/u
we found , h2 = 1.8 cm
Image features :- Virtual and erect and diminished . and at a distance of 8.4 cm from lens on the same side of object.

When the object is away from the lens then the image moves towards the focus of lens . And the size of image keeps on decreasing

Class 12 Physics NCERT Solutions Chapter 9 Question 9.10
What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

According to Question ,
f1 = 30cm
f2 = -20cm
f= ?

Using the formula,
1/f = 1/f1 + 1/f2
Putting all the values, we get
f = 60 cm

Class 12 Physics Ncert Solutions Chapter 9 Question 9.11 A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm.
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 (25cm), and
(b) at infinity?
What is the magnifying power of the microscope in each case?

According to question,
f0 = 2.0 cm
fe = 6.25 cm
u0 =?
(a) ve = -25 cm
Using the lens formulas we get ,
1/ve – 1/ue = 1/fe
Putting all the values , We get
ue = -5cm
There fore , v0 = 15- ue =10 cm
As distance b/w object and eye piece is 15 cm
therefore
1/uo = 1/v0 – 1/f0
Putting all values we get ,
u0 = -2.5 cm
Magnifying Power = [ v0 / |u0| ] [ 1+ d/fe ] ,
Putting all values we get ,
Magnifying Power = 20

(b) As ve = Infinite, -ue = fe = 6.25 cm
therefore , v0 = 16-6.25 = 8.75 cm uo = ?
Using Lens formula,
1/v0 – 1/u0 = 1/f0
Putting all the values we get ,
u0 =-25.9 cm
Magnifying Power

Class 12 Physics Ncert Solutions Chapter 9 Question 9.12 A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus.
What is the separation between the two lenses?
Calculate the magnifying power of the microscope.

According to question,
d = 25 cm
f0 = 8.0 mm = 0.8 cm
fe =2.5 cm
u0 = -9 m = -0.9 cm
ve = d = – 25 cm
Now,
1/ve – 1/ ue = 1/fe
Putting all the values we get ,
ue = -2.27 cm

Also,
1/v0 – 1/u0 = 1/f0
Putting all the values we get , v0 =7.2 cm

Separation b/w two lens = |ue| +v0 = 2.27 +7.2 = 9.47
Magnifying Power = [ v0 / |u0| ] [ 1+ d/fe ]
Putting all values we get ,
magnifying Power = 88 cm

Class 12 Physics NCERT Solutions Chapter 9 Question 9.13
A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

According to Question,
f1 = 144 cm
fe = 6cm
m = ?
L = ?
Now we know that
m = -f0 / fe = -144/6 = -24
Also
L = f0 + fe = 144+6 = 150 cm

Question 9.14
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm 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 3.48 × 106m, and the radius of lunar orbit is 3.8 × 108m.

According to question ,
f0 = 15m
fe =1 cm = 10-2 m
(a) Angular magnification is
m= -f0 / fe = -15 / 0.01 = -1500
(b) let d = diameter of image
Angle subtended by diameter of moon theta1 = l1/r1 = [3.48 * 106 ]/[3.8 * 108 ]
Angle subtended by image theta2 = l2 /r2 = d/f0 = d/15
Since
theta1 = theta2
Therefore ,
d/15 = [3.48 * 106 ] / [ 3.8 *108 ]
d = 1.73 cm

Class 12 Physics Ncert Solutions Chapter 9 Question 9.15 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.]

Mirror Equation : – 1/v 1/u = 1/f
(a) For Concave mirror ,
f < 0 i.e. f = -ve ,
u < 0 i.e. u = -ve
Also ATQ , Object is between F and 2F ,
Therefore ,
2f<u<f
=1/2f > 1/u > 1/f
= -1/2f <-1/u <-1/f
= 1/f – 1/2f < 1/f – 1/u < 0
= 1/2f < 1/v <0
This implies , 1/v is -ve , so v -s -ve.
Also v >2f, This implies image is beyond 2F

(b) For Convex Mirror
f = +ve , f >0
u = -ve i.e u < 0
As 1/v = 1/f – 1/u
1/v is +ve , v = +ve , Image is virtual

(c) For Convex Mirror f > 0 and u < 0
1/v + 1/ u = 1/f
1/v > 1/f
This implies v < f
Image is between pole and Focus . So image is diminished .

(d) For Concave Mirror,
f<0
object position i between pole and focus
This implies that f<u<0
Therefore , 1/f – 1/u > 0
1/f -1/u = 1/v > 0
v is +ve , Image is virtual
1/v<1/u This implies v> u
Therefore Image is enlarged.

Question 9.16 A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5.
Does the answer depend on the location of the slab?

Let Real Depth be x = 15 cm
Let apparent Depth be y = ?
Meu = 1.5
We know ,
Meu = x/y
Putting the known values we get ,
y = 10 cm
The Distance through which the pin is appeared to be raised = x-y = 15-10 = 5 cm

No, The answer does not depend upon the Location of Slab

Class 12 Physics NCERT Solutions Chapter 9 Question 9.17 (a) Figure 9.32 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?
Meu2 = 1.68
Meu1 =1.44
(a) Since Meu = Meu2 / Meu1 = 1/ Sin C
Sin C = Meu1 / Meu2 = 1.44 / 1.68 = 0.85
C = Sin-1 ( 0.85) = 59 degree.
i should be greater than 59 degree for internal reflection to take place ,
i> C
Also rmax = 90degree – C = 90degree – 59 degree = 31 degree
Sin imax / Sin rmax = 1.68
Putting values we get ,
imax =60degree
Hence all rays must be incident at an angle of angle i , 60 degree for total internal reflection

(b) If no outer coating is provided
therefore , Meu2 = 1.68
Meu1 = 1
Sin C’ = 1/ Meu = Meu1/ Meu2 = 1/1.68 = 0.5952
C’ = Sin-1( 0.59 ) = 36.5Degree
Now i = 90 degree will have r = 36.5 degree
therefore i’ = 90 – 36.5 = 53.5 degree . It is greater than C’.
so all rays incident at angle in zero to 90 degree will suffer total internal reflection.

Class 12 Physics NCERT Solutions Chapter 9 Question 9.18 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?

(a) Yes they produce real image. When incident rays converge behind the mirror , they are reflected to point on a screen in front of mirror. Hence a real image is formed. In this case the object is virtual
(b) No. Because Eye lens is convergent .
(c) Light travel from rare medium to denser medium ( from fisherman to diver in water ) , It will bends towards normal , and will take larger distance to reach the diver .So fisherman is looked taller by the diver in the water

(d) Yes the apparent depth is changed if viewed obliquely . The Apparent depth is decreased when seen obliquely as compared to depth when seen normally
(e) We Know Sin C = 1/Meu
Meudiamond > Mueglass
Critical angle for diamond is b/w 24 degree to 90 degree
This show that when lights enter diamond it suffers total internal reflection
This feature does not show that it is used as diamond cutter . It is the hardest so used as diamond cutter.

Class 12 Physics NCERT Solutions Chapter 9 Question 9.19 The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

For a real image minimum distance b/w object and image should be 4f.
Therefore , 4f = 3 m
f = 3/4 = 0.75 m

Class 12 Physics NCERT Solutions Chapter 9 Question 9.20 A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens.

Distance b/w object and screen , D = 90 cm
Distance b/w two locations of convex lens d = 20 cm
f = ?
We Know that
f = [D2 – d2 ] / 4D
Putting all value we get ,
f = 21.4 cm

Class 12 Physics NCERT Solutions Chapter 9 Question 9.21
(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm 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 40cm. Determine the magnification produced by the two-lens system, and the size of the image.

Answer And Explanation : –

Class 12 Physics NCERT Solutions Chapter 9 Question 9.22 At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face?
The refractive index of the material of the prism is 1.524.
i = ?
A = 60degree
Meu = = 1.524
Meu = 1/Sin C
C = r2
Sin C = Sin r2 = 1/ Meu = 1/ 1.524 = 0.6561
A = r1 + r2
r1 = A – r2 = 60 – 41 = 19 degree

Meu = Sin i1 / Sin r1
On calculating we get
i1 = 29 degree 45′

Class 12 Physics NCERT Solutions Chapter 9 Question 9.23
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.

(a) Area of each square = 1mm2
u = -9 cm
f = 10 cm

Using lens formula,
1/v – 1/ u = 1/f
Putting all values we get ,
v = – 90 cm
Magnification m = v /|u| = 90/9 = 10
Therefore area of each square in the virtual image is found to be 10*10*1 = 100 sq.mm

(b) Magnifying Power =d/u 25/9 = 2.8

(c) No, Magnification in (a) and (b) cannot be equal .
The reason is that image is located at the least distance of distinct vision.

Question 9.24
(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.

(a) v = -25 cm
f = 10 cm
u = ?
Using Lens formula , we get
u = -7.14 cm

(b) m = v/|u| = 25/7.14 = 3.5
Magnifying Power = d/u = 25/7.14 = 3.5

(c) Yes , magnifying power and magnification are same, because the image is formed at the least distance of distinct vision.

Question 9.25 What should be the distance between the object in Exercise 9.24 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 9.23 to 9.25 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Magnification in area = 6.25
Linear magnification m = sqroot(6.25) = 2.5
m = v/u
v =mu =2.5 u
Putting all the values in lens formula, we get
v = -15cm
u = -6 cm
As the virtual image is at 15 cm which is less than the distance of distinct vision which is 25 cm . So image cannot be seen distinctly by eyes

Class 12 Physics NCERT Solutions Chapter 9 Question 9.26 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?

By using magnifying glass , we keep the object far more closer to the eye than25 cm . The closer object has larger angular size than the same object at 25 cm . It is in this sense that angular magnification is achieved.

(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?
Yes the angular magnification changes 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?

it is difficult to grind lenses of very small focal lengths.

(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
This is because m = 1 +d/fe ,
As fe decrease the angular magnification increase.

(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?
This is because When the eye is too close to eye piece, field of view reduces and eye doesn’t collect much light .
The distance between eye and eyepiece depends on focal length of eye piece and separation of object and eyepiece

Question 9.27 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?
Angular magnification of eye piece = 1+d/fe = 1 + 25/5 = 6
According to question ;
Total magnification is 30 ,
magnification of objective lens is
m = 30/6 = 5
Also m = v0 / (- u0) = 5
Therefore , v0 = -5u0
Using mirror formula,
we get
1/v0 + – 1/u0 =1/f0
Putting all values we get ,
u0 = -1.5 cm
v0 = -5 * -1.5 cm
= 7.5 cm

Now again,
1/ve – 1/ue = 1/fe
ue = -4.14 cm
Separation b/w eye piece and objective lens = |ue |+|v0| = 4.17 +7.5 = 11.67 cm

Question 9.28 A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. 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 (25cm)?

According to Question,
f0 = 140 cm
fe = 5 cm
Magnifying power = ?
Magnifying Power = f0 / (-fe ) = 140/-5 = -28
(b) When final image is at the least distance of distinct vision,
Magnifying Power = [-f0 / fe ](1+fe / d )
= [ 140/-5] [1+5/25] = -33.6

Question 9.29
(a) For the telescope described in Exercise 9.28 (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 25cm?

(a) Separation b/w objective lens and eye lens = fo + fe = 140+5 = 145 cm
(b) Angle subtended by tower 100 m tall at 3 km
Alpha = 100 /(3* 1000 ) = 1/30 radian
If h= height of image formed by objective lens
alpha = h/f0 = h/140
h/140 = 1/30
h = 4.7 cm
(c) magnification produced by eye piece = (1+d/fe ) = 1+25/5 = 6
Therefore height of final image = 4.7 * 6 = 28.2 cm

Question 9.30 A Cassegrainian telescope uses two mirrors as shown in Fig. 9.30. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?
R1 = 220 mm
R2 =140 mm
f2 = R2 / 2 = 140/2 = 70mm
d = 20 mm
f1 = R1 / 2 = 220 / 2 = 110mm

For secondary mirror,
u=f1 – d = 110-20 = 90mm
Using mirror formula
1/v + 1/u = 1/f
Putting all values we get,
v =3.15 cm = 315 mm

Question 9.31 Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.33. A current in the coil produces a deflection of 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Let angle SMd be 2theta
Theta = 3.5degree
x = 1.5
d = ?
2Theta = 2*3.5degree = 7degree