An equi – convex lens of glass, refractive index 1.5 has focal length 0.4 m in air. Calculate the focal length of the lens when is immersed in a liquid of refractive index 1.7. What is the nature of the lens in the liquid?
Can someone plz give me the explanation as well?
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Answered by
1
Answer:
Correct option is
B
8
15
Given,
μ
1
=1.5
f
1
=0.2m
f
2
=(−0.5)
μ
2
=?
Using lens makers formula and taking ratios,
f
2
f
1
=
(μ
1
−1)
(μ
2
−1)
⇒
(−0.5)
0.2
=
(1.5−1)
(
μ
e
1.5
−1)
⇒
0.5
0.2×0.5
=(1−
μ
e
1.5
)
⇒
μ
e
1.5
=0.8
⇒μ
e
=
8
15
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