CBSE Class 12 Sample Paper 2025 Solution for Physics

Central Board of Secondary Education

SAMPLE PAPER
Class 12
2025

Solutions

MARKING SCHEME
PHYSICS
Subject Code – 042
CLASS – XII
Academic Session 2024 – 25

Maximum Marks:70 Time Allowed: 3hours

[SECTION – A]
Ans.1- (B) (1 mark)

VA> VB [VA = VC]
In the direction of electric field, the electric potential decreases.

Ans.2- (B) In the state of equilibrium, (1 mark)
The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere
kq1 kq2 q r
=  1 = 1
r1 r2 q2 r2

E1 q1 r22 r1 r22 r2
 = =  =
E 2 q2 r12 r2 r12 r1

Ans.3 - (C) (1 mark)

0 I  I
AtP2, B2 = = 0
  3a
3 a
2  
 2 
0 (I 4) 0 I
AtP1, B1 = =
2 ( a 2 ) 4a

 0 I 
B2  3a  B 4
 =  2 =
B1  0 I  B1 3
 4a 
 

Ans.4 - (D)Sound waves as well as light waves (1 mark)

Ans.5 -(A) (1 mark)

Ans.6 - (C)When all the given components are connected(1 mark)

IR = IXC = IXL = 10V
XC = XL = R
Z= R 2 + (X C − X L )2

Z= R 2 + (R − R)2
Z=R
VS = IZ = IR = 10 V
So, the source voltage is also 10 V
When the capacitor is short circuited then

Z= R 2 + (X L )2

= √𝑅 2 + 𝑅 2 = 𝑅√2
10
VL = I XL = R =5 2 V
2R

Ans.7 - (B)(1 mark)

Ans.8 - (B) The distance of closest approach(1 mark)
const
d= ...(1)
V12
d const
= ...(2)
2 V22
From equations (1) and (2),
V22
2= V2 = 2 V1
V12

 V2 = 2V Given, (V1 = V)

Ans.9 - (C)(1 mark)

𝑛2 𝑛1 𝑛2 − 𝑛1
− =
𝑣 𝑢 𝑅
1 3 [1 − 3 2]
− =
v 2[−6] −6

1 −3 1 −2 −1
= + = =
v 12 12 12 6
𝑣 = –6 cm
Ans.10 - (B)Diffraction (1 mark)
Ans.11- (A)doping level (1 mark)
Ans.12- (C)+0.4% (1 mark)
Ans.13- (A) (1 mark)
Ans.14- (A)(1 mark)
Ans.15- (D)(1 mark)
Ans.16- (A)(1 mark)

[SECTION – B]

Ans.17–
Given ∅𝟎 = 𝟓. 𝟔𝟑𝒆𝑽 = 𝟓. 𝟔𝟑 × 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱
𝝂 = 𝟏. 𝟔 × 𝟏𝟎𝟏𝟓 𝑯𝒛

𝒉𝒄
𝑲. 𝑬. = 𝒉𝝂 − ∅𝟎 = 𝝀 ½

𝒉𝒄
𝝀= ½
𝒉𝝂−∅𝟎

𝟔.𝟔𝟑 × 𝟏𝟎−𝟑𝟒 ×𝟑 ×𝟏𝟎𝟖
= 𝟔.𝟔𝟑 × 𝟏𝟎−𝟑𝟒 ×𝟏.𝟔×𝟏𝟎𝟏𝟓− 𝟓.𝟔𝟑 ×𝟏.𝟔×𝟏𝟎−𝟏𝟗 ½

𝟏𝟗. 𝟖𝟗 × 𝟏𝟎−𝟐𝟔
=
𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 (𝟔. 𝟔𝟑 − 𝟓. 𝟔𝟑)

𝟏𝟗.𝟖𝟗 × 𝟏𝟎−𝟐𝟔
= 𝟏.𝟔×𝟏𝟎𝟏𝟓
= 𝟏𝟐. 𝟒 × 𝟏𝟎−𝟕 𝒎 ½

Ans.18 - 𝜆1 = 4 × 10−7 𝑚𝜆2 = 6 × 10−7 𝑚
1 𝜆𝐷
Distance at which dark fringe is observed 𝑥 = (𝑛 + 2) 𝑑 ½
1 4×10−7
First Dark fringe for 𝜆1 𝑑1 = 2 10−2 𝑚 = 2 × 10−5 𝑚 ½

1 6×10−7
First Dark fringe for 𝜆2 𝑑2 = 𝑚 = 3 × 10−5 𝑚
2 10−2

First dark fringe will be the distance where both dark fringes will coincide i.e LCM of 𝑑1 &𝑑1 ½
i.e. 2 × 10−5 𝑚 × 3 × 10−5 𝑚
= 6 × 10−5 𝑚 ½

OR

(II) 𝑁𝑒𝑡 𝐼 = 𝐼 1 + 𝐼2 + 2√𝐼1 √𝐼2cosФ 0.5 M
Since, 𝐼 1 = 𝐼2 = 𝐼
Net I = I + I + 2 I cosФ
= 2I (1 + cosФ)
∅
= 2I (2 cos2 ) 0.5 M
2

For path difference λ/4 , phase difference is π/2 0.5 M
𝜋
Net I = 4 I cos24
Net I = 2 I 0.5 M

(2 Marks)
Ans.19 - (I)The direction of the magnetic field is perpendicular and inward into the plane of thepaper 0.5M
(II) For a head-on collision to take place, the radius of the path of each ion should be equal to 0.5
m.
mv
r= = 0.5 m 0.5M
qB

mv 4  10−26  2.4  105
B= = 0.5M
qr 4.8  10−19  0.5
B = 0.04 T 0.5M

For VI Candidate
2Л𝑚𝑣 𝑐𝑜𝑠Ɵ
(a) As Pitch (p)= 0.5M
𝑞𝐵
2 𝑋 3.14 𝑋 1.7𝑋10−27 𝑋 2 𝑋105 𝐶𝑜𝑠 300
Or, p= m
1.6 𝑋10−19 𝑋1.5
Or, P=7.7X10-3m 0.5M
(b)As, done by magnetic field is always zero K.E=1/2mv20.5M
KE=3.4 X 10-17J 0.5M
Ans.20 –(i) Nuclear fission –W
0.5M
Reason: As W has binding energy per nucleon less then Y and X and nucleus is larger
in size.
0.5M

(ii) Nuclear fusion-Z 0.5M
Reason: As Z has binding energy per nucleon more then Y and X and nucleus is smaller
in size. 0.5M
𝟏
Ans. 21 - (I) 𝑫𝒓𝒊𝒇𝒕 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 ∝ 𝑹𝒆𝒍𝒂𝒙𝒂𝒕𝒊𝒐𝒏 𝒕𝒊𝒎𝒆

1M

(II) Alternating current changes direction every half cycle. 0.5 M
So average drift velocity is zero 0.5 M
[SECTION – C]

(3 Marks)
Ans.22 -(I) X = Full wave rectifier½
Y = Filter½

(Output Waveform for X) ½

(Output Waveform for Y) ½

(ii) 1

For VI Candidates
Rectifier 0.5M
Underlying principle of Rectifier
The basic principle of the rectifiers is the transformation of current by changing the frequency of the
input signal, and diodes are used to do this.
0.5M

Working
In rectifier, one end of terminal which is connected to PN junction diode will never have negative
potential, as it allows current in forward biasing only. Hence potential difference across load resistor
will always be Positive or zero.
1M

For 60 Hz input of AC, output of
Half wave rectifier will be 60Hz 0.5M
Full wave rectifier will be 120 Hz 0.5M

Ans.23 - (I)The capacitance of a parallel plate capacitor with dielectric slab (t < d)

(3 Marks)

0.5M

+q, –q = the charges on the capacitor plates
+qi, – qi = Induced charges on the faces of the dielectric slab
E0 → electric field intensity in air between the plates
E → the reduced value of electric field intensityinside the dielectric slab.
When a dielectric slab of thickness t<d is introduced between the two plates of the capacitor the electric
field reduces to E due to the polarisation of the dielectric. The potential difference between the two plates is
given by
V = V1 + Vt + V2
V = E0d1 + Et + E0d2 … (1)0.5M
Here E is the reduced value of electric field intensity

E = E0 + Ei . Here Ei is the electric field due to the induced charges [+qi and – qi]

E = E02 + Ei2 + 2 E0 Ei cos180

( E0 − Ei )
2
=
E = E0 – Ei0.5M
Also the dielectric constant K is given by
E0
K= … (2)
E
 q
E0 = = … (3)
0 A0
From equations (1), (2) and (3)
𝐸
𝑉 = 𝐸0 [𝑑1 + 𝑑2 ] + 𝐾0 𝑡
𝑞 𝑡
𝑉 = 𝐴𝜀 [𝑑 − 𝑡 + 𝐾] … (4)
0

The capacitance of the capacitor on the introduction of the dielectric slab is
q
C= … (5)
V
From (4) and (5)
𝜀0 𝐴
𝐶= 𝑡 0.5M
𝑑−𝑡+
𝐾
0 A 0 A
If t = d, then C = K ⇒ C = KC0 Here C0 =
d d
Since K >1 therefore C > C0
𝜀0 𝐴
(II) For a metallic slab K is infinitely large, therefore 𝐶 = 1M
𝑑−𝑡

(3 Marks)
Ans.24 -(i)
2

(ii) 1
• It has mirror objective, which is free from chromatic and spherical aberrations.
• It can gather more light as objectives can be made larger, hence images can be brighter.
Any other two equivalent examples can be accepted.

For V.I Candidates

Objective mirror,
Radius of curvature, R1=200mm
Focal Length, f1=R1/2=100mm 0.5M
Secondary Mirror,
Radius of curvature, R1=150mm
Focal Length, f1=R1/2=75mm 0.5M

Distance between two mirror, x=20mm
For object at infinity, image is formed by objective lens will act as virtual object for secondary mirror
U2=(100-20)mm=80mm 0.5M
Applying, mirror formula for secondary mirror
𝟏 𝟏 𝟏
+ = 0.5M
𝒗𝟐 𝒖𝟐 𝒇𝟐
𝟏 𝟏 𝟏
Or, = -
𝒗𝟐 𝒇𝟐 𝒖𝟐
𝟏 𝟏 𝟏
= - = 0.5M
𝟕𝟓 𝟖𝟎 𝟏𝟐𝟎𝟎
V2=1200mm 0.5M

Ans.25 -

(a). T = 0 K

1M

(b) T = Room Temperature

1M

(ii) Answer will be (a) when switch is open 0.5M
as when switch is closed diode will be forward biased and current will by-pass the bulb. 0.5M

For V.I. Candidate
(i) A potential barrier is formed in a p-n junction due to the depletion layer, which is a layer of
unmovable positive and negative charges that develops on either side of the junction. The depletion
layer is created when holes move towards electrons, causing a layer of electrons on the p-type side
and a layer of holes on the n-type side. The potential difference across this region is called the barrier
potential 2M
(ii)(a) In forward biasing width of depletion region decreases. 0.5M
(b) In reverse biasing width of depletion region increases. 0.5M

Ans.26 - (3 Marks)
Given
𝐵 = 2 𝑇 , 𝑞 = 10𝑚𝐶 , 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙 = 10−2kg , 𝑔 = 9.8 𝑚⁄ 2
𝑠

Magnetic force (𝑞𝑣𝐵 sin 𝜃) = 𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 (𝑚𝑔)
𝑚𝑔
𝑣 = 𝑞𝐵 sin 𝜃 ½

For min. velocity sin 𝜃 = 1

𝑚𝑔 𝑚𝑔
𝑣 = 𝑞𝐵 sin 𝜃 = 𝑣 = 𝑞𝐵 ½

10−2 × 9.8
= m/s ½
10−2 × 2

=4.9m/s

𝑣 = 4.9 𝑚⁄ 2 ½
𝑠

As force is in upward direction so from Fleming’s Left-hand rule, magnetic field will be along North to
South. 1

(3 Marks)
Ans.27 - (I)Since the light ray enters perpendicular to the face AB, the angle of incidence onface AC will
be45° . 0.5M
So,
1
sin C =
n
1 1
𝑠𝑖𝑛 45° = 𝑛 = So, n = 2 0.5M
√2
𝑛𝑔 √2 √3
(II)In fig.2, the face AC of the prism is surrounded by a liquid so𝑛 = = 2 =
𝑛𝑙 ( ) √2
√3

1 2 √2
sin C = = 𝜃𝐶 = 𝑠𝑖𝑛−1 ( 3) = 54.6°
n 3 √

Since the angle of incidence on the surface AC is 45° , which is less than the critical angle for the
pair of media (glass and the liquid), the ray neither undergoes grazing along surface AC, nor does
it suffer total internal reflection 1M

Instead it passes through the surface AC and undergoes refraction into the liquid.
For refracting interface AC, n1 sin i = n2 sin r
2
𝑛1 . 𝑠𝑖𝑛45° = ( ) 𝑠𝑖𝑛 𝑟
√3
√3
𝑠𝑖𝑛 𝑟 = ⸫ 𝑟 = 60° .
2

1M

(3 Marks)
For V.I. candidates
(a) Let the angle of incidence of light at prism, I = x
So, angle of emergence as per question, e = x
4
Angle of prism, A = 𝑥
3
0.5M
Since prism Is equilateral
3A=1800
0.5M
Or, A= 600
Or, X=450
From prism formulaeδ
δ= i+e-A
0.5M
or, δ=45+45-60=300 0.5M

𝒔𝒊𝒏𝑨+𝜹𝒎
𝟐
(b)𝝁 = 0.5M
𝒔𝒊𝒏𝑨
𝟐
𝟔𝟎+𝟑𝟎
𝒔𝒊𝒏
𝟐
Or,𝝁= 𝟔𝟎
𝒔𝒊𝒏
𝟐
Or, 𝝁=√𝟐 0.5M
1
Ans.28– (I)Gauss’stheorem: The flux of electric field through any closed surface is times the total
0
charge enclosed by the closed surface.
q
= … (1)
0
By definition, the total electric flux through the closed surface is given by
⃗⃗⃗⃗
𝜙 = ∮ 𝐸⃗ . 𝑑𝑠 … (2)

∴ From (1) and (2), Gauss’s theorem may be expressed as follows
𝑞
𝜙 = ∮ 𝐸⃗ . ⃗⃗⃗⃗
𝑑𝑠 = 𝜀
0

1
∴ The surface integral of electric field over a closed surface is equal to times the total charge enclosed
0
by the surface. 1M

Application of Gauss’s theorem

To find electric field due to a line charge let us consider an infinitely long line charge placed along XX’ axis
with linear charge density λ. Our aim is to find electric field intensity at a point P distantr from the line
charge. We draw a cylindrical surface of radius r and length l coaxial with the line charge. The net flux
through the cylindrical gaussian surface i.e.

0.5M
⃗⃗⃗⃗ = ∫ 𝐸⃗ . 𝑑𝑠
𝜙 = ∮ 𝐸⃗ . 𝑑𝑠 ⃗⃗⃗⃗ + ∫ 𝐸⃗ . 𝑑𝑠
⃗⃗⃗⃗ + ∫ 𝐸⃗ . 𝑑𝑠
⃗⃗⃗⃗ 0.5M
𝐿𝐶𝐹 𝐶𝑆 𝑅𝐶𝐹
= ∫𝐿𝐶𝐹 𝐸𝑑𝑠 𝑐𝑜𝑠 9 0° + ∫𝐶𝑆 𝐸𝑑𝑠 𝑐𝑜𝑠 0 ° + ∫𝑅𝐶𝐹 𝐸𝑑𝑠 𝑐𝑜𝑠 9 0°0.5M
ϕ = ∫𝐶𝑆 𝐸𝑑𝑠 𝑐𝑜𝑠 0 ° = 𝐸. 2𝜋𝑟𝑙 … (1)
The charge enclosed by the gaussian surface is q = λl… (2)
Using Gauss’s theorem from equations(1) and (2)
l 
E ( 2rl ) = ⇒E = 0.5M
0 2 0 r
OR

(II) (a) Definition of electric flux and its SI unit 1M
(b)Electric field due to an infinite plane sheet of charge.
Let us consider an infinite thin plane sheet of positive charge having a uniform surface charge density .
Let P be the point where electric field E is to be found. Let us imagine a cylindrical gaussian surface
of length 2r and containing P as shown. The net flux through the cylindrical gaussian surface.

0.5M

 = ∮ 𝐸⃗ ⋅ ⃗⃗⃗⃗⃗
𝑑𝐴
=  E  dA +  E  dA +  E  dA 0.5M
RCF LCF CS

=  EdA cos 0 +  EdA cos 0 +  EdA cos90 0.5M
RCF LCF CS

= E A + EA + 0
 = 2 EA ...(1)
Here A is the area of cross-section of each circular facei.e. LCF and RCF.
The total charge enclosed by the gaussian cylinder
= A ...(2)0.5M
Using Gauss’s theorem, from (1) and (2),
A
2 EA =
ε0

E=
2ε 0

Ans.29 -I(B) II(C) III(B) IV (C) 0R IV (D) (4X1=4)

Ans.30 - I(D) II(C) III (A) IV(B) 0R IV (A) (4X1=4)

(5 Marks)
Ans.31–(I) (a) Kirchhoff’s I Law :The algebraic sum of all the currents meeting at a point in an electrical
circuit is always equal to zero.1M

[+I1] + [+I2] + [–I3] +[–I4] + [– I5] = 0
Or I1 + I2 = I3 + I4 + I5
Kirchhoff’s II Law :The algebraic sum of the changes in potential around any closed resistor loop must be
zero.1M

For closed mesh ABCFA
[+E1] [– I1R1] + [–E2] + [+I2R] = 0 … (1)

For closed mesh FCDEF
[+E2] + [-(I1+I2)]R3 + [–I2R2] = 0 … (2)

𝜀
(b).𝐼 = 𝑊ℎ𝑒𝑟𝑒 𝑅0 𝑖𝑠 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 𝑟𝑜𝑜𝑚 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 200 ½
𝑅0 + 𝑟

𝜀
⇒ 𝑅0 = 𝐼 -1

100
OR 𝑅0 = -1 = 𝑅0 = 9Ω ½
10

Now Final temperature is 3200C

So, 𝑅 = 𝑅0 (1 + 𝛼Δ𝑇) ½

= 9 ( (1 + 3.7 × 10−4 × 300)

= 10 Ohm ½

Power Consumed by cell (𝑃) = 𝑖 2 𝑟 ½
𝜀
= (𝑅+ 𝑟)2 × 𝑟 𝑊𝑎𝑡𝑡

100
= ( 11 )2 = 82.64 W ½

OR

(II)(a) The Wheatstone bridgeis as shown in the figure
1M

0.5M
Applying Kirchhoff’s II law to mesh ABDA
I1P + IgG – I2R = 0 …..(1) 0.5M
For the mesh BCDB
(I1 – Ig)Q + [–(I2 + Ig)S] + [–IgG] = 0 (2) 0.5M When the bridge is
balanced, no current flows through the galvanometer

i.e. Ig = 0 (3)

∴From equations (1) and (2) and (3)
I1P = I2R … (4)
I1Q = I2S … (5)
From equations (4)and (5), P/Q = R/S. 0.5M

(b).

This circuit is balanced wheat stone bridge that can be drawn as below,

As it is balanced wheatston bridge ,so circuit will be as below 1

8
𝑉𝐴𝐵 = 8𝑉 , ℎ𝑒𝑛𝑐𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝐴𝐷𝐵 = = 2𝐴 1
4

(for V.I. Candidates)

(II) (a) question is same
(b) The sensitivity of a Wheatstone bridge is the amount of deflection in the attached galvanometer for every
unit change in the unknown resistance 1M

A Wheatstone bridge is most sensitive when its four arms have resistances that are of the same order of
magnitude. This means that all four resistors provide the same output resistance. A Wheatstone bridge is in
a balanced state when its galvanomater shows zero deflection 1M

Ans.32 - (I) AC Generator (5 Marks)
It is a device used to convert mechanical energy into electrical energy

Principle: It is based on the principle of electromagnetic induction. When a closed coil is rotated
rapidly in a strong magnetic field, the magnetic flux linked with the coil changes continuously.
Hence an emf is induced in the coil and a current flows in it. In fact, the mechanical energy
expended in rotating the coil appears as electrical energy in the coil.
1M
Construction: Main Parts 1M
1. Armature: It is a rectangular coil ABCD having a large number of turns of insulated copper wire
wound on a soft-iron core. The use of soft-iron core increases the magnetic flux linked with the
armature.

2. Field Magnet: It a strong electromagnet having concave pole pieces N and S. The armature is
rotated between these pole pieces about an axis perpendicular to the magnetic field.

3. Slip Rings: The leads from the armature coil ABCD are connected to two copper rings R1 and R2
called the ‘slip rings’. These rings are concentric with the axis of the armature coil and rotate with it.

4. Brushes:These are two carbon pieces B1 and B2 called brushes which remain stationary pressing
against the slip rings R1 and R2 respectively. The brushes are connected to an external circuit.

Working Theory : When the coil ABCD is rotated inside the field, an emf is induced between its two
ends. Let the plane of the coil be at right angles to the magnetic field at t = 0 and angular speed of
the rotation of the coil be ω. Then at time t, θ = ωt. The magnetic flux linked with the coil at time t is
ϕ = n BA cosωt
−d  −d
Induced emf e = =  nBA cos t 
dt dt
⇒e = n BA ω sinωt
e = e0sinωt Where e0 = nBAω is the peak value of emf.
The current in the external load is given by
e sin t
i= 0
RL
i = i0sinωtHere i0 is the peak value of the current 1M

1M

1M
In an ac generator the source of electrical energy is the mechanical energy.

OR
(II)
(a)TRANSFORMER

Use: It is a device which converts low ac voltage at high current into high ac voltage at low current
and vice – versa.

Principle: It consists of two coils P and S wound on a closed soft iron core. The coil which is fed
from the ac supply is called primary coil (P) and the other connected to the load is called secondary
coil (S). The core of the transformer is made of soft -iron toreduce hysteresis loss and is laminated
to reduce eddy current losses.1M

Working: When an alternating emf ep is impressed on the primary winding it sends an ac current
through it which sets up an alternating magnetic flux in the core. This induces an alternating emf es
in the secondary. If NP and Ns are the number of turns in primary and secondary coil, their linkages
with the flux are

ϕP = NPBA B → Magnetic induction
ϕS = NsBA A → Area of cross section0.5 M
The magnitude of the emf induced in the secondary
d S dB
es = = NS A … (1)
dt dt
The changing flux also induces an emf in the primary, whose magnitude
d P dB
eP = = NP A … (2)
dt dt
From equations (1) and (2)
emf induced in secondary e N
= s = s … (3)0.5 M
voltage applied to primary eP N P
NS
= turns ratio or transformation ratio.
NP
If Ns> NP, eS>eP → Such a transformer is called step-up transformer
If NS< NP, es<eP → Such a transformer is called step-down transformer
In an ideal transformer
Instantaneous output power = instantaneous input power

esis = ePiP … (4)
From equations (3) and (4)
es i p N s
= = 0.5 M
e p is N p
In a step- up transformer Ns> Np, es>ep but is<ip
In a step-down transformer Ns< Np, es<ep but is>ip
At the generating station a step-up transformer is used for stepping up the voltage and at the
various receiving substations a step-down transformer is used

0.5M
(b) The two sources of energy losses are eddy current losses and flux leakage losses. 1M
(c)There is no violation of the principle of the conservation of energy in a stepup transformer. When
output voltage increases the output current decreases automatically keeping the power the
same.1M

(5 Marks)
Ans.33–(I) Given𝑓0=15m, 𝑓𝑒 =1cm=0.01m
𝑓0 15
(i) Angular magnification of the telescope M = = =1500 1M
𝑓𝑒 0.01
(ii) Let d be thediameter ofmoon᾿simage formed by the objective lens.

Therefore, Angle subtended by the moon at the objective lens
𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑜𝑛 3.48×106
α= = (1) 1.5M
𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑙𝑢𝑛𝑎𝑟 𝑜𝑟𝑏𝑖𝑡 3.8×108

Similarly, the angle subtended by moon᾿s image (formed by the objective) at the objective
𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑚𝑜𝑜𝑛᾿𝑠 𝑖𝑚𝑎𝑔𝑒 𝑑
α= = (2) 1.5M
𝑓0 15

Comparing equations (1) and (2) we have
𝑑 3.48×106
=
15 3.8×108

3.48×106
d= × 15=0.137m=13.7cm 1M
3.8×108

OR
(II) (a) For eyepiece,𝑣𝑒 =-25cm,𝑓𝑒 =6.25cm, 𝑢𝑒 =?
1 1 1
Using = -
𝑓 𝑣 𝑢

1 1 1 1 1 −1
= - = - = 0.5M
𝑢𝑒 𝑣𝑒 𝑓𝑒 −25 6.25 5

𝑢𝑒 =-5cm 0.5M
Therefore the image formed by the objective is formed at a distance of 10 cm towards the eyepiece.
Hence for the objective,𝑣0 =+10 cm,𝑓0=2cm, 𝑣0 =?
1 1 1 1 1
= - = - 0.5M
𝑢0 𝑣0 𝑓0 10 2

𝑢0 =-2.5cm 0.5M
𝑣0 𝐷 10 25
Therefore the magnifying power M= (1+ )= (1+ )=20 0.5M
|𝑢0 | 𝑓𝑒 2.5 6.25

(b) When the final image is formed at infinity the object for the eyepiece must lie at its principal
focus.Thereforethe distance of the image formed by the objective from its optical center,
𝑣0 =15-6.25=8.75cm 0.5M
1 1 1 1 1 6.75
= - = - = 0.5M
𝑢0 𝑣0 𝑓0 8.75 2 17.50
−17.5
𝑢0 = =-2.6cm 0.5M
6.75
𝑣0 𝐷 8.75 25
M= . = × =13.5 1M
|𝑢0 | 𝑓𝑒 2.6 6.25

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