| ⇦ | 
| ⇨ |
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 body covers 200 cm in the first 2 seconds and 220 cm in the next 4 seconds.
- A heavy uniform chain lies on a horizontal top table.If the coefficient of friction
- The surface tension of soap solution is 0.03 N/m. The work done in blowing to form a soap
- The numebr 0 (zero) is required for
- A bus is moving with a velocity of 5 ms⁻¹ towards a huge wall.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body covers 200 cm in the first 2 seconds and 220 cm in the next 4 seconds.
- A heavy uniform chain lies on a horizontal top table.If the coefficient of friction
- The surface tension of soap solution is 0.03 N/m. The work done in blowing to form a soap
- The numebr 0 (zero) is required for
- A bus is moving with a velocity of 5 ms⁻¹ towards a huge wall.
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