| ⇦ | 
| ⇨ |
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: - One-fourth length of a spring of force constant K is cut away. The force constant
- If the red light is replaced by blue light illuminating the object in a microscope
- On adjusting the P-N junction diode in forward bias,
- A string is stretched between fixed points seperated by 75 cm. It is observed
- In an LCR series A.C circuit, the voltage across each of the components, L, C and R is 50V
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- One-fourth length of a spring of force constant K is cut away. The force constant
- If the red light is replaced by blue light illuminating the object in a microscope
- On adjusting the P-N junction diode in forward bias,
- A string is stretched between fixed points seperated by 75 cm. It is observed
- In an LCR series A.C circuit, the voltage across each of the components, L, C and R is 50V
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