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A point object O is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image would be formed at (refractive index of glass = 1.5)

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30 cm left

Infinity

1 cm to the right

18 cm to the left

answer is A.

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Detailed Solution

By using formula: $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{v} - \frac{1}{- 15} = \frac{0.5}{30}$

$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15} = \frac{- 3}{60}$

$V = - 30 cm$

Negative sign shows that, image is obtained on the same side of object i.e. towards left.

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