| ⇦ | ⇨ |
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: - Two identical long conducting wires AOB and COD are placed at right angle
- “On flowing current in a conducting wire the magnetic field is produced around it”.
- Two balls of mass m₁ and m₂ are separated from each other by powder charge
- Two wheels having radii in the ratio 1:3 are connected by a common belt
- Two straight wires each 10cm long are parallel to one another and seperated by 2 cm.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two identical long conducting wires AOB and COD are placed at right angle
- “On flowing current in a conducting wire the magnetic field is produced around it”.
- Two balls of mass m₁ and m₂ are separated from each other by powder charge
- Two wheels having radii in the ratio 1:3 are connected by a common belt
- Two straight wires each 10cm long are parallel to one another and seperated by 2 cm.
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