The apparent thickness of a thick plano convex lens is measured once with the plane face upward and then with the convex face upward. The value will be

The correct option is C more in the second case
$\frac{1}{V_{1}} - \frac{μ}{- R} = \frac{1 - μ}{\infty} V_{1} = \frac{- R}{μ}$

$\frac{1}{V_{2}} - \frac{μ}{- R} = \frac{1 - μ}{- R} \frac{1}{V_{2}} + \frac{μ}{R} = \frac{μ - 1}{R} V_{2} = - R V_{2} > V_{1}$

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Similar questions

Q. A plano-convex lens has thickness 4 cm. When placed on a horizontal table with the curved surface in contact with it, the apparent depth of the bottom-most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face of the lens is found to be 25/8 cm. The focal length of the lens is

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The apparent thickness of a thick plano convex lens is measured once with the plane face upward and then with the convex face upward. The value will be

The correct option is C more in the second case1V1−μ−R=1−μ∞V1=−Rμ 1V2−μ−R=1−μ−R1V2+μR=μ−1RV2=−RV2>V1

The correct option is C more in the second case
$\frac{1}{V_{1}} - \frac{μ}{- R} = \frac{1 - μ}{\infty} V_{1} = \frac{- R}{μ}$

$\frac{1}{V_{2}} - \frac{μ}{- R} = \frac{1 - μ}{- R} \frac{1}{V_{2}} + \frac{μ}{R} = \frac{μ - 1}{R} V_{2} = - R V_{2} > V_{1}$