A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 cm

A Thin Equiconvex Lens Of Refractive Index 32 And Radius Physics Question

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:

  1. A certain metallic surface is illuminated with monochromatic light of wavelength
  2. A block slides from an inclination of 45°.If it takes time twice
  3. Which radiations are used in tratement of muscles ache?
  4. The displacement of particle is given by x=aₒ+(a₁t/2)-(a₂t²/3) what is acceleration?
  5. For a particle in a non-uniform accelerated circular motion correct statement is

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.

NEETLab Mobile App

Share this page with your friends

1 Comment on A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 cm

  1. 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

Leave a Reply

Your email address will not be published.


*