Question

A point object is placed at a distance of $20\text{cm}$ from a thin plano-convex lens of focal length $15\text{cm}(\mu =1.5)$. The curved surface is silvered. The final image will form at

• A. $60\text{cm}$ left of $\text{AB}$
• B. $30\text{cm}$ left of $\text{AB}$
• C. $\frac{20}{7}\text{cm}$ left of $\text{AB}$
• D. $\frac{20}{7}\text{cm}$ right of $\text{AB}$

Answer 1

The correct option is C $\frac{20}{7}\text{cm}$ left of $\text{AB}$

Given,
refractive index of air, ${\mu }_1=1$
refractive index of lens material, ${\mu }_2=1$
focal length of lens, $f=15\text{cm}$

Applying lens makers formula for the first refraction at plane surface:

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Substituting the values in the above equation,

$\Rightarrow \frac{1.5}{v_1}-\frac{1}{(-20)}=\frac{1.5-1}{\infty }$

$\Rightarrow \frac{1.5}{v_1}=\frac{-1}{20}$

$\Rightarrow v_1=-30\text{cm}$

Applying lens makers formula between the plane lense and concave mirror,

$\frac{1}{15}=(1.5-1)(\frac{1}{R}-\frac{1}{\infty })$

$\therefore R=7.5\text{cm}$

Now, applying mirror formula for reflection,

$\frac{1}{v}+\frac{1}{u}=\frac{2}{R}$

This image formed due to plane surface will act as an object for concave mirror.

$\Rightarrow \frac{1}{v_2}+\frac{1}{(-30)}=\frac{2}{-7.5}$

$\Rightarrow \frac{1}{v_2}=\frac{-7}{30}$

$\Rightarrow v_2=\frac{-30}{7}\text{cm}$

This image will act as object for second refraction.

For second refraction:

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{1}{v_3}-\frac{1.5}{(-\frac{30}{7})}=\frac{1-1.5}{\infty }$

$\Rightarrow \frac{1}{v_3}=\frac{-7}{20}$

$\Rightarrow v_3=\frac{-20}{7}\text{cm}$

Hence, final image is formed at $\frac{20}{7}\text{cm}$ left of $\text{AB}$.