⇦ | ⇨ |
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: - In old age arteries carrying blood in the human body become narrow resulting
- A,B and C are voltmeters of resistance R, 1.5 R and 3R respectively, as shown
- If the r.m.s. velocity of a gas is v, then
- A thin wire of resistance 4 ohm is bent to form a circle. The resistance across
- In a capillary tube water raises by 1.2 mm. The height of water that will rise
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In old age arteries carrying blood in the human body become narrow resulting
- A,B and C are voltmeters of resistance R, 1.5 R and 3R respectively, as shown
- If the r.m.s. velocity of a gas is v, then
- A thin wire of resistance 4 ohm is bent to form a circle. The resistance across
- In a capillary tube water raises by 1.2 mm. The height of water that will rise
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