a convex lens of focal length 0.2m is made of material of refractive index 7/2 . the lens is placed in a medium then its focal length is increased 0.35 what is its refractive index
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Class 12
>>Physics
>>Ray Optics and Optical Instruments
>>Combination of Mirror and Lenses
>>A convex lens of focal length 0.2 m and
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A convex lens of focal length 0.2 m and made of glass (
a
μ
g
=1.5) is immersed in water (
a
μ
w
=1.33). Find the change in the focal length of the lens.
Medium
Solution
We know ,
From Lens maker formula:
f
1
=(μ
g
−μ
m
)(
R
1
1
−
R
2
1
)
where,
f= focal length
μ
g
= refractive index of glass −1.5
μ
m
- refractive index of medium w.r.t air
R
1
and R
2
are radius of curvature
Refractive index of air is 1
Hence focal length of lens in air
f
a
1
=(1.5−1)(
R
1
1
−
R
2
1
)
Refractive index of water = 1.33
Hence focal length of water
f
w
1
=(1.5−1.33)(
R
1
1
−
R
2
1
)
From above two equation we get
f
a
f
w
=
1.5−1.33
1.5−1
⟹
f
a
f
w
=
0.17
0.5
=2.94
GIven, Focal length of lens in air, f
a
=0.2 m
∴ Focal length in water , f
w
=0.2×2.94=0.58 m