| ⇦ | ⇨ |
A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 cm is put in water (refractive index = 4/3) , its focal length is
Options
(a) 0.15 m
(b) 0.30 m
(c) 0.45 m
(d) 1.20 m
Correct Answer:
1.20 m
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A sphere of 4cm radius is suspended with in a hollow sphere of 6cm radius. If the inner
- The half life of a radioactive isotope ‘X’ is 20 years. It decays to another element Y
- In India, electricity is supplied for domestic use at 200V. It is supplied at 110 V
- Which of the following phenomena support the wave theory of light?
- Which of the following physical quantities has neither dimensions nor unit
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A sphere of 4cm radius is suspended with in a hollow sphere of 6cm radius. If the inner
- The half life of a radioactive isotope ‘X’ is 20 years. It decays to another element Y
- In India, electricity is supplied for domestic use at 200V. It is supplied at 110 V
- Which of the following phenomena support the wave theory of light?
- Which of the following physical quantities has neither dimensions nor unit
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

According to lens maker’s formula
1f=(μ−1)(1R1−1R2)
where μ=μLμM
Here, μL=32, μM=43, R1=+30cm, R2=−30cm
∴1f=(3243−1)(130−1−30)
=(18)(230)
1f=14×30=1120
or f=120cm=1.2m