Question

The convex surface of a thin concavo-convex lens of glass of refractive index $1.5$ has a radius of curvature $20cm$. The concave surface has a radius of curvature $60cm$. The convex side is silvered and placed on a horizontal surface.
(i) Where should a pin be placed on the optical axis such that its image is formed at the same place?
(ii) If the concave part is filled with water of refractive index $4/3$, find the distance through which the pin should be moved so that the image of the pin again coincidence with the pin.

Answer 1

Given,

Here ${\mu }_2=1.5,{\mu }_1=1,u=-x,R=-60cm$

a) Let the pin is at a distance of $x$ from the lens.

Then for $1^{st}$refraction, $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\therefore \frac{1.5}{v}-\frac{1}{-x}=\frac{0.5}{-60}$

$\Rightarrow 120(1.5x+v)=-vx$

$\Rightarrow v(120+x)=-180x$

$\Rightarrow v=\frac{-180x}{120+x}$

This image distance is again object distance for the concave mirror.

$u=\frac{-180x}{120+x},f=-10cm(\therefore f=R/2)$

$\therefore \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v_1}=\frac{1}{-10}-\frac{-(120+x)}{180x}$

$\Rightarrow \frac{1}{v_1}=\frac{120+x-18x}{180x}\Rightarrow v_1=\frac{180x}{120-17x}$

Again the image formed is refracted through the lens so that the image is formed

on the object taken in the $1^{st}$ refraction. So, for $2^{nd}$ refraction.

According to sign conversion $v=-x,{\mu }_2=1,{\mu }_1=1.5,R=-60$

Now, $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}[u=\frac{180x}{120-17x}]$

$\Rightarrow \frac{1}{-x}-\frac{1.5}{180x}(120-17x)=\frac{-0.5}{-60}$

$\Rightarrow \frac{1}{x}+\frac{120-17x}{120x}=\frac{-1}{120}$

Multiplying both sides with $120m,$we get

$\Rightarrow 16x=240\Rightarrow x=15cm$

$\therefore$ Object should be placed at 15$cm$ from the lens on the axis.