Question

If radii of curvature of both convex surfaces is $20\text{cm}$, then the focal length of the lens for an object placed far in air in the given arrangement is:
(consider lens to be thin)

• A. $10\text{cm}$
• B. $20\text{cm}$
• C. $40\text{cm}$
• D. $80\text{cm}$

Answer 1

The correct option is C $40\text{cm}$

Sign convention: Direction of incident ray taken as +ve.
Given that,
Refractive index of air, ${\mu }_1=1$
Refractive index of lens, ${\mu }_2=1.5$
Refractive index of third medium, ${\mu }_3=\frac{4}{3}$
​​​​​​
​​​Applying refraction at spherical interfaces and considering that object is placed at $\infty$ distance in medium ${\mu }_1=1$
$u=-\infty$
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
Here $R=+20\text{cm}$ at first interface
$\Rightarrow \frac{1.5}{v}-\frac{1}{-\infty }=\frac{1.5-1}{+20}$
or $\frac{1.5}{v}=\frac{0.5}{20}$
$\therefore v=+60\text{cm}$
Now the image at $v=+60\text{cm}$ will acts as object for refraction at right lens and third medium.
The new image distance is given by,
$\frac{{\mu }_3}{v_1}-\frac{{\mu }_2}{v}=\frac{{\mu }_3-{\mu }_2}{R}$
Here $R=-20\text{cm}$ according to sign convention.
$\Rightarrow \frac{4}{3v_1}-\frac{1.5}{60}=\frac{\frac{4}{3}-\frac{3}{2}}{-20}$
or, $\frac{4}{3v_1}-\frac{1}{40}=\frac{8-9}{6(-20)}$
or, $\frac{4}{3v_1}-\frac{1}{40}=\frac{1}{120}$
or, $\frac{4}{3v_1}=\frac{1}{120}+\frac{1}{40}=\frac{4}{120}$
$\therefore v_1=40\text{cm}$
Thus parallel rays are converged to produce a real image at $40\text{cm}$ from lens.
$f=+40$ cm

$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Since the medium of surrounding is not the same on both}\text{side of lens, use refraction at spherical surface}\text{to obtain focal length (take}u=-\infty )\end{array}$$