| ⇦ | ⇨ |
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 body of mass m is orbiting the earth at a radius r from the center of earth
- A geostationary satellite is orbiting the earth at a height of 5R above that surface
- A particle of mass m executes simple harmonic motion with amplitude
- Two identical charges repel each other with a force equal to 10 mgwt they are 0.6 m
- Temperature is a measurement of coldness or hotness of an object.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body of mass m is orbiting the earth at a radius r from the center of earth
- A geostationary satellite is orbiting the earth at a height of 5R above that surface
- A particle of mass m executes simple harmonic motion with amplitude
- Two identical charges repel each other with a force equal to 10 mgwt they are 0.6 m
- Temperature is a measurement of coldness or hotness of an object.
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