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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.
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- An alternating voltage e=200s in100t is applied to a series combination
- A thin rod of length L and mass M is held vertically with one end on the floor
- 300j work is done in sliding a 2kg block up an inclined plane of height 10m
- In the given figure, a diode D is connected to an external resistance R=100 Ω
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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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