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Question
A spherical marble, of refractive index 1.5 and curvature 1.5 cm, contains a tiny air bubble at its centre. Where will it appear when seen from outside?
Options
1 cm inside
at the centre
`5/3` cm inside
2 cm inside
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Solution
`bb(5/3)` cm inside
Explanation:
The refractive index of the marble is 1.5.
One can see the image formed due to the reflection in the spheres.
In order to determine the distance where the image will form, use the lens maker formula.
Hence, `mu_1/v - mu_2/u = (mu_2 - mu_1)/"R"`
Where, refractive index of air = 1, refractive index of glass = 1.5, Radius of the curvature = 1.5 cm
Substituting the values, we get,
`therefore 1/v - 1.5/0.75 = (1 - 1.5)/1.5`
`therefore 1/v - 2 = - 0.5/1.5`
`therefore 1/v - 2 = - 1/3`
`therefore 1/v = - 1/3 + 2`
`therefore 1/v = (- 1 + 6)/3`
`therefore 1/v = 5/3`
Therefore, the tiny bubble will be seen 5/3 cm inside when we see it from the outside.
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