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 transformer of 100% efficiency has 200 turns in the primary coil and 40000 turns
  2. If Q, E and W denote respectively the heat added, change in internal energy
  3. The physical quantities not having same dimensions are
  4. A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T.
  5. Zenor breakdown will occur if

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Your email address will not be published.


*