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
