JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
Region \(I\) and \(II\) are separated by a spherical surface of radius \(25\, {cm}\). An object is kept in region \(I\) at a distance of \(40\, {cm}\) from the surface. The distance of the image from the surface is \(.....\,cm\)
- A \(55.44\)
- B \(9.52\)
- C \(37.58\)
- D \(18.23\)
Answer & Solution
Correct Answer
(C) \(37.58\)
Step-by-step Solution
Detailed explanation
\(\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}\) \(\frac{1.4}{v}-\frac{1.25}{-40}=\frac{1.4-1.25}{-25}\) \(\frac{1.4}{v}=-\frac{0.15}{25}-\frac{1.25}{40}\) \(v=-37.58\, {cm}\)
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