TS EAMCET 6-Aug-2021 Shift 1 Question Paper

The given situation is shown below.Let u1= initial object distance from lens
f= focal length of lens
v1=60cm (initial image distance)
m=2 (lateral magnification)
Using, m=−‌

v1

u1

=2
‌

−60

u1

=2⇒u1=−30cm
After filling liquid into the tank.
Let u2= object distance
v2=120cm (final image distance)
In case 1 , (without liquid)
By lens formula,
‌

1

f

=‌

1

v1

−‌

1

u1

‌

1

f

=‌

1

60

−‌

1

(−30)

⇒f=20cm
In case 2, (with liquid)
By lens formula,
‌

1

f

‌‌=‌

1

v2

−‌

1

u2

⇒‌

1

20

=‌

1

120

−‌

1

u2

u2‌‌=24cm
Now, normal shift, OO′=H(1−‌

1

µ

)
30−24‌‌=24(1−‌

1

µ

)⇒6=24(1−‌

1

µ

)
[‌∵H=24cm‌ (height of liquid filled) ‌]
⇒‌‌µ=4∕3=1.33