| ⇦ | 
| ⇨ |
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: - Three identical spheres, each of mass 3 kg are placed touching each other
- Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons
- In Bohr model of hydrogen atom, the force on the electron depends on the principal
- Two simple harmonic motions of angular frequency 100 and 1000 rad s⁻¹ have the same
- An object is attached to the bottom of a light vertical spring and set vibrating.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Three identical spheres, each of mass 3 kg are placed touching each other
- Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons
- In Bohr model of hydrogen atom, the force on the electron depends on the principal
- Two simple harmonic motions of angular frequency 100 and 1000 rad s⁻¹ have the same
- An object is attached to the bottom of a light vertical spring and set vibrating.
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