⇦ | ⇨ |
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 tuning fork gives 4 beats with 50 cm length of a sonometer
- The motion of a particle along a straight line is described by equation
- The wavelength λₑ of an electron and λₚ of a photon are of same energy E are related by
- A current of 2 A is made to flow through a coil which has only one turn. The magnetic
- Two equal and opposite charges of masses m₁ and m₂ are accelerated
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A tuning fork gives 4 beats with 50 cm length of a sonometer
- The motion of a particle along a straight line is described by equation
- The wavelength λₑ of an electron and λₚ of a photon are of same energy E are related by
- A current of 2 A is made to flow through a coil which has only one turn. The magnetic
- Two equal and opposite charges of masses m₁ and m₂ are accelerated
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