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CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

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Question : 10 of 13
Marks: +1, -0
A point object in air is placed symmetrically at a distance of 60 cm60 \text{ cm} in front of a concave spherical surface of refractive index 1.5 . If the radius of curvature of the surface is 20 cm20 \text{ cm}, find the position of the image formed.
Solution:     👈: Video Solution
For refraction from concave spherical surface,
μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}
 or, 1.5v−1−60=1.5−1−20\text{ or, } \frac{1.5}{v} - \frac{1}{-60} = \frac{1.5-1}{-20}
 or, 1.5v+160=−0.520\text{ or, } \frac{1.5}{v} + \frac{1}{60} = -\frac{0.5}{20}
 or, 1.5v+160=−140\text{ or, } \frac{1.5}{v} + \frac{1}{60} = -\frac{1}{40}
 or, 1.5v=−140−160\text{ or, } \frac{1.5}{v} = -\frac{1}{40} - \frac{1}{60}
∴v=−36 cm\therefore v = -36 \text{ cm}
Image in formed in the medium where the object lies at a distance 36 cm36 \text{ cm} from point PP.
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