A planoconvex lens has thickness of 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 .

Question

A planoconvex lens has thickness of 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 is found to be 25 8 cm . The focal length of the lens is (assume thickness of lens to be negligible)

TestHub · Accepted Answer

In first case refraction of the rays, takes place from a plane surface, so we can use $d_{app} = \frac{d_{actual}}{μ}$

$3 = \frac{4}{μ} \Rightarrow μ = \frac{4}{3}$

where, $d_{app}$ and $d_{actual} =$ apparent and real depth, respectively.

Now, in second case, as from Fig. (b), refraction takes place from a spherical surface, so

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R m}$

or $\frac{1}{(- 25 / 8)} - \frac{4 / 3}{(- 4)} = \frac{1 - 4 / 3}{- R}$

or $\frac{1}{3 R} = \frac{1}{3} - \frac{8}{25} = \frac{1}{75}$

$R = 25 cm$

Now, using Lens' Makers formula, to calculate focal length

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f} = (\frac{4}{3} - 1) (\frac{1}{\infty} - \frac{1}{(- 25)}) = \frac{1}{75}$

$f = 75 cm$

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