32. An equiconvex lens is made of glass of
refractive index 1.5 and has a focal length
of 10 cm in air. The lens is cut into two equal
halves along a plane perpendicular to its
principal axis to yield two plano-convex
lenses. The two pieces are glued such that
the convex surfaces touch each other. If
this combiantion lens is immersed in water
of refractive index 4/3 its focal length (in
cm) is
Answers
ʟᴇɴ'ꜱ ᴍᴀᴋᴇʀ'ꜱ ꜰᴏʀᴍᴜʟᴀ:-
ꜰ
1
=(μ−1)(
ʀ
1
1
−
ʀ
2
1
)
ꜰᴏʀ ᴇQᴜɪ-ᴄᴏɴᴠᴇx ʟᴇɴꜱ:-
ꜰ
1
=(μ−1)(
ʀ
1
−
−ʀ
1
)
⟹
ꜰ
1
=(μ−1)
ʀ
2
ꜰᴏʀ, ᴛʜᴇ ᴘʟᴀɴᴏ-ᴄᴏɴᴠᴇx ʟᴇɴꜱ ꜰᴏʀᴍᴇᴅ:-
ꜰ
′
1
=(μ−1)(
ʀ
1
1
−
∞
1
)
ꜰ
′
1
=(μ−1)
ʀ
1
⟹ꜰ
′
=2ꜰ
If the combination of lens is immersed in water, its focal length is 40 cm
Explanation:
Given
The focal length of the lens cm
We know that when two lenses of focal length and are kept in contact, then the effective focal length is given by
cm
Therefore, when the two pieces are glued together, there is no change in focal length
Given the refractive index of glass w.r.t. air
Refractive index of water w.r.t. air
Refractive index of glass w.r.t. water
if focal length of the lens in water is then
cm
Hope this answer is helpful.
Know More:
Q: A glass convex lens of refractive index 1.5 has focal length 8 cm when in air what would be focal length of lens when immersed in water of refractive index 1.33.
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