A planoconvex lens has a thickness of = 4 cm. When placed on a horizontal surface (table), with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be $t_1$= 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the center of the plane surface is found to be $t_2$ = 25/8 cm. Find the focal length of the lens

75 cm

Refraction of flat surface:

Apparent depth= (Real depth)/ (R.I )

$⇒t_1=\frac{t}{n}$  . . .(1)

Refraction of curved surface

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

Here $n_1 = n, n_2 =1, u ≅ -t$, R is – R

$⇒\frac{1}{v}-\frac{n}{-t}=\frac{1-n}{-R}$

$⇒\frac{1}{v}=\frac{n-1}{R}-\frac{n}{t}$

Putting $v= t_2$ we obtain, $-\frac{1}{t_2}=\frac{n-1}{R}-\frac{n}{t}$  …(2)

$\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$, Putting $R_1 =R$ and $R_2= ∞$

we obtain, $f=\frac{R}{n-1}$  …(3)

Putting $\frac{n-1}{R}=\frac{1}{f}$ from (3) in (2) we obtain, $-\frac{1}{t_2}=\frac{1}{f}-\frac{n}{t}$

$⇒\frac{1}{f}=\frac{n}{t}-\frac{1}{t_2}⇒f=\frac{tt_2}{nt_2-t}$

Putting the value of n from equation (1)

$⇒f=\frac{t_1t_2}{\frac{t}{t_1}t_2-t}=\frac{tt_1t_2}{tt_2-tt_1}=75cm$