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
Answer & explanation
Correct answer: option 3
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)
$⇒f=\frac{t_1t_2}{\frac{t}{t_1}t_2-t}=\frac{tt_1t_2}{tt_2-tt_1}=75cm$