SikhLo - Current Affairs in Hindi, Daily current affairs, Current affairs

 Bihar Board Class 12th Physics Solutions Chapter 9 Ray Optics and Optical Instruments

Bihar Board Class 12 Physics Ray Optics and Optical Instruments Textbook Questions and Answers

Question 1
A small candle 2.5 cm in size is placed 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?
Answer:
Here, O = size of the object = 2.5 cm
u = distance of the object from the concave mirror = – 27 cm
R = radius of curvature of concave mirror = – 36 cm
∴ f =R/2 = – -−36236/2= -18 cm

(i) υ – distance of the image from the mirror = ?

υ = – 54 cm.
The negative sign shows that the image is formed in front of the mirror i.e., on the side of the object itself. Thus the screen must be placed at a distance of 54 cm in front of the mirror.

(ii) Nature and size of the image = ?, I = ?
Using the relation,
m = IO = –υu
We get,
I2.5=−(−54)(−27)
or I = (2.5) x (- 2) = – 5 cm.
Thus clearly the image is real, inverted and magnified.

(iii) If the candle is moved closer to the mirror, then the screen is to be moved farther and farther. But when the distance of the candle is lesser than the focal length of the mirror, as u → f, υ → ∞, for u < f, then the image would be virtual, so it cannot be collected on the screen i.e., no screen is required.

Question 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.
Answer:
Here, u = distance of the needle from the convex mirror = – 12 cm
f = focal length of the mirror – + 15 cm.
O = size of the object = 4.5 cm
v = location of the image = ?
Using the relation,
1f=1u+1v , we get
1v=1f⋅1u=115−(−1−12)
= 4+560=+960
∴ υ = 203 cm = + 6.67 cm. = 6.7 cm
The positive sign shows that the image is formed behind the mirror.
I = size of image = ?
using the relation,

The positive sign shows that the image formed is erect, virtual and of reduced size.
Also, magnification m is given by
m = 203
As the needle is moved farther from the mirror, the image moves towards the focus (upto the focus only) and gets smaller and smaller in size i.e. as u → ∞, υ →f (but never beyond) while m → O.

Question 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?
Answer:
Case I: When tank is filled with water
Here, real depth = 12.5 cm
Apparent depth = 9.4 cm
μ = Refractive index of water = ?
We know that

Case II. When the tank is rilled with the liquid :
Here, μ = R.I of the liquid = 1.63.
real depth = 12.5 cm
apparent depth = ?
real depth
∴ Using the relation,

the distance through which the microscope has to be moved up = 9.4 – 7.67 = 1.73 cm = 1.70 cm.

Question 4.
Figures (a) and (b) show refraction of an incident ray in air 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 at 45° with the normal to a water-glass interface [Fig. (c)].

Answer:
i = angle of incidence = 60°
r = angle of refraction = 35°
aμg= R.I. of glass w.r.t. air =?

Using the relation,

From fig. (b), here, i = 60°, r = 47°
aμw = R.I. of water w.r.t. air = ?
We know that

Question 5.
A small bulb is placed at the bottom of 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.)
Answer:
Here, source of light (O) is 80 cm below the surface of water i.e., OO’ = 80 cm = 0.80 m.
aμ2 = 1.33
The ray of light emitted from O will be refracted into air only if the angle of incidence is less than the critical angle C. When the angle of incidence is equal to critical angle C, the light will not be refracted along water surface into air, but it will graze the air water interface. Thus, the light will appear to come out of a cone having vertex angle 2C. If the angle of incidence at water-air interface is more than C, then the rays of light will be totally reflecfed.
∴When i = C, r = 90°
∴ Area of the surface of water through which light from the bulb can emerge out is the area of the circle of radius O’A = (O’B) i.e., r’ = AB2 = O’A = O’B


Now in ∆ OO’B,
tan C = O′BO′O
or O’B = OO’ tan C
= 0.80 x tan 48. 6°
= 0.8 x 1.1345 = 0.907 m = 90.7 cm
∴ Area of the surface of water through which light will emerge
= πra = 3.14 x (0.907)2
= 2.584 m2 = 2.6 m2

Question 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.
Answer:
Here, A = angle of prism = 60°
δm = angle of minimum deviation = 40°
μ = R.I. of the material of prism = ?
Using the relation,

Let δm be the new angle of minimum deviation of a parallel beam of light when the prism is placed in water. Here aμw = 1.33, aμg = 1.532

Question 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 20 cm?
Answer:
μ = R.I of glass = 1.55
f = focal length of the lens = 20 cm
R1 = R, R2 = – R where R1 and R2 are the radii of curvatures of the two faces forming the double convex lens. R = ?
Using the relation

or R = 1.10 x 20 = 22 cm.

Question 8.
A beam of light converges to a point P. Now a lens is placed in the path of the convergent beams 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm?
Answer:
Here, the point P on the right of the lens acts as a virtual object,
∴ u = + 12 cm, v = ?
(a) For convex lens, f = + 20 cm.

Using lens formula,

The image formed is 48 cm on the right of the lens where the beam would converge and is real.

Question 9.
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?
Answer:
Here, O = size of object = 3.0 cm.
v = distance of object from concave lens = -14 cm.
f = focal length of the concave lens = – 21 cm v = ?, I = ?
Using the relation

∴the image is virtual, erect and located at 8.4 cm from the lens on the same side as the object.
Also we know that

i.e., the image is of diminished size.
As the object is moved away from the lens, the virtual image moves towards the focus of the lens (but never beyond focus). The size of the image progressively goes on decreasing i.e. as u → ∞, v → f but never beyond f while m → 0.6

Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave length of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer:
Here, f1 = focal length of the convex lens = + 30 cm.
f2 = focal length of concave lens = – 20 cm
Let f = focal length of the combination of two lenses in contact = ?
We know that
1 F=1f1+1f2

As the focal length of the combination of two lenses is – ve, so the combination behaves as a diverging lens i.e., as a concave lens.

Question 11.
A compound microscope consists of an objective lens of focal length 2.0 cm and an eye-piece 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?
Answer:
Here, f0 = focal length of the objective lens = 2.0 cm
fe = focal length of the eye-piece = 6.25 cm
L = distance between two lenses = 15 cm.
(a) ve = – 25 cm = distance of the final image from E- lens
u0 = distance of the object from O-lens = ?

Using lens formula,
1f=1v=1u we get for eve-piece,

We know that the magniíyiig power of the compound microscope is given by

(b) The final image will be formed at infinity only if the image formed by the objective is in the focal plane of the eye piece i.e., at principal focus of the eye piece.
Thus here, ve = – ue = fe= 6.25 cm.
|v0| = L – |ue| = 15- 6.25 = 8.75 cm

Question 12.
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eye-piece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Answer:
Here, D = 25 cm
f0 = focal length of objective = 8.0 mm = 0.8 cm.
fe = focal length of eye-piece = 2.5 cm
u0 = – 9.0 mm = – 0.9 cm.
L = separation between two lenses = ?
M = ?
For eye piece, using lens formula, we get

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eye-piece of local length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?
Answer:
Here, f0 = focal length of the objective of the telescope = 144 cm.
fe = focal length of the objective of the telescope = 6.0 cm.
M = magnifying power of telescope = ?
L = separation between the objective and the eye-piece = ?
(i) In normal adjustment (i.e. when the final image is formed at co)
M = – f0fe = 1446 = -24 ,M = 24
∴ L = f0 + fe = 144 + 6 = 150 cm.

Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye-piece 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 3.48 x 108 m, and the radius of lunar orbit is 3.8 x 106 m.
Answer:
Here, f0 = focal length of the objective of the telescope = 15 m.
fe = focal length of the eye piece = 1.0 cm = 10-2 m.
(a) Angular magnification = ?
We know that Angular magnification = – f0fe
= – 1510−2 = – 1500
(b) Let d be the diameter of the image of the moon formed by the objective lens.
∴ angle subtended by the image = df0=d15 ….(i)
Also we know that angle subtended by the diameter of the moon

Question 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 to deduce algebraically properties of images that one obtains from explicit ray diagram.]
Answer:
(a) Here, f < u < 2f (given), υ = ?
Using mirror formula, 1u+1v=1f, we get [w and f both – ve and |u| > f]
Suppose u here is 32 f

∴ 2f < υ < ∞ ⇒ υ is between 2f and infinity which means that the image is formed beyond 2f.

(b) For convex mirror :

∴ uf is – ve and u – f is also – ve.
∴ υ is always + ve or image formed by convex mirror is always virtual and thus is independent of the location of the object. Now for convex mirror, f > 0. Also we have u < 0 (object lies on left)
∴ 1υ < 0 or υ > 0 or υ = + ve image is formed on right and is virtual whatever may be the value of u.

(c) From (b) υ = ufu−f
and m = −vu(−fu−f)
|u – f| is always grater than f ( ∵ u is – ve ) and ∣∣fu−f∣∣ < 1 and fu−f is -ve is sign. i.e., as f > 0 for convex mirror and u < 0, ∴ 1υ > f i.e., υ < f (image located between the pole and the focus)
Also υ < |u| (image diminished).
∴ m is + ve and less than one.
Therefore image produced by convex mirror is virtual and diminshed in size.

(d) From (a) for concave mirror
1υ = u–fuf
|u and f both – ve| i.e. f < 0 for concave mirror.
Here, u is less than f and is equal to f/2.
∴ u = −f2

or υ = f
∴ υ > |u| and υ is + ve. or υu > 1
∴ m = – υu = + ve and > 1
Thus the image formed is virtual and magnified.
Aliter :
f < u < 0 implies 1f – 1u > 0 or – 1υ > 0, or υ > 0 image is 1 1
formed on right and is virtual. Also 1u < 1|u|, ∴υ > |u| ⇒ image is enlarged.

Question 16.
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Answer:
Here, t = thickness of glass slab = 15 cm.
p = R.I of glass = 1.5.
The normal shift in the position of the pin is given by
d = t (1−1μ)
where t = real depth of pin = thickness of slab
Let x = apparent depth of pin
∴ d = Normal shift = t – x = ?
Also µ = tx or x = tµ
∴ d = normal shift = t – tu = t (1−1μ)
= 15 (1−11.5) = 15 (1−23)
= 15 x 13 = 5 cm,
i.e. the pin appears raised by 5 cm.
No, the answer does not depend upon the location of the slab for small angles of incidence,

Question 17.
(a) Figure 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?

Answer:
(a) Here aµg = Refractive index of the glass fibre w.r.t. air = 1.68.
aµc = Refractive index of the outer coating material w.r.t. air = 1.44.
If cµg be the Refractive index of the glass w.r.t. outer coating, then gµc = aμcaμg=1.441.68
∴ cµg = 1gμc=1.681.44 = 1.67
Let C be the critical angle for the fibre material w.r.t. the material of the outer coating.
∴ sin C = 1cμg=11.167 = 0.8569
∴ C = sin-1 (0.8569) = 58.97° = 59°
Total internal reflection will take place when i > C i.e. i > 58.97° = 59°
or when r < rmax
Where rmax = 90 – C = 90° – 59° = 31°.
Now refractive index of glass-fibre w.r.t. air is given by .

Thus all the rays which are incident in the range 0 < i < 60° will suffer total internal reflection in the pipe.

(b) If there is no outer coating of pipe, then refraction inside the pipe shall take place from glass to air.
∴ cµg = 1.68, aµa = 1.
∴ Sin C’ = 1μ=aμaaμg=11.68 = 0.5952
∴ C’ = 36.5°.
(∵ sini = µ = 1.68).
∴ rmax = 90° – 36.5° = 53.5°
which is greater than C’.
Thus all rays incident at angles in the range O to 900 with the axis
will suffer total internal reflection from inside the pipe.

Question 18.
Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real ¡mages 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 fisher man 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?
Answer:
(a) Yes, plane and convex mirrors can produce real images. When, the rays incident on them (plane or convex mirrors) are converging to a point behind the mirror, they are reflected to a point on a screen infront of the mirror. In other words, a plane or convex mirror can produce a real image if the object is virtual.

(b) No, there is no contradiction. It is due to the fact that the virtual image formed by a spherical mirror acts as virtual object for eye lens which in turn forms real image on the retina of the eye because eye lens is convergent lens.

(c) The man would look taller to the diver than what he actually is. As the man is in air, so light travels from rarer to denser medium and bends towards the normal and it appears to be coming from a larger distance as shown in the figure here. The size of light BP and BQ from the head (B) of the fisherman AB, on refraction at the water-air interface, bend towards the normal at points P and Q and appear to be coming from B’ to the diver. Clearly AB’ > AB. Where AB’ is the image of fisherman AB.

(d) Yes, the apparent depth of a tank of water changes if viewed obliquely. The apparent depth decreases when water tank is viewed obliquely as compared to the depth of the tank when seen normally.

(e) Yes. Refractive index of diamond is 2.4 which is much larger than the refractive index of ordinary glass (p = 1.5). The critical angle for diamond is 24.4° which is much smaller than for ordinary glass (= 90°), A skilled diamond cutter exploits the large range of angles of incidence in diamond 24° to 90° to ensure that light entering the diamond on any face at an angle of incidence more than 24°, it suffers total reflection from many faces before getting out. This produces sparkling effect in diamond.

Question 19.
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?
Answer:
Here, u + v = 3m, ∴ υ = 3 – u.
From lens formula,

For real solution, 9 -12 f should be positive i.e., 9 -12 f shouldbe positive
or 9 – 12 f > 0
9 > 12 f.
or f < 912 < 34 m.
∴ The maximum focal length of the lens required for the purpose is 34 m i.e. fmax = 0.75 m.
Aliter :
u + v = 3 or v = 3 – u = 3- (-v) = 3 + u ……(1) (using sign conventions)
From lens formula,

Question 20.
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.
Answer:
Here, O is position of object and I is position of image (screen). Distance Ol = 90 cm.
L1 and L2 are the two positions of the lens.
∴ Distance between L1 and L2 = O1 O2 = 20 cm.
For position L1 of the lens—Let x be the distance of the object from the lens.
∴ u1 = – x
∴ distance of the image from the lens, v1 = + (90 – x)

For position L2 of the lens—Let u2 and v2 be the distances of the object and image from the lens in this position.
∴ u2 = – (x + 20),
∴ υ2 = + [90 – (x + 20)] = + (70 – x)
Using lens formula,

Question 21.
(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 above 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.
Answer:
Here, as per Q. 9.10,
f1 = + 30 cm (focal length of convex lens)
f2 = – 20 cm (focal length of concave lens)
d = separation between the lenses = 8 cm
Let F = effective focal length of the combination = ?
Using the relation,

Thus as F is – ve, so the combination of these lenses acts as a diverging lens.

(i) Now let a parallel beam of light be incident on the lens of the combination.
∴ f1 = + 30 cm, u1 = – ∞, υ1 = ? (position of the image formed by convex lens).

Thus the parallel incident seam appears to diverge from a point (220 cm – 4) = 216 cm from the centre of two lens system.

(ii) Now Let the parallel beam be incident from Left on the concave Lens first,
∴ u1 = – oo, f1 = – 20 cm, υ1 = ?
∴ −1u+1v=1f gives

a distance of (420 – 4) = 416 cm from the centre of two lens system.
Thus we conclude that the answer depends on which side a parallel beam of light is incident as the distances are different in the two cases. Also the notion of effective focal length is not useful in this case.

(b) Here O = 1.5 cm, 1 = size of image = ?
u1 = distance between the object and convex lens = – 40 cm
M = magnification produced by the combination = ?
f1 = + 30 cm, υ1 = ?
∴ Using the relation, 1u+1v=1f, we get

If m1 be the magnification produced by the convex lens, then m1.
v1|u1|=12040 = 3
For concave lens :

If m2 be the linear magnification produced by concave lens, then

Also we know that
M = m1 x m2
= 3 x (523) = 1523 = 0.652
i.e. net magnitude of magnification = 0.652
Also M = IO
or I = M x O = 1523 x 1.5 = 4546 cm
= 0.98 cm.

Question 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.
Answer:
Here, A = angle of prism = 60°
μ = refractive index of the material of prism = 1.524.
A ray of light PQ incident on the face AB will suffer total internal reflection at the other face AC when it gets incident on the face AC at an angle of incidence equal to the critical angle for the material of the prism. Let C = critical angle for the material of the prism.
∴ sin C = 1μ = 11.524 = 0.6562
∴ C = sin-1 (0.6562) = 41°.
We know that for a prism,
A = r1 + r2 ”
Here r2 = C = 41°, A = 60°
∴ 60° = r1 + 41° or r1 = 60 – 41 = 19°.
For the face AB,
i = angle of incidence = ?
r1 = 19°, m = 1.524 .

Question 23.
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion, (b) disperse (and displace) a pencil of white light without much deviation.
Answer:
When a beam of white light is incident on a prism, the emergent beam is dispersed and is deviated from its original path. When two prisms of different material and angles are combined so as to produce dispersion without deviation then such a combination is called direct vision prism. On the other hand if the two prisms combined produce deviation without dispersion, then such a combination is called achromatic prism.

(a) Condition of deviation without dispersion for prism combination:
We know that θ = δv – δr = (μv – μr) A …..(i)
and θ’ = δ’v – δ’r = (μ’v – μ’r) A’ …..(ii)
For no dispersion produced by the combination θ + θ’ = 0 i.e., dispersion produced by the two prisms are equal and opposite.
or (μv – μr) A+ (μ’v – μ’r) A’ = 0.
or A′A=−μv−μrμv−μr ………(iii)
– ve sign shows that the two prisms have to be placed in opposite manner. Eqn. (iii) is the condition for no dispersion or condition for achromatism.
Eqn. (iii) can also be expressed as :

(b) The condition of dispersion without deviation by a combination of two prisms of different materials and different angles:
Let A, μ be the angle of prism and R. I. for mean colour of the crown glass.
Also A’ be the angle of prism for flint glass having R.I. μ’ for mean colour.
If δ, δ’ be the deviation suffered by mean light through crown and flint glass prisms respectively.
∴ δ = (μ – 1) A …(i)
and δ’ = (μ’ – 1) A’ …(ii)
If the combination does not produce any deviation, then δ + δ’ = 0 i.e., the deviations produced by the two prisms are equal and opposite,
or (μ – l)A + (μ’- l)A’= 0
A′A=−μ−1μ′−1]
The negative sign shows that the two prisms have to be placed in opposite manner.
eqn. (iii) is the required condition for no deviation.
Let θ and θ’ be the angular dispersions produced by crown and flint glasses respectively.
θ = δv – δr = (μv – μr) A
and θ’ = δ’v – δ’r =(μ’v – μ’r) A’
If θ + θ’ be the net dispersion produced, then

Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm 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 accomodation (i.e. the range of converging power of the eye-lens) of a normal eye.
Answer:
Power of cornea = 40 dioptre
And least converging power of eye lens = 20 dioptre
For far points situated at infinity least converging power = 40 + 20 = 60 dioptre.
f = focal length = 53 cm [∵fm = 1P=160 m = 10060
Here u = – ∞
By Lens formula – 1u+1v=1f
υ = 53
To focus the object at the near point :

∴ Power of eye lens = 64 – 40 = 24 dioptre.
Range of accommodation of eye lens = 20 dioptre to 24 dioptre.

Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accomodation? If not, what might cause these defects of vision?
Answer:
No, short – sightedness (Myopia) or long-sightedness (Hypermetropia) do not necessarily mean that eye has partially lost its
ability of accommodation. Myopia arises when the eye can see near by objects clearly but distant objects are not seen clearly due to the shift of the farthest point towards the eye and thus it becomes difficult to see beyond certain limit. This defect may be due to either the eye ball has become large or focal length of the eye lens has become too small. So the ray coming from infinity are focussed at a near point. In long sightedness, one cannot see near by objects clearly. This defects may be due to either the eye ball has become short or the focal length of the lens has become too large.

Question 26.
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.
Answer:
Here, P = -1D, ∴ f = 1p = 1−1 = – 100 cm
For normal vision, far point is at infinity.
∴ u = – ∞, υ = ?
Using lens formula,
1u+1v=1f
we get
−1−∞+1v=−1100
or 1υ = – 1100
∴ υ = -100 cm
Thus virtual image of the object at infinity is produced at 100 cm distance using spectacles.
To view objects at distance 25 cm to 100 cm, the person using ability of accomodation of eye which is partially lost in old age for which he needs another spectacles having power.
P = + 2 D,
∴ f = 1p = 12 = 0.5 m = 50 cm, v = 25 cm, u = ?
Using eqn. (1), we get

His near point shifts to 50 cm.

Question 27.
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
Answer:
The person looking at a mesh of crossed wires is able to see vertical wire more clearly than horizontal wires due to imperfect spherical nature of eye lens. As a result the focal length of this lens become different in two mutually perpendicular directions and eye cannot see objects in these two directions clearly simultaneously. This defect is directional and is removed by using cylindrical lenses in a particular direction. This defects is called astigmatism. The defect (called astigmatism) arises because the curvature of the cornea plus eye-lens refracting system is not the same in different planes.

[The eye- lens is usually spherical i.e., has the same curvature on different planes but the cornea is not spherical in case of an astigmatic eye.] In the present case, the curvature in the vertical plane is enough, so sharp images of vertical lines can be formed on the retina. But the curvature is insufficient in the horizontal plane, so horizontal lines appear blurred. The defect can be corrected by using a cylindrical lens with its axis along the vertical. Clearly, parallel rays in the vertical plane will suffer no extra refraction, but those in the horizontal plane can get the required extra convergence due to refraction by the curved surface of the cylindrical lens if the curvature of the cylindrical surface is chosen appropriately.

Question 28.
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?
Answer:
(a) u = ?
v = – 25 cm
f = 5 cm
Using lens formula, 1u+1v=1f , we get
−1u+125+15=1+525=625
∴ u = – 256 = – 4.167 cm. = – 4.2 cm
Thus closest distance at which he can read the book is – 4.2 cm. For the farthest distance,
u’ = ? v’ – oo, f = 5 cm
Using lens formula, we get’

which is the farthest distance.

(b) Maximum angular magnification (magnifying)
vu=d|u|=25256 = 6
(∴ υ = d = 25 cm = distance of distinct vision). Minimum angular magnification is at the farthest distance.
= v|v′|=255

Question 29.
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.
Answer:
Here, size of each square in the figure, = 1 mm2 = area of object
∴ O = size of object = 1mm
u = distance of object from convex lens = -9 cm
f = focal length of magnifying glass = +10 cm:

(a) m = magnification produced by the lens = ?
A = area of each square in virtual image = ?
For a lens,
M = υu ……..(1)
Now v can be calculated using lens formula,

∴ from (i),
m = −90−9 = 10
Now each square in the figure has area 1 mm2 i.e., 1 mm x 1 mm. Since the lens produces a linear magnification of 10, thus size of each square in the virtual image will appear as (10 x 1 mm) x (10 x 1 mm)
or A = 100 mm2.

Aliter :
m = IO,
∴ I = m x O = 10 x 1 mm2 = 10 mm.
∴ Area of each square in virtual image = (10 mm)2 = 100 mm2.

(b) Least distance of distinct vision, D = – 25 cm.
∴ M = Angular magnification – magnifying power = Du=−25..9
or M = 259 = 2.8

(c) No, the magnification produced in (a) (=vu) cannot be equal to the magnifying power (=Du)unless υ = D i.e., the image is located at the least distance of distinct vision.

Question 30.
(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.
Answer:
(a) Here, maximum magnifying power is obtained when the virtual image is formed at the least distance of distinct vision.
υ = – 25 cm, f = + 10 cm, u = ?
Using lens formula 1u+1v=1f , we get

(b) m – magnification = ?
Using the relation, m = υu, we get
m = −25(−507)=25×750
or m = 72 = 3.5

(c) Magnifying power, M = ?
When the object is placed, so that the image is formed at the least distance of distinct vision, then M is given by
m = 1 + Df = (1+2510) = 3.5
Also M = D|u|=25(507) = 3.5
Yes, the magnification is equal to magnifying power in this case because image is formed at the least distance of distinct vision.

Question 31.
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.]
Answer:
Here, area of object = 1 mm2, O = size of object = 1 mm.
Area of image = 6.25 mm2,
∴ I size of image = – 6.25−−−−√ mm.
m = linear magnification is given by,
m = IO=6.25√1 = 2.5
Also we know that m = υu
∴ 2.5 = υu
or υ = 2.5 u; f = 10 cm.
Using lens formula,

Hie image will be formed at 15 cm from the lens. Since the least distance of distinct vision is 25 cm so the square will not be seen distinctly by the eye when the eye is held close to the magnifier.

Question 32.
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 eye piece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eye piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eye-piece?
Answer:

(a) It is true that 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 i.e., angular size of the image is equal to the angular size of the object. By using magnifying glass, we keep the object far more closer to the eye than at 25 cm. The closer object has larger angular size than the same object at 25 cm. It is in this sense that a magnifying lens produces angular magnification.

(b) Yes, the angular magnification changes if the eye is moved back and it (angular magnification) decreases a little. It is because, now the angle subtended at the eye by the image is less than the angle subtended by the image at the lens. Tire angle subtended by the object at the eye is also less than that subtended at the lens but the difference is very small. Of course this effect is negligible when the image is at very-very large distance.

(c) Theoretically it is true, but when we decrease focal length, abberations both spherical as well as chromatic become larger. Moreover it is difficult to manufacture lenses of very-very short focal length.

(d) We know that the magnifying power of a compound microscope is given by
M = Lf0(1+Dfe)
where 1 + Dfe = magnification of eye lens
and Lf0=−v0u0 = magnification of objective.
D = least distance of distinct vision.
L = Distance between objective and eye-piece i.e., length of microscope tube f0 and f are the focal lengths of objective and eye piece. Thus for M to be large, both f0 and fe must be very small.

(e) The eye-piece produces the image of the object lens itself in the eye and is called eye-ring. An important point about the eye ring is that all the rays from the object refracted by the objective pass through it. Thus ideal position for our eyes for viewing is this eye ring only. The eye will receive all the rays from the object if it is placed at the position of the eye ring, provided area of the pupil of the eye is greater than or atleast equal to the area of the eye ring. In case eye is placed closed to the eye piece, then it would not collect all the rays from the object as the field of view will be reduced. The precise i.e. exact location of the eye ring would depend upon the sepration between the objective and eye piece and also on the focal length of the eye piece

Question 33.
An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eye-piece of focal length 5 cm. How will you set up the compound microscope?
Answer:
Here, f, = focal length of the objective = 1.25 cm.
fe, = focal length of the eye piece – 5 cm
M = angular magnification of the compound microscope = 30
In normal adjustment, the final image is formed at the least distance of distinct vision i.e., D = 25 cm.
∴ For eye piece, Me = 1 + Dfe = 1 + 255 = 1 + 5 = 6.
Let M0 = angular magnification of the objective.
∴ Using the relation, M = M0 x Me we get
M0 = MMe=306 = 5
∴ M0 = −v0−u0 gives −v0−u0 = 5 or υ0 = – 5 υ0
Using lens formula for objective,

∴ The compound microscope must be set up such that distance between objective and eyepiece is

Question 34.
A small telescope has an objective lens of focal length 140 cm and an eye-piece 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)?
Answer:
Here, f0 = focal length of objective = 140 cm.
fe = focal length of eyepiece = 5.0 cm
(a) When the telescope is in normal adjustment, the magnifying power is given by :
M = f0|fe|=+1405 = 28
(b) When the final image is formed at the least distance of distinct vision, then M is given by :

Question 35.
(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eye-piece?
(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?
Answer:
Here f0 = 140 cm
fe = 5 cm
(a) Separation between the objective and eye-lens is given by
L = f0 + fe = 140 + 5 = 145 cm.

(b) Let 0 be the angle made by 100 m tall tower AB at the point of observation O.
Here h = 100 m.
b = 3 km = 3,000 m.

∴ Using the relation, θ = 1003,000 = 130 radian. ……….(1)
Let y be the height of the image formed by the objective.
∴ AngIe subtended by the image produced by the objective is
yf0=y140 ………(ii)
Equating (i) and (ii), we get
yf0=y140
or y = 14030=143 cm = 4.7 cm

(c) υe = D = position of the final image formed by the telescope = 25 cm
if me be the magnification produced by the eye piece, then
me = 1 + yf0 = 1 + 255 = 1 + 5 = 6

Let I be the height of the final image of the tower =?
O = size of the object for eye piece
= size of the image formed by objective
= 143 cm.
∴ Using the relation,
m = IO, weget
me = IO
or I = me x O = 6 x 143 = 28cm,

Question 36.
A carsegrain telescope uses two mirrors as shown in Fig. given here. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of larger mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
Answer:
Here, R1 = radius of curvature of large mirror i.e. concave mirror = – 220 mm = – 22 cm
R2 = radius of curvature of small i.e. convex mirror = 140 mm = 14 cm
If f1, and f2 be the focal lengths of large and small mirror, then
f1 = R12=−222 = – 11cm and f2 = R22=142 = 7 cm
d = spacing between the mirrors = 20 mm = 2 cm
Using the sign conventions, f1, R1 are taken as – ve.

u – ∞ as object lies at infinity.
As per ray diagram final image of the object is formed at the back of objective mirror which is seen through eye piece.
Hie rays coming from object at ∞ tend to meet at principal focus of objective but before that they are intervened by the concave mirror of smaller focal length.
For objective :
u = -∞, f = f1 = 11 cm υ = ?
Using the relation,

Question 37.
Light incident normally on a plance mirror attached to a galvanometer coil retraces backwards as shown in Fig. here. A current in the coil produces 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?

Answer:
We know that the reflected rays of light get deflected by twice the angle of rotation of the mirror.
From fig., we see that when the mirror is turned from position M
to M’ through angle θ = 3.5°, then the reflected ray OB turns through
∠ 2θ = 2 x 3.5° = 7° = ∠AOB .
Here, OA = 1.5m,
∴Displacement, AB = d = ?
Now in rt. ∠d ∆AOB,
tan 2θ = ABAO
or tan 7° = d1.5
or d = 1.5 x tan 7°
= 1.5 x 0.1228 = 0.1842 m
= 18.42 cm

Question 38.
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a place 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?

Answer:
Here let ft = focal length of double convex lens L = 30 cm.
Let F = Focal length of the combination of double-convex lens L and plano-concave liquid lens L’
∴ F = 45.0 cm, μ = 1.5 = 32
Also let f2 be the focal length of the plano-concave lens made of liquid between the convex lens and plane mirror.

∴ radii of curvature of the two surfaces of plano-concave lens of liquid formed between foci convex lens and plane mirror are -R and oo, where R is the radius of curvature of the surface of equi-convex lens.

Also, we know that the object uncoincide with its image only if the ray of light retrace their paths i.e. they fall normally on the plane mirror and it can be possible only if the object lies at the focus of the combined lens system.
For lens L,