Question

The rear face of an equiconvex lens of focal length 30 cm is silvered so that it behaves like a concave mirror. An object is placed at a distance of 7.5 cm in front of this silvered lens. Find distance (in cm) of image from pole. $({\mu }_g=\frac{3}{2})$

Answer 1

For an equiconvex lens, the radius of curvature is same for both the surfaces.
Let $R_1=R,R_2=-R$
By lens maker’s formula
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{30}=(\frac{3}{2}-1)\times (\frac{1}{R}-\frac{1}{-R})$
$\Rightarrow \frac{1}{30}=\frac{1}{2}\times \frac{2}{R}$
$\Rightarrow R=30cm$
The rays from the object are first refracted through the lens. Next, they suffer reflection from the concave mirror. After reflection from the mirror, rays again suffer refraction from the lens. The three steps are shown one by one with calculations.



$u=–7.5cm,f=+30cm,u=–10cm,f=–15cm\text{Here,}{I}_2$ is at focal distance 30 cm
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}(\because f=\frac{R}{2})\Rightarrow$ The rear face of image is at infinity
$Rightarrowv=-10cm\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow$ The rear face of $v=+30cm$
Alternate solution :
Since it is observed that the rays of light undergo two refractions through the same lens and are reflection at the mirror, we can use the formula for effective focal length in case of lenses in contact (with a slight modification) as
$\frac{1}{f_{\text{eff}}}=\frac{-2}{f_1}+\frac{1}{f_m}$
where $f_{\text{eff}}$ is the focal length of spherical mirror now effectively formed, $f_l$ is the focal length of the lens and $f_m$ is the focal length of the mirror formed by the silvered surface
Here,
$\frac{1}{f_1}=(\frac{3}{2}-1)(\frac{1}{30}-\frac{1}{-30})$
or $f_1=+30cm$
$f_m=-15cm$
Now $\frac{1}{f_{\text{eff}}}=\frac{-2}{+30}+\frac{1}{-15}$
or $f_{\text{eff}}=-7.5cm$
(The negative sign implies that the effective mirror formed is a concave mirror). In this case the object is placed at the focus, hence the image is formed at infinity.