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.
Where should a pin be placed on the optic axis such that its image is formed at the same place?

• A. $x=5cm$
• B. $x=20cm$
• C. $x=15cm$
• D. $x=25cm$

Answer 1

The correct option is C $x=15cm$

Method 1 : The optical arrangement is equivalent to the concave mirror of focal length $F$ given by
$\frac{1}{F}=\frac{1}{f_g}+\frac{1}{f_m}+\frac{1}{f_g}$
where $f_x$ is the focal length of the lens without silvering and $f_m$ is the focal length of the mirror.
$\frac{1}{f_g}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$=(1.5-1)(\frac{1}{20}-\frac{1}{+60})=\frac{1}{60}$
$f_g=60cm$
$f_2=R_1/2=20/2=10cm$
$\frac{1}{F}=\frac{1}{60}+\frac{1}{10}+\frac{1}{60}=\frac{8}{60}$
$F=\frac{60}{9}=7.5cm$
For the image to be formed at the place of the object,
$X=R=2F=7.5\times 2=15cm$
Method 2 : We use the relation
$\frac{n_2}{x_2}-\frac{n_1}{x_1}=\frac{n_2-n_1}{R}$
For the object and the image to coincide, the rays fall normal on the reflecting surface, i.e., on the silvered face of the lens.
Then, the rays retrace backward and meet at the object point again (optical reversibility).
For the refraction at the upper surface of the lens,
$n_1=1.0,n_2=1.5,x_1=20,R=+60$
$(x_2=+20$ ensures that the rays fall on the silvered face normally.)
$\frac{1.5}{20}-\frac{1.0}{x_1}=\frac{1.5-1.0}{+60}$
$\frac{1.0}{x_1}=\frac{1.5}{20}-\frac{0.5}{60}=\frac{3.0}{60}$
$x_1=15cm$
Method 3 : We use lensmaker's formula and the equation
$\frac{1}{f}=\frac{1}{x^2}-\frac{1}{x_1}$
The given optical arrangement can be visualised as a convex lens of focal length $60cm$ and a concave mirror of focal length $10cm$ kept in contact as shown in the figure.
If the rays fall normally on the mirror after the refraction through the lens, they will retrace backward and meet at the point of the pin again.
For the lens, $x_1=?$
$x_2=+20$ (for normal incidence on the mirror)
$f=-60$ (using cartesian-coordinate sign convention)
$\frac{1}{-60}=\frac{1}{+20}-\frac{1}{x_1}$.

$x_1=15cm$