Chapter 19 Optical Instruments free study material by TEACHING CARE online tuition and coaching classes
Human Eye.
 Eye lens : Over all behaves as a convex lens of m = 437
 Retina : Real and inverted image of an object, obtained at retina, brain sense it
 Yellow spot : It is the most sensitive part, the image formed at yellow spot is
 Blind spot : Optic nerves goes to brain through blind It is not sensitive for light.
 Ciliary muscles – Eye lens is fixed between these It’s both radius of curvature can be changed by applying pressure on it through ciliary muscles.
 Power of accomodation : The ability of eye to see near objects as well as far objects is called power of
Note : @When we look distant objects, the eye is relaxed and it’s focal length is largest.
 Range of vision : For healthy eye it is 25 cm (near point) to ¥ (far point).
A normal eye can see the objects clearly, only if they are at a distance greater than 25 cm. This distance is called Least distance of distinct vision and is represented by D.
 Persistence of vision : Is 1/10 sec. e. if time interval between two consecutive light pulses is lesser than
0.1 sec., eye cannot distinguish them separately.
 Binocular vision : The seeing with two eyes is called binocular
 Resolving limit : The minimum angular displacement between two objects, so that they are just
_{‘} æ 1 ö^{o}
resolved is called resolving limit. For eye it is 1 = ç 60 ÷ .
è ø
(11) Defects in eye
Presbyopia : In this defect both near and far objects are not clearly visible. It is an old age disease and it is due to the loosing power of accommodation. It can be removed by using bifocal lens.
Astigmatism : In this defect eye cannot see horizontal and vertical lines clearly, simultaneously. It is due to imperfect spherical nature of eye lens. This defect can be removed by using cylindrical lens (Torric lenses).
Microscope.
It is an optical instrument used to see very small objects. It’s magnifying power is given by
m = Visual angle with instrument(b ) Visual angle when object is placed at least distance of distinct vision (a)
(1) Simple miscroscope
 It is a single convex lens of lesser focal
 Also called magnifying glass or reading
 Magnification’s, when final image is formed at D and ¥
(i.e. mD and m¥ )
æ D ö
mD = ç1 + f ÷
and
æ D ö
m¥ = ç f ÷
è ømax è ømin
Note : @ mmax . – mmin . = 1
@ If lens is kept at a distance a from the eye then mD
(2) Compound microscope
 Consist of two converging lenses called objective and eye
= 1 + D – a
f
and m¥
= D – a
f
feye lens > fobjective and
(diameter)eye lens > (diameter)objective
 Final image is magnified, virtual and
 u_{0} = Distance of object from objective (o),
v_{0} = Distance of image (A¢B¢) formed by objective from objective, ue =
Distance of
A¢B¢
from eye lens, v_{e} = Distance
of final image from eye lens, f_{0} = Focal length of objective, f_{e} = Focal length of eye lens.
v_{0} æ D ö
f_{0} æ D ö
(v_{0} – f_{0} ) æ D ö
Magnification :
mD = – u
ç1 +
0 è
÷ = –
fe ø (u0
 f0
ç1 +
) è
÷ = –
fe ø f0
ç1 +
è
÷
fe ø
v_{0} D
 f0
æ D ö (v_{0} – f_{0} ) D
m¥ = – u . F
= (u –
ç ÷ = – .
f F
0 e 0
f0 ) è fe ø 0 e
Length of the tube (i.e. distance between two lenses)
When final image is formed at D ;
L = v + u =
u0 f0
 fe D
D 0 e
u0 – f0
fe + D
When final images is formed at ¥ ;
L¥ = v0 + fe =
u0 f0 u0 – f0
 fe
(Do not use sign convention while solving the problems)
Note : @ m
= (L¥ – f0 – fe )D

f0 fe
@ For maximum magnification both f_{0}
and
fe must be less.
@ m = mobjective ´ meye lens
@ If objective and eye lens are interchanged, practically there is no change in magnification.
 Resolving limit and resolving power : In reference to a microscope, the minimum distance between two lines at which they are just distinct is called Resolving limit (RL) and it’s reciprocal is called Resolving power (RP)
R.L. = l
2m sinq
and R.P. = 2m sinq
l
Þ R.P. µ 1
l
l = Wavelength of light used to illuminate the object,
m = Refractive index of the medium between object and objective,
q = Half angle of the cone of light from the point object, m sinq = Numerical aperture.
Note : @Electron microscope : electron beam(l » 1Å) than that of ordinary microscope (l » 5000 Å)
is used in it so it’s R.P. is approx 5000 times more
Telescope.
By telescope distant objects are seen.
(1) Astronomical telescope
 Used to see heavenly
fobjective
 feyelens and dobjective >deye lens .
 Intermediate image is real, inverted and
 Final image is virtual, inverted and small.
 Magnification : m
= – f0 æ1 + fe ö and m = – fo
D f ç D ÷ ¥ f
e è ø e
 Length : L
= f + u
= f +
fe D
and L
= f + f

D 0 e
fe + D
¥ 0 e
(2) Terrestrial telescope
 Used to see far off object on the
 It consists of three converging lens : objective, eye lens and erecting
 It’s final image is virtual erect and
 Magnification : m
= f0 æ1 + fe ö and m = f0
_{D} ç ÷


fe è ø fe
 Length :
LD = f0
 4 f + ue
= f0
+ 4 f +
fe D fe + D
and
L¥ = f0
 4 f + fe
(3) Galilean telescope
 It is also a terrestrial telescope but of much smaller field of
 Objective is a converging lens while eye lens is diverging
 Magnification : m
= f_{0} æ1 – f_{e} ö
and m
= f0
_{D} ç ÷


fe è ø fe
 Length :
LD = f0 – ue
and L¥
= f0 – fe
(4) Resolving limit and resolving power
Smallest angular separations (dq) between two distant objects, whose images are separated in the telescope is called resolving limit. So resolving limit dq = 1.22l
a
and resolving power (RP) = 1
dq
= a
1.22l
Þ R.P. µ 1
l
where a = aperture of objective.
Note : @Minimum separation (d) between objects, so they can just resolved by a telescope is – d = r
R.P.
where r = distance of objects from telescope.
(5) Binocular
If two telescopes are mounted parallel to each other so that an object can be seen by both the eyes simultaneously, the arrangement is called ‘binocular’. In a binocular, the length of each tube is reduced
by using a set of totally reflecting prisms which provided intense, erect image free from lateral inversion. Through a binocular we get two images of the same object from different angles at same time. Their superposition gives the perception of depth also along with length and breadth, i.e., binocular vision gives proper threedimensional (3D) image.
Concepts
As magnifying power is negative, the image seen in astronomical telescope is truly inverted, i.e., left is turned right with upside down simultaneously. However, as most of the astronomical objects are symmetrical this inversion does not affect the observations.
Objective and eye lens of a telescope are interchanged, it will not behave as a microscope but object appears very small.
In a telescope, if field and eye lenses are interchanged magnification will change from (f_{o} / f_{e}) to (f_{e} / f_{o}), i.e., it will change from m to (1/m), i.e., will become (1/m^{2}) times of its initial value.
As magnification for normal setting as (f_{o} / f_{e}), so to have large magnification, f_{o} must be as large as practically possible and f_{e} small. This is why in a telescope, objective is of large focal length while eye piece of small.
In a telescope, aperture of the field lens is made as large as practically possible to increase its resolving power as resolving power of a telescope µ (D/l)^{*}. Large aperture of objective also helps in improving the brightness of image by gathering more light from distant object. However, it increases aberrations particularly spherical.
For a telescope with increase in length of the tube, magnification decreases. In case of a telescope if object and final image are at infinity then :
m = fo = D
fe d
If we are given four convex lenses having focal lengths
f1 > f2 > f3 > f4 . For making a good telescope and microscope. We
choose the following lenses respectively. Telescope
f_{1}(o), f_{4} (e)
Microscope
f_{4} (o), f_{3} (e)
If a parrot is sitting on the objective of a large telescope and we look towards (or take a photograph)of distant astronomical object (say moon) through it, the parrot will not be seen but the intensity of the image will be slightly reduced as the parrot will act as obstruction to light and will reduce the aperture of the objective.
Example: 1 A man can see the objects upto a distance of one metre from his eyes. For correcting his eye sight so that he can see an object at infinity, he requires a lens whose power is
or
A man can see upto 100 cm of the distant object. The power of the lens required to see far objects will be
[MP PMT 1993, 2003]
(a) +0.5 D (b) +1.0 D (c) +2.0 D (d) –1.0 D
Solution: (d) f = –(defected far point) = – 100 cm. So power of the lens P = 100 =
f
100
– 100
= 1D
Example: 2 A man can see clearly up to 3 metres. Prescribe a lens for his spectacles so that he can see clearly up to 12 metres
[DPMT 2002]
(a) – 3/4 D (b) 3 D (c) – 1/4 D (d) – 4 D
Solution: (c) By using f = xy Þ f = 3 ´ 12 = 4m . Hence power P = 1 = – 1 D
x – y 3 – 12 f 4
Example: 3 The diameter of the eyeball of a normal eye is about 2.5 cm. The power of the eye lens varies from
 2 D to 10 D (b) 40 D to 32 D (c) 9 D to 8 D (d) 44 D to 40 D
Solution: (d) An eye sees distant objects with full relaxation so 1 – 1 = 1 or P = 1 = 1 = 40D
2.5 ´ 10 ^{–}^{2} – ¥ f f 25 ´ 10 ^{–}^{2}
An eye sees an object at 25 cm with strain so 1 – 1 = 1 or P = 1 = 40 + 4 = 44 D
2.5 ´ 10^{–}^{2} – 25 ´ 10^{–}^{2} f f
Example: 4 The resolution limit of eye is 1 minute. At a distance of r from the eye, two persons stand with a lateral separation of 3 metre. For the two persons to be just resolved by the naked eye, r should be
(a) 10 km (b) 15 km (c) 20 km (d) 30 km
d æ 1 ö ^{o} æ 1 ö p
Solution: (a) From figure q = r ; where q = 1′ = ç 60 ÷ = ç 60 ÷ ´ 180 rad
è ø è ø
Þ 1 ´ 1 ´ p = 3 Þ r = 10 km
60 180 r
Example: 5 Two points separated by a distance of 0.1 mm can just be resolved in a microscope when a light of wavelength 6000 Å is used. If the light of wavelength 4800 Å is used this limit of resolution becomes
[UPSEAT 2002]
(a) 0.08 mm (b) 0.10 mm (c) 0.12 mm (d) 0.06 mm
Solution: (a) By using resolving limit (R.L.) µ l Þ
(R.L.)_{1}
= l_{1}
Þ 0.1 = 6000
Þ (R.L.)
= 0.08 mm .
(R.L.)2 l_{2}
(R.L.)_{2}
4800 ^{2}
Example: 6 In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm an object is placed at 2 cm
form objective and the final image is formed at 25 cm from eye lens. The distance between the two lenses is
[EAMCET (Med.) 2000]
(a) 6.00 cm  (b) 7.75 cm  (c) 9.25 cm  (d) 11.00 cm  
Solution: (d)  It is given that f_{o} = 1.5 cm,  f_{e} = 6.25 cm, u_{o} = 2 cm 
When final image is formed at least distance of distinct vision, length of the tube LD = uo fo + fe D
Þ LD = 2 ´ 1.5 + 6.25 ´ 25 = 11 cm .
uo – fo
fe + D
(2 – 1.5) (6.25 + 25)
Example: 7 The focal lengths of the objective and the eyepiece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eyepiece is 15.0 cm. The final image formed by the eyepiece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively [IITJEE 1995]
(a) 2.4 and 12.0 (b) 2.4 and 15.0 (c) 2.3 and 12.0 (d) 2.3 and 3.0
Solution: (a) Given that
fo = 2 cm ,
fe = 3 cm , L¥ = 15 cm
By using
L¥ = vo + fe
Þ 15 = vo + 3
Þ vo = 12 cm . Also
vo = vo – fo Þ 12 = 12 – 2
Þ uo = 2.4 cm .
uo fo uo 2
Example: 8 The focal lengths of the objective and eyelens of a microscope are 1 cm and 5 cm respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is [CPMT 1979]
(a) 30 cm (b) 25 cm (c) 15 cm (d) 12 cm
Solution: (c) Given that
fo = 1 cm ,
fe = 5 cm , m¥ = 45
By using m¥
= (L¥ – fo – fe )
fo fe
Þ 45 = (L¥ – 1 – 5)´ 25
1´ 5
Þ L¥
= 15 cm
Example: 9 If the focal lengths of objective and eye lens of a microscope are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from the objective lens and the final image is formed at infinity, then magnifying power of the microscope is [CBSE PMT 1999]
(a) 150 (b) 200 (c) 250 (d) 400
Solution: (b) Given that
fo = 1.2 cm ,
fe = 3 cm , uo = 1.25 cm
By using m¥ = – fo × D Þ m = – 1.2 ´ 25 = 200 .
(u_{o} – f_{o} ) f_{e} ^{¥} (1.25 – 1.2) 3
Example: 10 The magnifying power of an astronomical telescope is 8 and the distance between the two lenses is 54cm. The focal length of eye lens and objective lens will be respectively [MP PMT 1991; CPMT 1991; Pb. PMT 2001]
(a) 6 cm and 48 cm (b) 48 cm and 6 cm (c) 8 cm and 64 cm (d) 64 cm and 8 cm
Solution: (a) Given that m¥ = 8 and L¥ = 54
By using m¥ = fo
fe
and L¥ = fo + fe
we get
fo = 6 cm and
fe = 48 cm .
Example: 11 If an object subtend angle of 2° at eye when seen through telescope having objective and eyepiece of focal
length
fo = 60 cm and
fe = 5 cm
respectively than angle subtend by image at eye piece will be [UPSEAT 2001]
(a) 16° (b) 50° (c) 24° (d) 10°
Solution: (c) By using
b = fo
a fe
Þ b = 60
20 5
Þ b = 24°
Example: 12 The focal lengths of the lenses of an astronomical telescope are 50 cm and 5 cm. The length of the telescope when the image is formed at the least distance of distinct vision is [EAMCET (Engg.) 2000]
(a) 45 cm (b) 55 cm (c)
275 cm (d)
6
325 cm
6
Solution: (d) By using
LD = fo + ue = fo + fe D = 50 + 5 ´ 25 = 325 cm
fe + D
(5 + 25) 6
Example: 13 The diameter of moon is
3.5 ´ 10^{3} km
and its distance from the earth is
3.8 ´ 10^{5} km . If it is seen through a
telescope whose focal length for objective and eye lens are 4 m and 10 cm respectively, then the angle subtended by the moon on the eye will be approximately [NCERT 1982; CPMT 1991] (a) 15° (b) 20° (c) 30° (d) 35°
3.5 ´ 10^{3} 3.5 _{–}_{2}
Solution: (b) The angle subtended by the moon on the objective of telescope a = 3.8 ´ 105 = 3.8 ´ 10
rad
Also m = fo = b
Þ 400 = b
Þ b = 40 a
Þ b = 40 ´ 3.5 ´ 103 ´ 180 = 20°
fe a
10 a
3.8 ´ 10^{5} p
Example: 14 A telescope has an objective lens of 10 cm diameter and is situated at a distance one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of [CBSE PMT 2004]
(a) 0.5 m (b) 5 m (c) 5 mm (d) 5cm
Solution: (b) Suppose minimum distance between objects is x and their distance from telescope is r
So Resolving limit dq = 1.22l = x Þ x = 1.22l ´ r = 1.22 ´(5000 ´ 1010 )´(1´ 103 ) = 6.1´ 10^{–}^{3}m = 6.1 mm
a r a
Hence, It’s order is » 5 mm.
(0 – 1)
Example: 15 A compound microscope has a magnifying power 30. The focal length of its eyepiece is 5 cm. Assuming the final image to be at the least distance of distinct vision. The magnification produced by the objective will be
(a) +5 (b) – 5 (c) +6 (d) – 6
Solution (b) Magnification produced by compound microscope m = mo ´ me
æ D ö 25
where m_{o} = ? and me = ç1 + f
÷ = 1 +
5
= 6 Þ 30 = –mo ´ 6 Þ mo = 5 .

è e ø
Tricky example: 1 
A man is looking at a small object placed at his least distance of distinct vision. Without changing his position and that of the object he puts a simple microscope of magnifying power 10 X and just sees the clear image again. The angular magnification obtained is
(a) 2.5 (b) 10.0 (c) 5.0 (d) 1.0 Solution : (d) Angular magnification = b = tan b = I / D = I a tan a O / D O Since image and object are at the same position, I = v = 1 Þ Angular magnification = 1 O u 
Tricky example: 2 
A compound microscope is used to enlarge an object kept at a distance 0.03m from it’s objective which consists of several convex lenses in contact and has focal length 0.02m. If a lens of focal length 0.1m is removed from the objective, then by what distance the eyepiece of the microscope must be moved to refocus the image
(a) 2.5 cm (b) 6 cm (c) 15 cm (d) 9 cm Solution : (d) If initially the objective (focal length F ) forms the image at distance v then v = uo fo = 3 ´ 2 = 6 cm o o o u – f 3 – 2 o o ü Now as in case of lenses in contact 1 = 1 + 1 + 1 + ….. = 1 + 1 ìwhere 1 = 1 + 1 + ….. Fo f1 f2 f3 f1 Fo¢ í F_{o}¢ f_{2} f_{3} ý î þ So if one of the lens is removed, the focal length of the remaining lens system 
1 = 1 – 1 = 1 – 1 Þ Fo¢ = 2.5 cm
Fo¢ F0 f1 2 10 This lens will form the image of same object at a distance v¢ such that v¢ = uo Fo¢ = 3 ´ 2.5 = 15 cm ^{o} ^{o} uo – Fo¢ (3 – 2.5) So to refocus the image, eyepiece must be moved by the same distance through which the image formed by the objective has shifted i.e. 15 – 6 = 9 cm. 