33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Telegram @unacademyplusdiscounts
Ray Optics and Optical Instruments
87
63. (b) : Here f o = 40 cm, f e = 4 cm
Tube length(l) = Distance between lenses
= v o + f e
For objective lens,
u o = –200 cm, v o = ?
1 1 1 1 1 1
− = or − =
v u f v − 200 40
o o o o
or
1 1 1 4
= − =
vo
40 200 200
∴ v0
= 50 cm
\ l = 50 + 4 = 54 cm
64. (b) :
f e
(f e + f o )
For eye-piece lens,
f h
fe
I
m =
f + u
= 1
h
⇒ =
0 fe − ( fo + fe
) L
fo
L
⇒ =− = Magnification of the telescope
fe
I
65. (d) : Magnifying power of a microscope,
L D
m = ⎛ ⎞ ⎛ ⎞
⎝ ⎜ fo
⎠
⎟
⎝
⎜ fe
⎠
⎟
where f o and f e are the focal lengths of the objective and
eyepiece respectively and L is the distance between their
focal points and D is the least distance of distinct vision.
If f o increases, then m will decrease.
Magnifying power of a telescope, m
f o
=
fe
where f o and f e are the focal lengths of the objective and
eyepiece respectively.
If f o increases, then m will increase.
66. (a) : Converging power of cornea,
P c = + 40 D
Least converging power of eye lens, P e = + 20 D
f o
Power of the eye-lens, P = P c + P e
= 40 D + 20 D = 60 D
Power of the eye lens
1
P =
Focal length of the eye lens ( f )
1 1 1 100 5
f = = = m= cm=
cm
P 60 D 60 60 3
Distance between the retina and cornea-eye lens = Focal
length of the eye lens
5
= cm=
167 . cm
3
67. (c) : Magnifying power, m f o
= =9 ...(i)
fe
where f o and f e are the focal lengths of the objective and
eyepiece respectively
Also, f o + f e = 20 cm
...(ii)
On solving (i) and (ii), we get
f o = 18 cm, f e = 2 cm
68. (d) : Length of astronomical telescope (f o + f e ) = 44
cm and ratio of focal length of the objective lens to that
of the eye piece f o
= 10
fe
From the given ratio, we find that f o = 10 f e .
Therefore 10f e + f e = 44 or f e = 4 cm
and focal length of the objective (f o )
= 44 – f e = 44 – 4 = 40 cm
69. (c) : Time of exposure t ∝ (f – number) 2
∴ = ⎛ 2
t
⎛
⎝ ⎜ ⎞ ⎝ ⎜ 56 . ⎞
⎠
⎟ = 4 or t
1 28
= 0.
02 s
.
⎠
⎟
200
70. (a) : Magnifying power of telescope,
M = f o /f e
To produce largest magnifications f o > f e and f o and f e both
should be positive (convex lens).
Therefore f e = +15 cm.
vvv