(a) OP is the incidence ray on the prism and QR emergent ray.
∠i1‌=‌ angle if incidence ‌
∠i2‌=‌ angle of emergence ‌
A‌=‌ angle of the prism ‌
µ‌=‌ refractive index of the material of the prism. ‌
δ‌=‌ angle of deviation ‌
For minimum deviation, ∠r1=∠r2=∠r
‌A=∠r1+∠r2
So, ‌A=∠r+∠r=∠2r
‌∠r=2A
Also, ∠i1+∠i2‌=∠i
A+δm‌=∠i1+∠i2
So, i=‌

(A+δm)

2

Now, from snell's law,
∝=‌

sin‌i

sin‌r

µ=‌

sin‌ A+δm 2

sin‌‌ A 2

(b)(i)The critical angle =sin‌−1‌

1

1.5

=41.3∘.
So, the ray which is incident on AB surface will be reflected making an angle 45∘.
The angle of incidence on AC surface is also 45∘; so, the ray will be reflected making an angle 45∘.
The ray is incident normally on the surface BC. So, there will be no deviation due to refraction.
(ii) If µ=1.4, then the critical angle = sin‌−1‌

1

1.4

=45.23∘.
So, the ray will be refracted out from the AB face.
Angle of incidence =45∘
∴µ‌=‌

sin‌i

sin‌r

‌ Or, ‌‌‌‌

1

1.4

‌=‌

sin‌45∘

sin‌r

‌ Or, ‌sin‌r‌=1.4×sin‌45∘=0.99
∴‌‌r‌=‌ angle of refraction ‌=81.9∘
So, the ray will be refracted out making an angle of refraction 81.9∘.
OR
(a) Consider two thin lenses L1 and L2 of focal length f1 and f2 are placed coaxially in contact with each other.
The lenses are so thin that their optical centers are assumed to coincide at point P.

An object is placed at O on the common principal axis. The lens L1 produces an image at I1 and this image acts as the object for the second lens L2. The final image is produced at l as shown in figure.
PO=u, object distance for the first lens (L1),
PI=v, final image distance and
PI1=v1, image distance for the first lens (L1)= object distance for second lens (L2).
For the image I1 produced by the first lens L1
‌

1

v1

−‌

1

u

=‌

1

f1

......(1)
For the final image I, produced by the second lens L2,
‌

1

v

−‌

1

v1

=‌

1

f2

.......(2)
Adding equations (1) and (2),
‌

1

v

−‌

1

u

=‌

1

f1

+‌

1

f2

.......(3)
If the combination is replaced by a single lens of focal length f such that it forms the image of O at the same position I, then
‌

1

v

−‌

1

u

=‌

1

f

......(4)
Comparing equations (3) and (4),
‌

1

f1

+‌

1

f2

=‌

1

f

(b) Refractive index of the medium of lens =1.5
Power of the lens =−10D
Focal length of the lens =f‌air ‌=‌

1

10

=0.1m=10‌cm
In liquid, its focal length =f‌liquid ‌=−50‌cm
According to lens makers' formula
‌

1

f‌air ‌

=[‌

n2

n1

−1][‌

1

R1

−‌

1

R2

]
Or, ‌

1

10

=[‌

1.5

1

−1][‌

1

R1

−‌

1

R2

]
∴‌

1

10

=0.5×[‌

1

R1

−‌

1

R2

] .......(1)
In the liquid,
‌

1

f‌liquid ‌

=[‌

n2

n1

−1][‌

1

R1

−‌

1

R2

]
Or, ‌

1

−50

=[‌

1.5

n1

−1][‌

1

R1

−‌

1

R2

] .......(2)
Dividing eqn (1) by eqn (2),
−‌

‌ 1 10

1 50

=‌

0.5

1.5 n1 −1

Or, ‌

1.5

n1

−1=‌

1

10

∴n1= Refractive index of the liquid medium =1.36