33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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82 NEET-AIPMT Chapterwise Topicwise Solutions Physics
sini
=µ
sinr R
sini
=µ
sinr G
...(i)
...(ii)
sinrp
=µ ...(iii)
sinip
From equations (i), (ii) and (iii), we get
i = i R = i G
Thus two point propagation in two different parallel
direction.
10. (b) : Frequency remains same.
11. (c) : Total apparent depth,
y = y 1 + y 2 = 5 + 2 = 7 cm.
If x is real depth = thickness of slab, then as
real depth
µ= =
apparent depth
x
y
or, x = my = 1.5 × 7 = 10.5 cm.
12. (a) : When the angle of refraction is equal to 90°,
the angle of incidence is called the critical angle.
13. (b) : Difference between apparent and real depth of
a pond is due to refraction. Other three are due to total
internal reflection.
14. (c) : For total internal reflection, sini > sinC
where, i = angle of incidence, C = critical angle
But,
1
µ>
sin 45°
µ> 2
sinC= 1 ∴ sini> 1 or µ >
1
µ µ sini
Hence, option (c) is correct.
( i = 45°
( Given) )
15. (c) : Refractive index for medium M 1 is
8
c 3×
10
µ 1 = = = 2
v 8
1 15 . × 10
Refractive index for medium M 2 is
8
c 3×
10 3
µ 2 = = =
v 8
2 20 . × 10 2
For total internal reflection, sini ≥ sinC
where i = angle of incidence, C = critical angle
But sinC = µ 2
µ 1
µ 2 3/
2
−1
⎛ 3 ⎞
∴ sini≥ ≥ ⇒ i≥
sin
⎝
⎜
⎠
⎟
µ 1 2
4
16. (d) :
From figure,sinC =
where C is the critical angle.
Also,sinC = l µ a
3 3
=
2 2
( 4) + () 3
5
1 ⎡
l 1 ⎤
sinC = ⎢since
µ
a a = ⎥
a
µ l ⎣⎢
µ l ⎦⎥
a velocity of light in air ( c)
Also µ l =
velocity of light in liquid ( v)
∴ sinC = v v
=
c 3 × 10 8
8 3
8 −1
or , v = 3× 10 × = 18 . × 10 m s .
5
17. (c) : Applying Snell’s law of refraction at A, we get
sini
sin 45°
µ= =
sinr
sinr
∴ sin r = 1/
2µ
⎛ ⎞
−
∴ r = sin 1 1
⎜ ⎟
⎝ 2 µ ⎠
...(i)
Applying the condition of total internal reflection at B,
we get
i c = sin –1 (1/m)
... (ii)
where i c is the critical angle.
Now, r + i c = 90° = p/2
−1 1 π −1
1
∴ sin = − sin
2 µ 2 µ
−1 1 −1
1
or,sin = cos
2 µ µ
1
∴ =
2 µ
∴ µ =
18. (a)
3/
2
2
µ −1 1 2
or = µ −1
µ 2
19. (b) : q is the critical angle.
\ q = sin –1 (1/m) = sin –1 (3/5)
or, sinq = 3/5
\ tanq = 3/4 = r/4
or, r = 3 m.